Question

1. let (f(x)=|x - 2|). (a) use the definition of derivative to determine if (f(2))exists. (b) find (f(x)) and sketch the graph of (f(x)). (c) is (f(x)) continuous at (x = 2), justify your answer.

Explanation:

Step1: Recall definition of derivative

The definition of the derivative of a function $y = f(x)$ at $x = a$ is $f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. For $f(x)=|x - 2|$ and $a = 2$, we have $f(2)=|2 - 2|=0$, and $f(2 + h)=|(2 + h)-2|=|h|$. Then $\lim_{h
ightarrow0}\frac{f(2 + h)-f(2)}{h}=\lim_{h
ightarrow0}\frac{|h|}{h}$.

Step2: Find left - hand and right - hand limits

The left - hand limit as $h
ightarrow0^{-}$: $\lim_{h
ightarrow0^{-}}\frac{|h|}{h}=\lim_{h
ightarrow0^{-}}\frac{-h}{h}=- 1$. The right - hand limit as $h
ightarrow0^{+}$: $\lim_{h
ightarrow0^{+}}\frac{|h|}{h}=\lim_{h
ightarrow0^{+}}\frac{h}{h}=1$. Since $\lim_{h
ightarrow0^{-}}\frac{|h|}{h}
eq\lim_{h
ightarrow0^{+}}\frac{|h|}{h}$, $f^{\prime}(2)$ does not exist.

Step3: Find $f^{\prime}(x)$ for $x

eq2$
When $x>2$, $f(x)=x - 2$, and $f^{\prime}(x)=\frac{d}{dx}(x - 2)=1$. When $x<2$, $f(x)=-(x - 2)=2 - x$, and $f^{\prime}(x)=\frac{d}{dx}(2 - x)=-1$. So $f^{\prime}(x)=

$$\begin{cases}-1, & x<2\\\text{does not exist},&x = 2\\1,&x>2\end{cases}$$

$

Step4: Check continuity at $x = 2$

We know that $\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{-}}(2 - x)=0$, $\lim_{x
ightarrow2^{+}}f(x)=\lim_{x
ightarrow2^{+}}(x - 2)=0$, and $f(2)=0$. Since $\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{+}}f(x)=f(2)$, $f(x)$ is continuous at $x = 2$.

Answer:

(a) $f^{\prime}(2)$ does not exist because $\lim_{h
ightarrow0^{-}}\frac{f(2 + h)-f(2)}{h}=-1$ and $\lim_{h
ightarrow0^{+}}\frac{f(2 + h)-f(2)}{h}=1$ and the left - hand and right - hand limits are not equal.
(b) $f^{\prime}(x)=

$$\begin{cases}-1, & x<2\\\text{does not exist},&x = 2\\1,&x>2\end{cases}$$

$. To sketch $y = f^{\prime}(x)$, draw a horizontal line $y=-1$ for $x<2$, a hole at the point $(2,-1)$ and $(2,1)$, and a horizontal line $y = 1$ for $x>2$.
(c) $f(x)$ is continuous at $x = 2$ since $\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{+}}f(x)=f(2)=0$.