Session 01 - Using Functions to Understand and Predict Systems

📅 Micro-Lecture

1. Functions, Domain, and Range

Mathematical Foundation: A function maps inputs (domain) to outputs (range). Domain restrictions occur when mathematical operations become undefined, such as division by zero ( $q(x) \neq 0$ in $\frac{p(x)}{q(x)}$ ) or even roots of negative numbers ( $g(x) \ge 0$ in $\sqrt{g(x)}$ ).
Engineering Context: Domain restrictions define physical operating limits and safety envelopes, such as non-negative mass, maximum material stress constraints, or operational temperature ranges. Piecewise functions model sudden system state transitions, such as active heating/cooling thresholds in thermal control systems or changing structural load regimes.

2. Function Composition and Inversion

Mathematical Foundation: Composition $(f \circ g)(x) = f(g(x))$ chains operations sequentially. An inverse function $f^{-1}(x)$ reverses input-output mappings such that $f(f^{-1}(x)) = x$ , reflected across the line $y = x$ .
Engineering Context: Composition represents multi-stage physical processes, such as converting pressure into mechanical displacement and subsequently into an electrical voltage signal. Inverses enable back-calculating raw physical inputs from sensor measurements, such as extracting precise temperatures from thermistor resistance readings.

3. Trigonometry and Unit Circle Dynamics

Mathematical Foundation: Angles $\theta$ map to points $(x, y) = (\cos\theta, \sin\theta)$ on the unit circle $x^2 + y^2 = 1$ . Key identity:
$\sin^2\theta + \cos^2\theta = 1$
Inverse trigonometric functions ( $\arcsin, \arccos, \arctan$ ) require restricted ranges to remain single-valued functions.
Engineering Context: Critical for vector force decomposition in structural statics, AC power phase calculations, wave optics, and spatial coordinate transformations in robotics and flight kinematics.

4. Exponential and Logarithmic Systems

Mathematical Foundation: Logarithms inverse exponential relationships ( $y = b^x \iff \log_b(y) = x$ ). Key identity rules:
$\ln(AB) = \ln A + \ln B$

$\ln\left(\frac{A}{B}\right) = \ln A - \ln B$

$\ln(A^k) = k\ln A$
Engineering Context: Logarithms linearize wide dynamic ranges into manageable scales (e.g., decibels in signal attenuation or acoustic pressure). Exponential models govern natural decay and growth dynamics, including RC circuit transient responses, heat dissipation via Newton's Law of Cooling, and radioactive half-life calculations.

Segment 1 of 8

Skill Block 1

Real-Life Engineering Applications

🤖

Robotics Engineering

Lidar Safety Envelopes (Robotics & Computer Science)

In autonomous vehicle design, a lidar sensor scans the surroundings by emitting light pulses across a specific field of view. The domain of the range-finding function is restricted by the sensor's physical limits: the horizontal scanning angle $\theta$ is restricted to $\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]$ radians, and the radial distance $d$ is restricted to $[0.1, 150]$ meters.

To determine how the vehicle should respond to obstacles, software engineers program a piecewise control function $A(d)$ to evaluate safety zones:

A(d) = \begin{cases} \text{Full Emergency Braking} & : 0.1 \le d < 3 \\ \text{Proportional Deceleration} & : 3 \le d < 15 \\ \text{Maintain Cruise Speed} & : 15 \le d \le 150 \end{cases}

Determine the vehicle's response and safety state if an obstacle is detected at a distance of:

$d = 2\text{ meters}$
$d = 10\text{ meters}$
$d = 80\text{ meters}$

🎯 Why This Matters:

& Real-Life Application: Identifying domain constraints and evaluating piecewise functions in real-time allows the vehicle's onboard computer to calculate safe velocity profiles without executing undefined operations or misinterpreting out-of-range sensor noise. For example, if an obstacle is detected outside the sensor's domain ( $d > 150$ or $d < 0.1$ ), the system ignores the signal or triggers a sensor fault handler.

💡 Show Worked Derivation

In autonomous vehicle design, a lidar sensor scans the surroundings by emitting light pulses across a specific field of view. The domain of the range-finding function is restricted by the sensor's physical limits: the horizontal scanning angle $\theta$ is restricted to $\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]$ radians, and the radial distance $d$ is restricted to $[0.1, 150]$ meters.

To determine how the vehicle should respond to obstacles, software engineers program a piecewise control function $A(d)$ to evaluate safety zones:

A(d) = \begin{cases} \text{Full Emergency Braking} & : 0.1 \le d < 3 \\ \text{Proportional Deceleration} & : 3 \le d < 15 \\ \text{Maintain Cruise Speed} & : 15 \le d \le 150 \end{cases}

Determine the vehicle's response and safety state if an obstacle is detected at a distance of:

$d = 2\text{ meters}$
$d = 10\text{ meters}$
$d = 80\text{ meters}$

Step-by-Step Solution:

For $d = 2\text{ m}$ , since $0.1 \le 2 < 3$ , we use the first piecewise condition: $A(2) = \text{Full Emergency Braking}$
For $d = 10\text{ m}$ , since $3 \le 10 < 15$ , we use the second piecewise condition: $A(10) = \text{Proportional Deceleration}$
For $d = 80\text{ m}$ , since $15 \le 80 \le 150$ , we use the third piecewise condition: $A(80) = \text{Maintain Cruise Speed}$

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

General Overview Skill Block 1 establishes core fluency in function evaluation, algebraic expression substitution, domain/range identification, and piecewise logic analysis. Mastered operations include evaluating functions at numeric and variable inputs, determining mathematical boundaries, and applying conditionally defined function rules.

Conceptual Framework

Variable Substitution & Expansion: Function evaluation replaces the independent variable $x$ with a given input or algebraic expression (e.g., $f(a+1)$ ), requiring polynomial expansion and combination of like terms.
Domain Restrictions (Allowed Inputs): The domain includes all real numbers except those that cause non-real outputs or undefined operations. Primary red flags include dividing by zero and taking even roots of negative numbers.
Range Restrictions (Possible Outputs): The range represents all output values produced by valid domain inputs. Determining range relies on structural properties like non-negative outputs from squares ( $x^2 \ge 0$ ) or principal square roots ( $\sqrt{\text{expression}} \ge 0$ ).
Piecewise Conditional Logic: Piecewise functions partition the domain into distinct intervals, where the input value $x$ determines which specific sub-function governs the output.

Key Equations & Structural Rules

Polynomial Functions:
- Domain: $(-\infty, \infty)$
- Example: $f(x) = x^2 + 1 \implies \text{Domain: } (-\infty, \infty), \text{Range: } [1, \infty)$
Radical Functions (Square Roots):
- Rule: $\sqrt{g(x)} \implies g(x) \ge 0$
- Example: $f(x) = \sqrt{x+3} \implies x + 3 \ge 0 \implies \text{Domain: } [-3, \infty), \text{Range: } [0, \infty)$
Rational Functions:
- Rule: $\frac{p(x)}{q(x)} \implies q(x) \neq 0$
- Example: $f(x) = \frac{9}{x-2} \implies x - 2 \neq 0 \implies \text{Domain: } (-\infty, 2) \cup (2, \infty), \text{Range: } (-\infty, 0) \cup (0, \infty)$
Piecewise Evaluation Rule:
$h(x) = \begin{cases} 1 - x^5 & : x < 3 \\ e^x & : x = 3 \\ 8 & : x > 3 \end{cases}$
- Select the single matching condition for input $x$ before evaluating the corresponding sub-function.

