Session 01 - Using Functions to Understand and Predict Systems

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

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

Substitute and distribute terms:

Combine like terms:

Domain Solution: Identify all structural restrictions simultaneously:

1. Radicand Restriction: The expression inside the square root must be non-negative:
   $$x + 4 \ge 0 \implies x \ge -4$$
2. 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:

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:

• (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:
  $$k(9) = \sqrt{9} + 5 = 3 + 5 = 8$$

(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:

First, expand $(a+1)^2$: Now substitute that back and distribute the $3$, as well as the negative sign: Combine like terms ($6a - a$ and $3 - 1$):

(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: $[1, \infty)$

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(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.

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: $[0, \infty)$

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(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:

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 $y$:

This shows that $y$ cannot equal $0$, as it would cause division by zero.

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

(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 $8$.

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(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 $1 - x^5$:

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(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 $e^x$:

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

Simplify the radicand:

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

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

• 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 $2$ ($5 = \frac{10}{2}$):
    $$y = -\frac{3}{4}x - \frac{3}{2} + \frac{10}{2}$$
    $$y = -\frac{3}{4}x + \frac{7}{2}$$

• (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: $\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$

(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 $f(x)$:

Substitute the given equations: Optional Simplification (Rationalizing or Factoring): Notice that $x^2 - 1 = (x - 1)(x + 1)$. For $x > 1$, you can rewrite the expression as:

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

Substitute $(x + 1)$ into $f(x) = \sqrt{x^2 - 1}$: Expand the squared binomial $(x + 1)^2 = x^2 + 2x + 1$: Simplify by canceling out $+1$ and $-1$:

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(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 $f(x)$:

Step 2: Plug this result into $g(x)$ Now find $g(\sqrt{3})$ by substituting $\sqrt{3}$ into $g(x) = x + 1$: Therefore:

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:

Substitute $m = \frac{2}{3}$, $x_1 = 1$, and $y_1 = 7$: Distribute the $\frac{2}{3}$ on the right side: Add $7$ to both sides to isolate $y$: To combine $-\frac{2}{3}$ and $7$, find a common denominator ($3$):

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

Substitute the known slope $m = \frac{2}{3}$ and the coordinates of the point $x = 1, y = 7$ to solve for the $y$-intercept ($b$): Subtract $\frac{2}{3}$ from both sides: Substitute $m$ and $b$ back into the slope-intercept equation:

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

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

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

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 $(4 + x)$: Step 3: Combine like terms on both sides. Step 4: Isolate the variable $x$. Add $x$ to both sides: Add $4$ to both sides: Divide by $3$:

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(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 ($ax^2 + bx + c = 0$):

Step 2: Factor the trinomial. We need two numbers that multiply to $4$ and add up to $5$. Those numbers are $1$ and $4$: Step 3: Apply the Zero Product Property. Set each factor equal to zero and solve: Final Answer:

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(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:

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:

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:

• 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 $r = 5$:
  $$\sin(\theta) = \frac{y}{r} = -\frac{4}{5}$$
  $$\tan(\theta) = \frac{y}{x} = \frac{-4}{-3} = \frac{4}{3}$$
  $$\csc(\theta) = \frac{r}{y} = -\frac{5}{4}$$
  $$\sec(\theta) = \frac{r}{x} = -\frac{5}{3}$$
  $$\cot(\theta) = \frac{x}{y} = \frac{-3}{-4} = \frac{3}{4}$$

• (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.

• (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.

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.

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

Use the Pythagorean theorem ($a^2 + b^2 = c^2$) to find the adjacent side ($x$): Since the angle is in Quadrant II, the adjacent side along the $x$-axis must be negative:

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

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

Tangent ($\tan(\theta)$): $\frac{\text{Opposite}}{\text{Adjacent}}$ Cosecant ($\csc(\theta)$): $\frac{\text{Hypotenuse}}{\text{Opposite}}$ (or the reciprocal of $\sin(\theta)$) Secant ($\sec(\theta)$): $\frac{\text{Hypotenuse}}{\text{Adjacent}}$ (or the reciprocal of $\cos(\theta)$) Cotangent ($\cot(\theta)$): $\frac{\text{Adjacent}}{\text{Opposite}}$ (or the reciprocal of $\tan(\theta)$)

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

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

Step 2: Swap the positions of $x$ and $y$ to set up the inverse relation. Step 3: Solve for $y$. Multiply both sides by the denominator $(2y + 3)$ to eliminate the fraction: Distribute the $x$: Move all terms containing $y$ to one side and constants/other variables to the opposite side. Subtract $4y$ and $3x$ from both sides: Factor out the common variable $y$ on the left side: Isolate $y$ by dividing both sides by $(2x - 4)$: Optional simplification: Factor out a $-1$ from the numerator and denominator to eliminate leading negatives: Step 4: Rewrite as $f^{-1}(x)$.

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

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

Step 2: Swap the positions of $x$ and $y$. Step 3: Solve for $y$. To undo the natural logarithm ($\ln$), convert the equation to exponential form by base-$e$ exponentiating both sides ($e^{\ln(\text{something})} = \text{something}$): Add $1$ to both sides: Isolate $y$ by dividing by $5$: Step 4: Rewrite as $f^{-1}(x)$.