Question

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 \to 2} f(x)
eq 0 ).
e ( x = 2 ) is a vertical asymptote of the graph of ( f ).

Explanation:

Step1: Check Continuity at \( x = 2 \)

To check continuity, we need to verify \( \lim_{x \to 2} f(x) = f(2) \).
First, \( f(2)=\sqrt{|2 - 2|}=\sqrt{0}=0 \).
Now, find the limit as \( x \to 2 \). For \( x \to 2 \), \( |x - 2| \to 0 \), so \( \sqrt{|x - 2|} \to \sqrt{0}=0 \). Thus, \( \lim_{x \to 2} f(x)=0 = f(2) \), so \( f \) is continuous at \( x = 2 \).

Step2: Check Differentiability at \( x = 2 \)

Differentiability at a point requires the left - hand derivative and right - hand derivative to exist and be equal.
The derivative of a function \( y = f(x) \) at \( x = a \) is given by \( f^\prime(a)=\lim_{h\to0}\frac{f(a + h)-f(a)}{h} \).
For \( x = 2 \), the left - hand derivative (LHD) is \( \lim_{h\to0^-}\frac{f(2 + h)-f(2)}{h}=\lim_{h\to0^-}\frac{\sqrt{|(2 + h)-2|}-0}{h}=\lim_{h\to0^-}\frac{\sqrt{|h|}}{h} \). Since \( h\to0^- \), \( |h|=-h \), so \( \frac{\sqrt{-h}}{h}=\frac{\sqrt{-h}}{h}=-\frac{1}{\sqrt{-h}} \) (for \( h
eq0 \)). As \( h\to0^- \), \( \sqrt{-h}\to0^+ \), so \( \lim_{h\to0^-}\frac{\sqrt{|h|}}{h}=-\infty \) (the left - hand derivative does not exist as a finite number).
The right - hand derivative (RHD) is \( \lim_{h\to0^+}\frac{f(2 + h)-f(2)}{h}=\lim_{h\to0^+}\frac{\sqrt{|(2 + h)-2|}-0}{h}=\lim_{h\to0^+}\frac{\sqrt{h}}{h}=\lim_{h\to0^+}\frac{1}{\sqrt{h}}=\infty \) (the right - hand derivative also does not exist as a finite number). Since the left - hand and right - hand derivatives do not exist (or are infinite), \( f \) is not differentiable at \( x = 2 \).

Step3: Analyze Other Options

• Option B: We just showed \( f \) is not differentiable at \( x = 2 \), so B is false.
• Option C: We showed \( f \) is continuous at \( x = 2 \), so C is false.
• Option D: We found \( \lim_{x\to2}f(x)=0 \), so D is false.
• Option E: A vertical asymptote at \( x = a \) means \( \lim_{x\to a^+}f(x)=\pm\infty \) or \( \lim_{x\to a^-}f(x)=\pm\infty \). But \( \lim_{x\to2}f(x)=0 \), so \( x = 2 \) is not a vertical asymptote, E is false.

Answer:

A. \( f \) is continuous but not differentiable at \( x = 2 \).