Worked Example A: Algebraic Expression Evaluation

Given $f(x) = 2x^2 - 3x + 1$ , find and simplify $f(2x - 3)$ .

💡 Reveal Worked Solution

Substitute the entire expression $(2x - 3)$ into every instance of $x$ :

f(2x - 3) = 2(2x - 3)^2 - 3(2x - 3) + 1

Expand the binomial term $(2x - 3)^2$ :

(2x - 3)^2 = 4x^2 - 12x + 9

Substitute and distribute terms:

f(2x - 3) = 2(4x^2 - 12x + 9) - 6x + 9 + 1

f(2x - 3) = 8x^2 - 24x + 18 - 6x + 9 + 1

Combine like terms:

f(2x - 3) = 8x^2 - 30x + 28

Worked Example B: Combined Domain and Range Determination

Find the domain and range for the combined rational and radical function $g(x) = \frac{\sqrt{x + 4}}{x - 2}$ .

💡 Reveal Worked Solution

Domain Solution: Identify all structural restrictions simultaneously:

Radicand Restriction: The expression inside the square root must be non-negative: $x + 4 \ge 0 \implies x \ge -4$
Denominator Restriction: The denominator cannot equal zero: $x - 2 \neq 0 \implies x \neq 2$

Combine restrictions on a number line to state the valid input set:

\text{Domain: } [-4, 2) \cup (2, \infty)

Range Solution: Analyze output behavior across the continuous intervals of the domain:

At the lower boundary $x = -4$ : $g(-4) = \frac{\sqrt{0}}{-6} = 0$ .
Approaching the vertical asymptote at $x = 2$ from the left ( $x \to 2^-$ ), the numerator approaches $\sqrt{6} > 0$ while the denominator approaches $0$ through negative values, driving $g(x) \to -\infty$ .
Approaching $x = 2$ from the right ( $x \to 2^+$ ), the numerator approaches $\sqrt{6} > 0$ while the denominator approaches $0$ through positive values, driving $g(x) \to +\infty$ .

Since the function outputs span smoothly from negative infinity to positive infinity, every real output is covered:

\text{Range: } (-\infty, \infty)

Worked Example C: Piecewise Function Evaluation

Given the piecewise function:

k(x) = \begin{cases} x^2 - 4 & : x < -1 \\ 2x + 1 & : -1 \le x \le 4 \\ \sqrt{x} + 5 & : x > 4 \end{cases}

Evaluate (a) $k(-3)$ , (b) $k(-1)$ , and (c) $k(9)$ .

💡 Reveal Worked Solution

(a) Evaluate $k(-3)$ : The input $x = -3$ satisfies the condition $-3 < -1$ . Use the first sub-function: $k(-3) = (-3)^2 - 4 = 9 - 4 = 5$
(b) Evaluate $k(-1)$ : The input $x = -1$ satisfies the condition $-1 \le -1 \le 4$ . Use the second sub-function: $k(-1) = 2(-1) + 1 = -2 + 1 = -1$
(c) Evaluate $k(9)$ : The input $x = 9$ satisfies the condition $9 > 4$ . Use the third sub-function: $k(9) = \sqrt{9} + 5 = 3 + 5 = 8$

Practice Problem 1

Let $f(x) = 3x^2 - x$ . Find the following:

(a) $f(1)$
(b) $f(-4)$
(c) $f(a+1)$

Practice Space Solve this problem on paper or tablet.

Substitute the input value or algebraic expression in place of every $x$ in the function, using parentheses to ensure correct distribution and expansion. For (c), expand the squared binomial $(a+1)^2 = a^2 + 2a + 1$ before distributing the coefficient.

Answer: * (a) $2$

(b) $52$
(c) $3a^2 + 5a + 2$

Step-by-Step Execution

(a) Find $f(1)$ Substitute $x = 1$ into the function: $f(1) = 3(1)^2 - (1)$ $f(1) = 3(1) - 1$ $f(1) = 3 - 1$ $f(1) = 2$

(b) Find $f(-4)$ Substitute $x = -4$ into the function: $f(-4) = 3(-4)^2 - (-4)$ $f(-4) = 3(16) + 4$ $f(-4) = 48 + 4$ $f(-4) = 52$

(c) Find $f(a+1)$ Substitute the expression $(a+1)$ everywhere there is an $x$ in the function:

f(a+1) = 3(a+1)^2 - (a+1)

First, expand

(a+1)^2

:

(a+1)^2 = a^2 + 2a + 1

Now substitute that back and distribute the

3

, as well as the negative sign:

f(a+1) = 3(a^2 + 2a + 1) - a - 1

f(a+1) = 3a^2 + 6a + 3 - a - 1

Combine like terms (

6a - a

and

3 - 1

):

f(a+1) = 3a^2 + 5a + 2

Practice Problem 2

State the domain and range of the following functions: How Domain and Range Work

(a) $f(x) = x^2 + 1$
(b) $f(x) = \sqrt{x+3}$
(c) $f(x) = \frac{9}{x-2}$

Practice Space Solve this problem on paper or tablet.

For domain, watch out for dividing by zero (rational denominators cannot equal 0) and taking even roots of negative numbers (radicands must be $\ge 0$ ). For range, look at the structural limits of outputs (e.g., $x^2 \ge 0$ , $\sqrt{\text{expression}} \ge 0$ , and non-zero constants in a numerator mean the fraction cannot equal 0).

Answer: * (a) Domain: $(-\infty, \infty)$ ; Range: $[1, \infty)$

(b) Domain: $[-3, \infty)$ ; Range: $[0, \infty)$
(c) Domain: $(-\infty, 2) \cup (2, \infty)$ ; Range: $(-\infty, 0) \cup (0, \infty)$

Step-by-Step Execution

(a) $f(x) = x^2 + 1$

Domain:

This is a polynomial function (a parabola). There are no fractions with variables in the denominator and no square roots to restrict our inputs. You can plug in any real number.

Domain: $(-\infty, \infty)$

Range:

The term $x^2$ is always greater than or equal to $0$ for any real number $x$ . Adding $1$ shifts the entire graph up by $1$ unit, meaning the minimum value the function can output is $1$ . Range: $[1, \infty)$

(b) $f(x) = \sqrt{x+3}$

Domain: The expression inside a square root (the radicand) must be greater than or equal to $0$ to stay within the real number system.

x + 3 \geq 0

x \geq -3

Domain:

[-3, \infty)

Range: The principal (positive) square root function $\sqrt{\text{anything}}$ always outputs values greater than or equal to $0$ . As $x$ goes from $-3$ to infinity, the outputs go from $0$ to infinity.

Range: $[0, \infty)$

(c) $f(x) = \frac{9}{x-2}$

Domain: A rational function is undefined when its denominator is equal to $0$ . We must exclude any values of $x$ that cause division by zero:

x - 2 = 0 \implies x = 2

Therefore,

x

can be any real number except

2

.

Domain: $(-\infty, 2) \cup (2, \infty)$

Range: To find the range, we can set $y = \frac{9}{x-2}$ and see what values $y$ cannot take. Notice that the numerator is a constant ( $9$ ). A fraction with a non-zero numerator can never equal $0$ , because the only way a fraction equals $0$ is if its numerator is $0$ . Therefore, $y \neq 0$ . Alternatively, solving for $x$ in terms of $y$ :

y(x - 2) = 9

x - 2 = \frac{9}{y}

x = \frac{9}{y} + 2

This shows that

y

cannot equal

0

, as it would cause division by zero.

