11. Let $f$ be the function defined by $f(x)=\sqrt{|x-2|}$ for all $x$. Which of the following statements is true?
(A) $f$ is continuous but not differentiable at $x=2$.
(B) $f$ is differentiable at $x=2$.
(C) $f$ is not continuous at $x=2$.
(D) $\lim _{x ightarrow 2} f(x)
eq 0$
(E) $x=2$ is a vertical asymptote of the graph of $f$.

Question

11. Let $f$ be the function defined by $f(x)=\sqrt{|x-2|}$ for all $x$. Which of the following statements is true?
(A) $f$ is continuous but not differentiable at $x=2$.
(B) $f$ is differentiable at $x=2$.
(C) $f$ is not continuous at $x=2$.
(D) $\lim _{x ightarrow 2} f(x) 
eq 0$
(E) $x=2$ is a vertical asymptote of the graph of $f$.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We start with the function $f(x) = \sqrt{|x-2|}$. ‖ ‖ step 2 ⋮ To determine continuity at $x=2$, we check the limit: $\lim_{x o 2} f(x) = \sqrt{|2-2|} = 0$. Since $f(2) = 0$, $f$ is continuous at $x=2$. ‖ ‖ step 3 ⋮ To determine differentiability at $x=2$, we need to check the left-hand and right-hand derivatives. ‖ ‖ step 4 ⋮ For $x > 2$, $f(x) = \sqrt{x-2}$, so $f'(x) = \frac{1}{2\sqrt{x-2}}$. As $x o 2^+$, $f'(x) o \infty$. ‖ ‖ step 5 ⋮ For $x < 2$, $f(x) = \sqrt{2-x}$, so $f'(x) = -\frac{1}{2\sqrt{2-x}}$. As $x o 2^-$, $f'(x) o -\infty$. ‖ ‖ step 6 ⋮ Since the left-hand and right-hand derivatives are not equal, $f$ is not differentiable at $x=2$. ‖ ***A*** ∻Key Concept∻ ⚹Continuity and Differentiability⚹ ∻Explanation∻ ⚹A function can be continuous at a point but not differentiable if the left-hand and right-hand derivatives are not equal.⚹◊What is the definition of a function being differentiable at a specific point?⍭ Generate me a similar question◊

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