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Piecewise Functions

Formulas & Definitions
Core Definition

A piecewise function is a function whose mathematical rule is defined by multiple sub-functions, each applied over a specific sub-interval of the domain:

f(x)={f1(x):xโˆˆD1f2(x):xโˆˆD2โ‹ฎf(x) = \begin{cases} f_1(x) & : x \in D_1 \\ f_2(x) & : x \in D_2 \\ \vdots & \end{cases}

Concept Explanation

Piecewise functions model systems whose underlying physics or operating rules change depending on the input condition. Examples include tax brackets, step voltages, traffic lights, and structural loads along non-uniform beams.

Evaluating Limits of Piecewise Functions

To evaluate limโกxโ†’af(x)\lim_{x \to a} f(x) for a piecewise function:

  1. Interior Points: If aa lies strictly inside one sub-interval (e.g. a<ca < c), evaluate the limit using the corresponding sub-function rule directly.
  2. Boundary Points: If aa is a transition threshold between two pieces (e.g. x=cx = c), you must evaluate both directional limits:
    • Left-hand limit: limโกxโ†’cโˆ’f(x)\lim_{x \to c^-} f(x) using the formula for x<cx < c.
    • Right-hand limit: limโกxโ†’c+f(x)\lim_{x \to c^+} f(x) using the formula for x>cx > c.
  3. Existence Check: Apply the Existence Theorem:
    • If limโกxโ†’cโˆ’f(x)=limโกxโ†’c+f(x)=L\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L, then limโกxโ†’cf(x)=L\lim_{x \to c} f(x) = L.
    • If they disagree, the limit is DNE due to a Jump Discontinuity.

Example Problems

Example Problem 1: Three-Piece Piecewise Function at Boundaries

Consider the function:

f(x)={0:x<โˆ’525โˆ’x2:โˆ’5โ‰คx<45x:xโ‰ฅ4f(x) = \begin{cases} 0 & : x < -5 \\ \sqrt{25 - x^2} & : -5 \le x < 4 \\ 5x & : x \ge 4 \end{cases}

Evaluate: (a) limโกxโ†’โˆ’5f(x)\lim_{x \to -5} f(x) and (b) limโกxโ†’4f(x)\lim_{x \to 4} f(x).

Solution

Part (a) โ€” Boundary at x=โˆ’5x = -5

  • Left limit (x<โˆ’5x < -5): limโกxโ†’โˆ’5โˆ’(0)=0\lim_{x \to -5^-} (0) = 0
  • Right limit (x>โˆ’5x > -5): limโกxโ†’โˆ’5+25โˆ’(โˆ’5)2=0=0\lim_{x \to -5^+} \sqrt{25 - (-5)^2} = \sqrt{0} = 0
  • Since both one-sided limits equal 00:
    limโกxโ†’โˆ’5f(x)=0\lim_{x \to -5} f(x) = 0

Part (b) โ€” Boundary at x=4x = 4

  • Left limit (x<4x < 4): limโกxโ†’4โˆ’25โˆ’(4)2=9=3\lim_{x \to 4^-} \sqrt{25 - (4)^2} = \sqrt{9} = 3
  • Right limit (x>4x > 4): limโกxโ†’4+(5x)=5(4)=20\lim_{x \to 4^+} (5x) = 5(4) = 20
  • Since 3โ‰ 203 \neq 20, the directional limits disagree:
    limโกxโ†’4f(x)=DNE(Jumpย Discontinuity)\lim_{x \to 4} f(x) = \text{DNE} \quad (\text{Jump Discontinuity})

Example Problem 2: Function Value vs Limit Value in Piecewise Definitions

Evaluate limโกxโ†’1g(x)\lim_{x \to 1} g(x) and g(1)g(1) for:

g(x)={2x+1:x<16:x=1x2+2:x>1g(x) = \begin{cases} 2x + 1 & : x < 1 \\ 6 & : x = 1 \\ x^2 + 2 & : x > 1 \end{cases}

Solution

Step 1: Left limit (xโ†’1โˆ’x \to 1^-)

limโกxโ†’1โˆ’(2x+1)=2(1)+1=3\lim_{x \to 1^-} (2x + 1) = 2(1) + 1 = 3

Step 2: Right limit (xโ†’1+x \to 1^+)

limโกxโ†’1+(x2+2)=12+2=3\lim_{x \to 1^+} (x^2 + 2) = 1^2 + 2 = 3

Step 3: Compare with function value

  • Limit: limโกxโ†’1g(x)=3\lim_{x \to 1} g(x) = 3 (because both sides agree).
  • Function value: g(1)=6g(1) = 6.

Conclusion: The value of g(1)=6g(1) = 6 does not alter the limit value limโกxโ†’1g(x)=3\lim_{x \to 1} g(x) = 3. The function exhibits a removable discontinuity (hole) at (1,3)(1, 3).

Connections & References