Range: $(-\infty, 0) \cup (0, \infty)$

Practice Problem 3

Use the function given to evaluate the following:

h(x) = \begin{cases} 1 - x^5 & : x < 3 \\ e^x & : x = 3 \\ 8 & : x > 3 \end{cases}

(a) $h(5)$
(b) $h(-1)$
(c) $h(3)$

Practice Space Solve this problem on paper or tablet.

Compare the input value $x$ to the inequalities on the right side of the piecewise function first to determine which specific sub-function formula to evaluate.

Answer: * (a) $8$

(b) $2$
(c) $e^3$

Step-by-Step Execution

(a) Find $h(5)$

Determine the condition: The input is $x = 5$ . Since $5 > 3$ , we use the third piece of the function. Evaluate: The function states that for any $x > 3$ , the output is a constant $8$ .

h(5) = 8

(b) Find $h(-1)$

Determine the condition: The input is $x = -1$ . Since $-1 < 3$ , we use the first piece of the function.

Evaluate: Substitute $x = -1$ into $1 - x^5$ :

h(-1) = 1 - (-1)^5

h(-1) = 1 - (-1)

h(-1) = 1 + 1

h(-1) = 2

(c) Find $h(3)$

Determine the condition: The input is $x = 3$ . Since $3 = 3$ , we use the second piece of the function. Evaluate: Substitute $x = 3$ into $e^x$ :

h(3) = e^3

Segment 2 of 8

📅 Regroup 1

Review domain restrictions (denominator != 0, radicand >= 0).
Preview Block 2

Segment 3 of 8

Skill Block 2

Real-Life Engineering Applications

⚡

Electrical Engineering

Sensor Signal Calibration (Electrical & Systems Engineering)

Engineers design sensor systems (like pressure transducers or temperature sensors) by chaining multiple physical operations together. For example, an industrial pressure sensor converts a physical pressure input $P$ (in psi) into a voltage signal $V$ (in volts) using a linear calibration function:

V(P) = 0.08P + 0.5

This voltage is then fed into an Analog-to-Digital Converter (ADC) that maps the voltage to a 10-bit digital value $D$ represented by the function:

D(V) = 204.6V

Find the composite function $(D \circ V)(P)$ that directly maps physical pressure $P$ to the digital output $D$ .
Solve for the exact pressure $P$ (in psi) that will cause the digital output to reach a threshold alarm value of $D = 800$ .

🎯 Why This Matters:

& Real-Life Application: Function composition allows systems engineers to create direct mathematical mappings from raw physical phenomena (pressure) to digital representations in firmware. Solving algebraic equations on this composition enables calibration of safety thresholds—ensuring the system triggers an emergency shutdown at exactly $42.63\text{ psi}$ when the ADC reads a digital value of $800$ .

💡 Show Worked Derivation

Engineers design sensor systems (like pressure transducers or temperature sensors) by chaining multiple physical operations together. For example, an industrial pressure sensor converts a physical pressure input $P$ (in psi) into a voltage signal $V$ (in volts) using a linear calibration function:

V(P) = 0.08P + 0.5

This voltage is then fed into an Analog-to-Digital Converter (ADC) that maps the voltage to a 10-bit digital value $D$ represented by the function:

D(V) = 204.6V

Find the composite function $(D \circ V)(P)$ that directly maps physical pressure $P$ to the digital output $D$ .
Solve for the exact pressure $P$ (in psi) that will cause the digital output to reach a threshold alarm value of $D = 800$ .

Step-by-Step Solution:

Substitute $V(P)$ into the digital output function $D(V)$ : $(D \circ V)(P) = D(V(P)) = 204.6(0.08P + 0.5)$ $(D \circ V)(P) = 16.368P + 102.3$
To find the pressure corresponding to $D = 800$ , set $(D \circ V)(P) = 800$ and solve for $P$ : $16.368P + 102.3 = 800$ $16.368P = 697.7$ $P = \frac{697.7}{16.368} \approx 42.63\text{ psi}$

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Skill Block 2 covers structural function operations (quotients and compositions), linear equation representations, and multi-step root-finding for algebraic and trigonometric equations.

1. Function Composition and Combination

Conceptual Framework:
- Quotients: Division combines functions into a single rational relation $\left(\frac{g}{f}\right)(x) = \frac{g(x)}{f(x)}$ , restricted to inputs where $f(x) \neq 0$ .
- Composition: Composition $(f \circ g)(x) = f(g(x))$ feeds the output of the inner function $g(x)$ as the input to the outer function $f(x)$ .
Step-by-Step Process for Composition $f(g(x))$ :
1. Identify the inner expression $g(x)$ and the outer structure $f(x)$ .
2. Replace every occurrence of $x$ in $f(x)$ with the full expression of $g(x)$ .
3. Expand algebraic expressions (e.g., binomial squares) and simplify like terms.
Step-by-Step Process for Point Evaluation $(g \circ f)(c)$ :
1. Compute the inner output value: $v = f(c)$ .
2. Plug $v$ into the outer function: $g(v)$ .

2. Linear Modeling Forms

Conceptual Framework: Linear relations define constant rates of change ( $m = \frac{\Delta y}{\Delta x}$ ). Conversion between forms simplifies graph analysis or algebraic manipulation.
Key Formulas:
- Point-Slope Form: $y - y_1 = m(x - x_1)$
- Slope-Intercept Form: $y = mx + b$
Step-by-Step Process (Point to Slope-Intercept Form):
1. Identify given point $(x_1, y_1)$ and slope $m$ .
2. Substitute values into point-slope form: $y - y_1 = m(x - x_1)$ .
3. Distribute slope $m$ across $(x - x_1)$ .
4. Isolate $y$ by adding $y_1$ to both sides, combining constants over a common denominator.

3. Multi-Type Equation Solving

Algebraic & Polynomial Equations:
- Linear Parentheses Reduction: Distribute negative signs from inner to outer grouping symbols, combine like terms on each side, and isolate the variable $x$ .
- Quadratic Factoring: Rearrange to standard form $ax^2 + bx + c = 0$ . Identify factors $r_1, r_2$ satisfying $(x - r_1)(x - r_2) = 0$ and invoke the Zero Product Property ( $A \cdot B = 0 \implies A = 0 \text{ or } B = 0$ ).
Trigonometric Equations:
- Factoring Common Terms: Extract common variable factors (e.g., $2\theta \cos\theta + \theta = 0 \implies \theta(2\cos\theta + 1) = 0$ ).
- Splitting Equations: Set each factor equal to zero independently using the Zero Product Property.
- Unit Circle Evaluation: Isolate the trigonometric function (e.g., $\cos\theta = -\frac{1}{2}$ ) and identify all reference angles within the specified interval $[0, 2\pi]$ using quadrant sign rules.

Worked Example A: Function Composition and Evaluation

Given $f(x) = \sqrt{x - 2}$ and $g(x) = 2x + 3$ :

(a) Find $f(g(x))$ and simplify.
(b) Evaluate $(g \circ f)(11)$ .

💡 Reveal Worked Solution

(a) Find $f(g(x))$ : Substitute the expression for $g(x)$ into $f(x)$ everywhere $x$ appears:

f(g(x)) = f(2x + 3) = \sqrt{(2x + 3) - 2}

Simplify the radicand:

f(g(x)) = \sqrt{2x + 1}

(b) Evaluate $(g \circ f)(11)$ : Evaluate from the inside out: $g(f(11))$ .

