Find each x-value at which f is discontinuous and

Questions: Find each x-value at which f is discontinuous and for each x-value, determine whether f is continuous from the right, or from the left, or neither. f(x)= - x+4 if x ≤ 1 - 1 / x if 1<x<2 - sqrt(x-2) if x ≥ 2 x= (smaller value) (smaller value) continuous from the right continuous from the left neither x= (larger value) continuous from the right continuous from the left neither Sketch the graph of f.

Transcript text: Find each $x$-value at which $f$ is discontinuous and for each $x$-value, determine whether $f$ is continuous from the right, or from the left, or neither. \[ \begin{array}{l} f(x)=\left\{\begin{array}{ll} x+4 & \text { if } x \leq 1 \\ 1 / x & \text { if } 1

Solution

Solution Steps

Step 1: Identify points of discontinuity

The function $ f(x) $ is defined piecewise: \[ f(x) = \begin{cases} x + 4 & \text{if } x \leq 1 \\ \frac{1}{x} & \text{if } 1 < x < 2 \\ \sqrt{x - 2} & \text{if } x \geq 2 \end{cases} \]

To find points of discontinuity, we need to check the boundaries of the piecewise function: $ x = 1 $ and $ x = 2 $.

Step 2: Check continuity at $ x = 1 $

For $ x \leq 1 $, $ f(x) = x + 4 $.
For $ 1 < x < 2 $, $ f(x) = \frac{1}{x} $.

Evaluate the left-hand limit and right-hand limit at $ x = 1 $: \[ \lim_{{x \to 1^-}} f(x) = 1 + 4 = 5 \] \[ \lim_{{x \to 1^+}} f(x) = \frac{1}{1} = 1 \]

Since the left-hand limit (5) does not equal the right-hand limit (1), $ f(x) $ is discontinuous at $ x = 1 $.

Step 3: Check continuity at $ x = 2 $

For $ 1 < x < 2 $, $ f(x) = \frac{1}{x} $.
For $ x \geq 2 $, $ f(x) = \sqrt{x - 2} $.

Evaluate the left-hand limit and right-hand limit at $ x = 2 $: \[ \lim_{{x \to 2^-}} f(x) = \frac{1}{2} = 0.5 \] \[ \lim_{{x \to 2^+}} f(x) = \sqrt{2 - 2} = 0 \]

Since the left-hand limit (0.5) does not equal the right-hand limit (0), $ f(x) $ is discontinuous at $ x = 2 $.