Calculate inner function value $f(11)$ : $f(11) = \sqrt{11 - 2} = \sqrt{9} = 3$
Substitute $3$ into outer function $g(x)$ : $g(3) = 2(3) + 3 = 6 + 3 = 9$ $(g \circ f)(11) = 9$

Worked Example B: Finding Equations of Lines

Find the equation of the line passing through the point $(-2, 5)$ with slope $m = -\frac{3}{4}$ in slope-intercept form.

💡 Reveal Worked Solution

Step 1: Apply Point-Slope Form
$y - y_1 = m(x - x_1)$
- Substitute $m = -\frac{3}{4}$ , $x_1 = -2$ , and $y_1 = 5$ : $y - 5 = -\frac{3}{4}(x - (-2))$ $y - 5 = -\frac{3}{4}(x + 2)$
Step 2: Distribute Slope
$y - 5 = -\frac{3}{4}x - \frac{6}{4}$
- Simplify fraction: $y - 5 = -\frac{3}{4}x - \frac{3}{2}$
Step 3: Convert to Slope-Intercept Form ( $y = mx + b$ )
- Add $5$ to both sides, expressing $5$ with a common denominator of $2$ ( $5 = \frac{10}{2}$ ): $y = -\frac{3}{4}x - \frac{3}{2} + \frac{10}{2}$ $y = -\frac{3}{4}x + \frac{7}{2}$

Worked Example C: Solving Equations (Algebraic and Trigonometric)

Solve the following equations:

(a) $2x^2 + 5x - 3 = 0$
(b) $2\sin(\theta)\cos(\theta) - \sqrt{3}\sin(\theta) = 0$ on the interval $[0, 2\pi]$

💡 Reveal Worked Solution

(a) Algebraic Solution (Quadratic Factoring): Factor the trinomial into two binomials:
$(2x - 1)(x + 3) = 0$
- Apply the Zero Product Property: $2x - 1 = 0 \implies x = \frac{1}{2}$ $x + 3 = 0 \implies x = -3$ $\text{Final Answer: } x = \frac{1}{2}, \quad x = -3$
(b) Trigonometric Solution (Factoring and Unit Circle): Factor out the common term $\sin(\theta)$ :
$\sin(\theta)(2\cos(\theta) - \sqrt{3}) = 0$
- Apply the Zero Product Property to set up two independent equations:
  1. $\sin(\theta) = 0$
  2. $2\cos(\theta) - \sqrt{3} = 0 \implies \cos(\theta) = \frac{\sqrt{3}}{2}$
Determine all solutions on $[0, 2\pi]$ :
- From $\sin(\theta) = 0$ : $\theta = 0, \pi, 2\pi$
- From $\cos(\theta) = \frac{\sqrt{3}}{2}$ (positive cosine in Quadrants I and IV with reference angle $\frac{\pi}{6}$ ):
  - Quadrant I: $\theta = \frac{\pi}{6}$
  - Quadrant IV: $\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$

\text{Final Answer: } \theta = 0, \quad \frac{\pi}{6}, \quad \pi, \quad \frac{11\pi}{6}, \quad 2\pi

Practice Problem 4

Let $f(x) = \sqrt{x^2 - 1}$ and $g(x) = x + 1$ .

Find the following:

(a) $\left(\frac{g}{f}\right)(x)$
(b) $f(g(x))$
(c) $(g \circ f)(2)$

Practice Space Solve this problem on paper or tablet.

For (a), construct the ratio $\frac{g(x)}{f(x)}$ . For (b), substitute the entire expression $g(x)$ in place of $x$ inside $f(x)$ , then expand and simplify. For (c), evaluate from the inside out: first calculate $f(2)$ , then substitute that numerical result into $g(x)$ .

Answer: * (a) $\frac{x+1}{\sqrt{x^2 - 1}}$ (or $\sqrt{\frac{x+1}{x-1}}$ for $x > 1$ )

(b) $\sqrt{x^2 + 2x}$
(c) $\sqrt{3} + 1$

Step-by-Step Execution

(a) Find $\left(\frac{g}{f}\right)(x)$ The notation $\left(\frac{g}{f}\right)(x)$ represents the division of function $g(x)$ by function $f(x)$ :

\left(\frac{g}{f}\right)(x) = \frac{g(x)}{f(x)}

Substitute the given equations:

\left(\frac{g}{f}\right)(x) = \frac{x + 1}{\sqrt{x^2 - 1}}

Optional Simplification (Rationalizing or Factoring): Notice that

x^2 - 1 = (x - 1)(x + 1)

. For

x > 1

, you can rewrite the expression as:

\frac{x + 1}{\sqrt{(x - 1)(x + 1)}} = \frac{\sqrt{x + 1}\sqrt{x + 1}}{\sqrt{x - 1}\sqrt{x + 1}} = \frac{\sqrt{x + 1}}{\sqrt{x - 1}} = \sqrt{\frac{x + 1}{x - 1}}

(b) Find $f(g(x))$ The composite notation $f(g(x))$ means you substitute the entire function $g(x)$ into every $x$ in $f(x)$ :

f(g(x)) = f(x + 1)

Substitute

(x + 1)

into

f(x) = \sqrt{x^2 - 1}

:

f(g(x)) = \sqrt{(x + 1)^2 - 1}

Expand the squared binomial

(x + 1)^2 = x^2 + 2x + 1

:

f(g(x)) = \sqrt{(x^2 + 2x + 1) - 1}

Simplify by canceling out

+1

and

-1

:

f(g(x)) = \sqrt{x^2 + 2x}

(c) Find $(g \circ f)(2)$ The composition notation $(g \circ f)(2)$ means $g(f(2))$ . Work from the inside out.

Step 1: Find $f(2)$ Substitute $x = 2$ into $f(x)$ :

f(2) = \sqrt{(2)^2 - 1}

f(2) = \sqrt{4 - 1}

f(2) = \sqrt{3}

Step 2: Plug this result into $g(x)$ Now find

g(\sqrt{3})

by substituting

\sqrt{3}

into

g(x) = x + 1

:

g(f(2)) = g(\sqrt{3}) = \sqrt{3} + 1

Therefore:

(g \circ f)(2) = \sqrt{3} + 1

Practice Problem 5

Find the equation of the line through the point $(1, 7)$ with slope $\frac{2}{3}$ . Present your answer in slope-intercept form.

How Point-Slope Form Works

Practice Space Solve this problem on paper or tablet.

Use the point-slope form $y - y_1 = m(x - x_1)$ with the point $(1, 7)$ and slope $m = \frac{2}{3}$ , then distribute the slope and isolate $y$ to convert it to the slope-intercept form $y = mx + b$ .

Answer: * $y = \frac{2}{3}x + \frac{19}{3}$

Step-by-Step Execution

Step 1: Choose a linear equation form You can use either the point-slope form or the slope-intercept form to find the equation. Both methods yield the same result.

Method 1: Using Point-Slope Form The point-slope form of a line is:

y - y_1 = m(x - x_1)

Substitute

m = \frac{2}{3}

,

x_1 = 1

, and

y_1 = 7

:

y - 7 = \frac{2}{3}(x - 1)

Distribute the

\frac{2}{3}

on the right side:

y - 7 = \frac{2}{3}x - \frac{2}{3}

Add

7

to both sides to isolate

y

:

y = \frac{2}{3}x - \frac{2}{3} + 7

To combine

-\frac{2}{3}

and

7

, find a common denominator (

3

):

7 = \frac{21}{3}

y = \frac{2}{3}x - \frac{2}{3} + \frac{21}{3}

y = \frac{2}{3}x + \frac{19}{3}

Method 2: Using Slope-Intercept Form Directly The slope-intercept form is:

y = mx + b

Substitute the known slope

m = \frac{2}{3}

and the coordinates of the point

x = 1, y = 7

to solve for the

y

-intercept (

b

):

7 = \frac{2}{3}(1) + b

7 = \frac{2}{3} + b

Subtract

\frac{2}{3}

from both sides:

7 - \frac{2}{3} = b

\frac{21}{3} - \frac{2}{3} = b

b = \frac{19}{3}

Substitute

m

and

b

back into the slope-intercept equation:

y = \frac{2}{3}x + \frac{19}{3}

Final Answer The equation of the line in slope-intercept form is:

y = \frac{2}{3}x + \frac{19}{3}

Practice Problem 6

Find the solutions to the following:

(a) $1 - (4 - (5 - x)) = 3x - (4 + x)$
(b) $x^2 + 5x = -4$
(c) $2\theta \cos(\theta) + \theta = 0$ on $[0, 2\pi]$

Practice Space Solve this problem on paper or tablet.

For (a), distribute negative signs from the inside-out and combine like terms before isolating $x$ . For (b), set the quadratic to zero ( $x^2 + 5x + 4 = 0$ ) and factor. For (c), factor out $\theta$ to obtain $\theta(2\cos\theta + 1) = 0$ , apply the zero product property, and find the angles in $[0, 2\pi]$ where $\cos\theta = -\frac{1}{2}$ .

Answer: * (a) $x = 2$

(b) $x = -1, \ -4$
(c) $\theta = 0, \ \frac{2\pi}{3}, \ \frac{4\pi}{3}$

Step-by-Step Execution

(a) Solve: $1 - (4 - (5 - x)) = 3x - (4 + x)$

Step 1: Simplify the innermost parentheses on the left side. Distribute the negative sign into $(5 - x)$ :

1 - (4 - 5 + x) = 3x - (4 + x)

1 - (-1 + x) = 3x - (4 + x)

Step 2: Distribute the remaining negative signs on both sides. On the left, distribute the negative sign into

(-1 + x)

. On the right, distribute the negative sign into

(4 + x)

:

1 + 1 - x = 3x - 4 - x

Step 3: Combine like terms on both sides.

2 - x = 2x - 4

Step 4: Isolate the variable $x$ . Add

x

to both sides:

2 = 3x - 4

Add

4

to both sides:

6 = 3x

Divide by

3

:

x = 2

(b) Solve: $x^2 + 5x = -4$

Step 1: Set the quadratic equation to zero. Add $4$ to both sides to write it in standard form ( $ax^2 + bx + c = 0$ ):

x^2 + 5x + 4 = 0

Step 2: Factor the trinomial. We need two numbers that multiply to

4

and add up to

5

. Those numbers are

1

and

4

:

(x + 1)(x + 4) = 0

Step 3: Apply the Zero Product Property. Set each factor equal to zero and solve:

x + 1 = 0 \implies x = -1

x + 4 = 0 \implies x = -4

Final Answer:

x = -1, \quad x = -4

(c) Solve: $2\theta \cos(\theta) + \theta = 0$ on $[0, 2\pi]$

Step 1: Factor out the common term. Both terms share a $\theta$ , so factor it out:

\theta(2\cos(\theta) + 1) = 0

Step 2: Apply the Zero Product Property. Set each factor equal to zero: 1.

\theta = 0

2.

2\cos(\theta) + 1 = 0

Step 3: Solve the trigonometric equation. Isolate $\cos(\theta)$ in the second equation:

2\cos(\theta) = -1

\cos(\theta) = -\frac{1}{2}

Step 4: Find solutions within the interval $[0, 2\pi]$ . * From the first factor:

\theta = 0

is within the interval. * From the second factor:

\cos(\theta) = -\frac{1}{2}

when

\theta

is in Quadrant II and Quadrant III with a reference angle of

\frac{\pi}{3}

: * Quadrant II:

\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}

* Quadrant III:

\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}

Final Answer:

\theta = 0, \quad \frac{2\pi}{3}, \quad \frac{4\pi}{3}

Segment 4 of 8

📅 Regroup 2

Highlight slope-intercept vs point-slope forms.
Preview Block 3

Segment 5 of 8

Skill Block 3

Real-Life Engineering Applications

🚀

Aerospace Engineering

Robotic Arm Path Planning (Mechanical & Aerospace Engineering)

In robotics, the forward kinematics function maps the joint angles of a robotic arm to a physical coordinate $(x, y)$ in space. However, to control the robot, engineers must solve the inverse kinematics problem: given a target coordinate $(x, y)$ , calculate the joint angle $\theta$ required to reach it.

For a single-link arm of length $L = 5\text{ units}$ , the position is given by $x = 5\cos\theta$ and $y = 5\sin\theta$ .

Reconstruct the trigonometric relationship to show that $\theta = \arctan\left(\frac{y}{x}\right)$ .
If the arm needs to target the coordinate $(x, y) = (-3, 4)$ in Quadrant II, evaluate the joint angle $\theta$ in radians, accounting for quadrant corrections.

🎯 Why This Matters:

& Real-Life Application: Standard inverse trigonometric functions like $\arctan(z)$ are mathematically restricted to Quadrants I and IV ( $(-\frac{\pi}{2}, \frac{\pi}{2})$ ). In robotics and path planning, software engineers use quadrant dynamics to reconstruct the true angle in 2D space (often using functions like atan2(y, x)) to prevent the arm from moving to the incorrect mirror-image location.

💡 Show Worked Derivation

In robotics, the forward kinematics function maps the joint angles of a robotic arm to a physical coordinate $(x, y)$ in space. However, to control the robot, engineers must solve the inverse kinematics problem: given a target coordinate $(x, y)$ , calculate the joint angle $\theta$ required to reach it.

For a single-link arm of length $L = 5\text{ units}$ , the position is given by $x = 5\cos\theta$ and $y = 5\sin\theta$ .

Reconstruct the trigonometric relationship to show that $\theta = \arctan\left(\frac{y}{x}\right)$ .
If the arm needs to target the coordinate $(x, y) = (-3, 4)$ in Quadrant II, evaluate the joint angle $\theta$ in radians, accounting for quadrant corrections.

Step-by-Step Solution:

Using the given coordinates: $\frac{y}{x} = \frac{5\sin\theta}{5\cos\theta} = \tan\theta$ Taking the inverse tangent of both sides: $\theta = \arctan\left(\frac{y}{x}\right)$
For $(x, y) = (-3, 4)$ : $\theta_{\text{ref}} = \arctan\left(\left|\frac{4}{-3}\right|\right) = \arctan\left(\frac{4}{3}\right) \approx 0.927\text{ rad}$ Since $x < 0$ and $y > 0$ , the target lies in Quadrant II. The true angle $\theta$ is: $\theta = \pi - \theta_{\text{ref}} = \pi - \arctan\left(\frac{4}{3}\right) \approx 3.142 - 0.927 = 2.215\text{ rad}$

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Skill Block 3 covers foundational precalculus operations: reference right-triangle trigonometry, inverse algebraic/logarithmic functions, exponential and logarithmic equations, log properties, inverse trigonometric evaluations, and algebraic function symmetry.

1. Trigonometric Ratios and Quadrant Dynamics

Conceptual Framework: Given a trigonometric ratio and quadrant constraints, reference right triangles translate spatial relationships across coordinate axes. The standard ratios on a right triangle with adjacent side $x$ , opposite side $y$ , and hypotenuse $r = \sqrt{x^2 + y^2}$ are defined as: $\sin\theta = \frac{y}{r}, \quad \cos\theta = \frac{x}{r}, \quad \tan\theta = \frac{y}{x}$ $\csc\theta = \frac{r}{y}, \quad \sec\theta = \frac{r}{x}, \quad \cot\theta = \frac{x}{y}$
Quadrant Sign Conventions (CAST Rule):
- Quadrant I ( $0 < \theta < \frac{\pi}{2}$ ): All functions positive ( $x > 0, y > 0$ ).
- Quadrant II ( $\frac{\pi}{2} < \theta < \pi$ ): Sine and Cosecant positive ( $x < 0, y > 0$ ).
- Quadrant III ( $\pi < \theta < \frac{3\pi}{2}$ ): Tangent and Cotangent positive ( $x < 0, y < 0$ ).
- Quadrant IV ( $\frac{3\pi}{2} < \theta < 2\pi$ ): Cosine and Secant positive ( $x > 0, y < 0$ ).
Step-by-Step Triangle Reconstruction Process:
1. Determine signs of $x$ and $y$ based on the given quadrant interval.
2. Use the Pythagorean theorem ( $x^2 + y^2 = r^2$ ) to solve for the missing side length.
3. Assign the correct algebraic sign ( $+$ or $-$ ) to $x$ or $y$ .
4. Construct the remaining 5 trigonometric ratios.

2. Finding Inverse Functions $f^{-1}(x)$

Algebraic Inverses:
1. Replace $f(x)$ with $y$ .
2. Swap $x$ and $y$ variables.
3. Isolate $y$ algebraically (e.g., clearing rational denominators or exponentiating both sides).
4. Replace $y$ with $f^{-1}(x)$ .
Exponential/Logarithmic Inverse Pairing:
- To undo $\ln(g(y))$ , exponentiate both sides with base $e$ : $e^{\ln(g(y))} = g(y)$ .
- To undo $e^{g(y)}$ , take the natural logarithm of both sides: $\ln(e^{g(y)}) = g(y)$ .

3. Exponential and Logarithmic Equations

Common Base Method: If $b^{g(x)} = b^k$ , then $g(x) = k$ .
Quadratic-Form Exponentials: Equations like $a\cdot e^{2x} + b\cdot e^x + c = 0$ transform into standard quadratics $au^2 + bu + c = 0$ using $u = e^x$ . Discard extraneous negative roots since $e^x > 0$ for all real $x$ .
Logarithmic Properties & Condensing Rules:
- Product Rule: $\log_b(A) + \log_b(B) = \log_b(A \cdot B)$
- Quotient Rule: $\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)$
- Power Rule: $k \cdot \log_b(A) = \log_b(A^k)$
- Extraneous Solution Check: Verify that inputs into any logarithmic term satisfy $\text{Argument} > 0$ .

4. Inverse Trigonometric Functions

Restricted Domain & Output Range:
- $\arcsin(x)$ or $\sin^{-1}(x)$ : Output range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
- $\arccos(x)$ or $\cos^{-1}(x)$ : Output range $[0, \pi]$
- $\arctan(x)$ or $\tan^{-1}(x)$ : Output range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
Evaluation Framework: Rephrase $\theta = f^{-1}(k)$ as finding an angle $\theta$ inside the restricted range such that $f(\theta) = k$ .

5. Function Symmetry (Even, Odd, Neither)

Algebraic Test: Evaluate $f(-x)$ and compare with $f(x)$ :
- Even Function: $f(-x) = f(x)$ $\implies$ Symmetric across the $y$ -axis.
- Odd Function: $f(-x) = -f(x)$ $\implies$ Symmetric about the origin.
- Neither: $f(-x) \neq f(x)$ and $f(-x) \neq -f(x)$ .

Worked Example A: Reference Triangle Trigonometry

Given $\cos(\theta) = -\frac{3}{5}$ and $\pi < \theta < \frac{3\pi}{2}$ , find the exact values of the remaining 5 trigonometric functions.

💡 Reveal Worked Solution

Step 1: Identify Quadrant and Signs The interval $\pi < \theta < \frac{3\pi}{2}$ indicates Quadrant III. In Quadrant III, $x < 0$ and $y < 0$ . Cosine is defined as $\frac{x}{r} = -\frac{3}{5}$ . Assign $x = -3$ and radius $r = 5$ .
Step 2: Solve for Missing Side ( $y$ ) Using $x^2 + y^2 = r^2$ : $(-3)^2 + y^2 = 5^2 \implies 9 + y^2 = 25 \implies y^2 = 16 \implies y = -4 \quad (\text{since } y < 0)$
Step 3: Construct Trigonometric Functions With $x = -3$ , $y = -4$ , and $r = 5$ : $\sin(\theta) = \frac{y}{r} = -\frac{4}{5}$ $\tan(\theta) = \frac{y}{x} = \frac{-4}{-3} = \frac{4}{3}$ $\csc(\theta) = \frac{r}{y} = -\frac{5}{4}$ $\sec(\theta) = \frac{r}{x} = -\frac{5}{3}$ $\cot(\theta) = \frac{x}{y} = \frac{-3}{-4} = \frac{3}{4}$

Worked Example B: Inverses, Exponentials, and Logarithmic Equations

(a) Find $f^{-1}(x)$ for $f(x) = \frac{3x + 2}{x - 4}$ .
(b) Solve $\log_2(x) + \log_2(x - 2) = 3$ for $x$ .

💡 Reveal Worked Solution

(a) Finding $f^{-1}(x)$ : Replace $f(x)$ with $y$ and swap variables $x$ and $y$ :
$x = \frac{3y + 2}{y - 4}$
- Multiply by $(y - 4)$ and distribute $x$ : $x(y - 4) = 3y + 2 \implies xy - 4x = 3y + 2$
- Isolate $y$ -terms on one side: $xy - 3y = 4x + 2$
- Factor out $y$ and solve: $y(x - 3) = 4x + 2 \implies y = \frac{4x + 2}{x - 3}$ $f^{-1}(x) = \frac{4x + 2}{x - 3}$
(b) Solving Logarithmic Equation: Apply the product rule to condense logarithms:
$\log_2(x(x - 2)) = 3 \implies \log_2(x^2 - 2x) = 3$
- Convert to exponential form ( $2^3 = \text{argument}$ ): $x^2 - 2x = 2^3 \implies x^2 - 2x = 8 \implies x^2 - 2x - 8 = 0$
- Factor the quadratic equation: $(x - 4)(x + 2) = 0 \implies x = 4 \quad \text{or} \quad x = -2$
- Check for extraneous solutions (arguments must be $>0$ ):
  - Testing $x = -2$ : $\log_2(-2)$ is undefined. Reject $x = -2$ .
  - Testing $x = 4$ : $\log_2(4)$ and $\log_2(2)$ are both valid.

\text{Final Answer: } x = 4

Worked Example C: Inverse Trig Evaluation and Function Symmetry

(a) Evaluate $\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ .
(b) Determine whether $f(x) = \frac{x^3}{x^2 + 1}$ is even, odd, or neither.

💡 Reveal Worked Solution

(a) Inverse Trig Evaluation: Let $\theta = \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ . Rephrase as $\sin(\theta) = -\frac{\sqrt{3}}{2}$ where $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . Since the sine value is negative, $\theta$ must lie in Quadrant IV (negative angle range $\left[-\frac{\pi}{2}, 0\right]$ ). The reference angle yielding $\frac{\sqrt{3}}{2}$ is $\frac{\pi}{3}$ . $\theta = -\frac{\pi}{3}$
(b) Algebraic Function Symmetry Test: Substitute $-x$ into $f(x)$ :
$f(-x) = \frac{(-x)^3}{(-x)^2 + 1}$
- Simplify exponents (odd powers retain negative sign, even powers cancel negative sign): $f(-x) = \frac{-x^3}{x^2 + 1}$
- Factor out the negative sign: $f(-x) = -\left(\frac{x^3}{x^2 + 1}\right) = -f(x)$
- Since $f(-x) = -f(x)$ , the function is Odd.

Practice Problem 7

Given $\sin(\theta) = \frac{12}{13}$ and $\frac{\pi}{2} < \theta < \pi$ , find the other 5 trigonometric values. Trigonometry Review

Practice Space Solve this problem on paper or tablet.

The interval tells you $\theta$ is in Quadrant II, meaning the adjacent side $x$ is negative while the opposite side $y$ is positive. Set up a reference triangle where $\text{opposite} = 12$ and $\text{hypotenuse} = 13$ , and use the Pythagorean theorem to find the adjacent side.

Answer: * $\cos\theta = -\frac{5}{13}$

$\tan\theta = -\frac{12}{5}$
$\csc\theta = \frac{13}{12}$
$\sec\theta = -\frac{13}{5}$
$\cot\theta = -\frac{5}{12}$

Step-by-Step Execution

Problem Analysis

We are given: * $\sin(\theta) = \frac{12}{13}$ * $\frac{\pi}{2} < \theta < \pi$ (This indicates that $\theta$ lies in Quadrant II)

In Quadrant II: * Sine ( $\sin$ ) and Cosecant ( $\csc$ ) are positive. * Cosine ( $\cos$ ), Secant ( $\sec$ ), Tangent ( $\tan$ ), and Cotangent ( $\cot$ ) are negative.

Step 1: Find the missing side of the reference triangle Sine is defined as the ratio of the opposite side to the hypotenuse:

\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{12}{13}

Use the Pythagorean theorem (

a^2 + b^2 = c^2

) to find the adjacent side (

x

):

x^2 + 12^2 = 13^2

x^2 + 144 = 169

x^2 = 25

x = \pm 5

Since the angle is in Quadrant II, the adjacent side along the

x

-axis must be negative:

\text{Adjacent} = -5

Step 2: Evaluate the remaining 5 trigonometric functions Using our values ( $\text{Opposite} = 12$ , $\text{Adjacent} = -5$ , $\text{Hypotenuse} = 13$ ):

Cosine ( $\cos(\theta)$ ): $\frac{\text{Adjacent}}{\text{Hypotenuse}}$

\cos(\theta) = -\frac{5}{13}

Tangent ( $\tan(\theta)$ ):

\frac{\text{Opposite}}{\text{Adjacent}}

\tan(\theta) = -\frac{12}{5}

Cosecant ( $\csc(\theta)$ ):

\frac{\text{Hypotenuse}}{\text{Opposite}}

(or the reciprocal of

\sin(\theta)

)

\csc(\theta) = \frac{13}{12}

Secant ( $\sec(\theta)$ ):

\frac{\text{Hypotenuse}}{\text{Adjacent}}

(or the reciprocal of

\cos(\theta)

)

\sec(\theta) = -\frac{13}{5}

Cotangent ( $\cot(\theta)$ ):

\frac{\text{Adjacent}}{\text{Opposite}}

(or the reciprocal of

\tan(\theta)

)

\cot(\theta) = -\frac{5}{12}

Practice Problem 8

Find the inverse, $f^{-1}(x)$ , of the following functions:

(a) $f(x) = \frac{4x-1}{2x+3}$
(b) $f(x) = \ln(5x-1)$

Practice Space Solve this problem on paper or tablet.

Set $y = f(x)$ , swap $x$ and $y$ , and solve for $y$ . For (a), clear the denominator and group all $y$ terms on one side to factor them. For (b), exponentiate both sides with base $e$ to undo the natural log.

Answer: * (a) $f^{-1}(x) = \frac{3x+1}{4-2x}$ (or $\frac{-3x-1}{2x-4}$ )

(b) $f^{-1}(x) = \frac{e^x+1}{5}$

Step-by-Step Execution

(a) $f(x) = \frac{4x-1}{2x+3}$

Step 1: Replace $f(x)$ with $y$ .

y = \frac{4x-1}{2x+3}

Step 2: Swap the positions of $x$ and $y$ to set up the inverse relation.

x = \frac{4y-1}{2y+3}

Step 3: Solve for $y$ . Multiply both sides by the denominator

(2y + 3)

to eliminate the fraction:

x(2y + 3) = 4y - 1

Distribute the

x

:

2xy + 3x = 4y - 1

Move all terms containing

y

to one side and constants/other variables to the opposite side. Subtract

4y

and

3x

from both sides:

2xy - 4y = -3x - 1

Factor out the common variable

y

on the left side:

y(2x - 4) = -3x - 1

Isolate

y

by dividing both sides by

(2x - 4)

:

y = \frac{-3x - 1}{2x - 4}

Optional simplification: Factor out a

-1

from the numerator and denominator to eliminate leading negatives:

y = \frac{-(3x + 1)}{-(4 - 2x)} = \frac{3x + 1}{4 - 2x}

Step 4: Rewrite as $f^{-1}(x)$ .

f^{-1}(x) = \frac{3x + 1}{4 - 2x} \quad \left(\text{or } \frac{-3x - 1}{2x - 4}\right)

(b) $f(x) = \ln(5x-1)$

Step 1: Replace $f(x)$ with $y$ .

y = \ln(5x-1)

Step 2: Swap the positions of $x$ and $y$ .

x = \ln(5y-1)

Step 3: Solve for $y$ . To undo the natural logarithm (

\ln

), convert the equation to exponential form by base-

e

exponentiating both sides (

e^{\ln(\text{something})} = \text{something}

):

e^x = e^{\ln(5y-1)}

e^x = 5y - 1

Add

1

to both sides:

e^x + 1 = 5y

Isolate

y

by dividing by

5

:

y = \frac{e^x + 1}{5}

Step 4: Rewrite as $f^{-1}(x)$ .

f^{-1}(x) = \frac{e^x + 1}{5}

Practice Problem 9

Solve the following equations:

(a) $4^{2x-3} = 64$
(b) $4e^{2x} - 7e^x = 15$
(c) $\log_3 x + \log_3(x-6) = 3$

Practice Space Solve this problem on paper or tablet.

For (a), write 64 as $4^3$ to equate exponents. For (b), let $u = e^x$ , solve the resulting quadratic equation, and discard any negative values for $e^x$ . For (c), use the product property to condense the logs, rewrite in exponential form ( $3^3 = \dots$ ), solve the quadratic, and check for extraneous solutions.

Answer: * (a) $x = 3$

(b) $x = \ln(3)$
(c) $x = 9$ (note: $x = -3$ is extraneous)

Step-by-Step Execution

(a) Solve: $4^{2x-3} = 64$

Step 1: Rewrite both sides with a common base. Notice that $64$ is a power of $4$ ( $4^3 = 64$ ):

4^{2x-3} = 4^3

Step 2: Set the exponents equal to each other. Since the bases are identical, their powers must be equal:

2x - 3 = 3

Step 3: Solve for $x$ . Add

3

to both sides:

2x = 6

Divide by

2

:

x = 3

(b) Solve: $4e^{2x} - 7e^x = 15$

Step 1: Set the equation to zero to form a quadratic-like structure. Subtract $15$ from both sides:

4e^{2x} - 7e^x - 15 = 0

Step 2: Use $u$ -substitution to make factoring clearer. Let

u = e^x

. Since

(e^x)^2 = e^{2x}

, then

u^2 = e^{2x}

. Substitute these into the equation:

4u^2 - 7u - 15 = 0

Step 3: Factor the quadratic equation. We look for two numbers that multiply to

4 \times (-15) = -60

and add up to

-7

. Those numbers are

-12

and

5

. Split the middle term:

4u^2 - 12u + 5u - 15 = 0

Factor by grouping:

4u(u - 3) + 5(u - 3) = 0

(4u + 5)(u - 3) = 0

Step 4: Solve for $u$ .

4u + 5 = 0 \implies u = -\frac{5}{4}

u - 3 = 0 \implies u = 3

Step 5: Substitute $e^x$ back in and solve for $x$ .

Case 1: $e^x = -\frac{5}{4}$ An exponential function with a positive base ( $e \approx 2.718$ ) can never produce a negative output. Therefore, this equation has no real solution.
Case 2: $e^x = 3$ Take the natural logarithm ( $\ln$ ) of both sides to isolate $x$ : $\ln(e^x) = \ln(3)$ $x = \ln(3)$

(c) Solve: $\log_3 x + \log_3(x-6) = 3$ Step 1: Condense the logarithms using logarithmic properties. Use the product property ( $\log_b A + \log_b B = \log_b(AB)$ ):

\log_3(x(x - 6)) = 3

\log_3(x^2 - 6x) = 3

Step 2: Convert the logarithmic equation into exponential form. Rewrite the equation as

\text{base}^{\text{exponent}} = \text{argument}

:

3^3 = x^2 - 6x

27 = x^2 - 6x

Step 3: Move all terms to one side to solve the quadratic equation. Subtract

27

from both sides:

x^2 - 6x - 27 = 0

Step 4: Factor the trinomial. Find two numbers that multiply to

-27

and add to

-6

. Those numbers are

-9

and

3

:

(x - 9)(x + 3) = 0

Step 5: Find the potential solutions.

x - 9 = 0 \implies x = 9

x + 3 = 0 \implies x = -3

Step 6: Check for extraneous solutions. The argument of a logarithm must always be strictly greater than zero (

\text{argument} > 0

).

Test $x = 9$ : $\log_3(9)$ and $\log_3(9-6)=\log_3(3)$ are both valid inputs.
Test $x = -3$ : $\log_3(-3)$ is undefined because you cannot take the logarithm of a negative number. Thus, $x = -3$ is an extraneous solution.

Final Answer:

x = 9

Practice Problem 10

Using the properties of logarithms, express the given quantity as a single logarithm:

\ln(a+b) + \ln(a-b) - 2\ln c

Logarithmic Review

Practice Space Solve this problem on paper or tablet.

First, use the power property to move the coefficient in $2\ln c$ to the exponent ( $\ln(c^2)$ ). Then, use the product property for addition and the quotient property for subtraction to condense into a single log.

Answer: * $\ln\left(\frac{a^2-b^2}{c^2}\right)$ or $\ln\left(\frac{(a+b)(a-b)}{c^2}\right)$

Step-by-Step Execution

Express as a single logarithm: $\ln(a+b) + \ln(a-b) - 2\ln c$ To combine these terms into a single logarithm, we will apply the properties of logarithms step by step.

Step 1: Apply the Power Property to the last term. The power property states that $n\ln(x) = \ln(x^n)$ . Move the coefficient $2$ to the exponent of $c$ :

\ln(a+b) + \ln(a-b) - \ln(c^2)

Step 2: Apply the Product Property to the first two terms. The product property states that $\ln(A) + \ln(B) = \ln(A \cdot B)$ . Combine the first two added logarithms:

\ln[(a+b)(a-b)] - \ln(c^2)

Optional Identity Step: Recognize that

(a+b)(a-b)

is a difference of squares, which simplifies to

a^2 - b^2

:

\ln(a^2 - b^2) - \ln(c^2)

Step 3: Apply the Quotient Property to combine the remaining terms. The quotient property states that $\ln(A) - \ln(B) = \ln\left(\frac{A}{B}\right)$ . Bring the subtracted term into the denominator:

\ln\left(\frac{(a+b)(a-b)}{c^2}\right)

Final Answer Both forms are mathematically correct single logarithms:

\ln\left(\frac{a^2 - b^2}{c^2}\right) \quad \text{or} \quad \ln\left(\frac{(a+b)(a-b)}{c^2}\right)

Segment 6 of 8

📅 Regroup 3

Q/A for skills today

Segment 7 of 8

📅 Synthesis Wrap-up

Recap key skills
Q/A about how the course works

Segment 8 of 8

Using Functions to Understand and Predict Systems

📅 Micro-Lecture

Skill Block 1

Real-Life Engineering Applications

Lidar Safety Envelopes (Robotics & Computer Science)

Step-by-Step Solution:

Worked Example A: Algebraic Expression Evaluation

Worked Example B: Combined Domain and Range Determination

Worked Example C: Piecewise Function Evaluation

Practice Problem 1

Step-by-Step Execution

(a) Find f(1)f(1)f(1) Substitute x=1x = 1x=1 into the function: f(1)=3(1)2−(1)f(1) = 3(1)^2 - (1)f(1)=3(1)2−(1) f(1)=3(1)−1f(1) = 3(1) - 1f(1)=3(1)−1 f(1)=3−1f(1) = 3 - 1f(1)=3−1 f(1)=2f(1) = 2f(1)=2

(b) Find f(−4)f(-4)f(−4) Substitute x=−4x = -4x=−4 into the function: f(−4)=3(−4)2−(−4)f(-4) = 3(-4)^2 - (-4)f(−4)=3(−4)2−(−4) f(−4)=3(16)+4f(-4) = 3(16) + 4f(−4)=3(16)+4 f(−4)=48+4f(-4) = 48 + 4f(−4)=48+4 f(−4)=52f(-4) = 52f(−4)=52

Practice Problem 2

Step-by-Step Execution

Practice Problem 3

Step-by-Step Execution

📅 Regroup 1

Skill Block 2

Real-Life Engineering Applications

Sensor Signal Calibration (Electrical & Systems Engineering)

Step-by-Step Solution:

Worked Example A: Function Composition and Evaluation

Worked Example B: Finding Equations of Lines

Worked Example C: Solving Equations (Algebraic and Trigonometric)

Practice Problem 4

Step-by-Step Execution

Practice Problem 5

Step-by-Step Execution

Practice Problem 6

Step-by-Step Execution

📅 Regroup 2

Skill Block 3

Real-Life Engineering Applications

Robotic Arm Path Planning (Mechanical & Aerospace Engineering)

Step-by-Step Solution:

Worked Example A: Reference Triangle Trigonometry

Worked Example B: Inverses, Exponentials, and Logarithmic Equations

Worked Example C: Inverse Trig Evaluation and Function Symmetry

Practice Problem 7

Step-by-Step Execution

Practice Problem 8

Step-by-Step Execution

Practice Problem 9

Step-by-Step Execution

Practice Problem 10

Step-by-Step Execution

📅 Regroup 3

📅 Synthesis Wrap-up

(a) Find $f(1)$ Substitute $x = 1$ into the function: $f(1) = 3(1)^2 - (1)$ $f(1) = 3(1) - 1$ $f(1) = 3 - 1$ $f(1) = 2$

(b) Find $f(-4)$ Substitute $x = -4$ into the function: $f(-4) = 3(-4)^2 - (-4)$ $f(-4) = 3(16) + 4$ $f(-4) = 48 + 4$ $f(-4) = 52$