Question

show that the function ( f(x) = |x - 2| ) is not differentiable at 2.

we have
f(x) = |x - 2| = \begin{cases} square & \text{if } x geq 2 \\ square & \text{if } x < 2 end{cases}

the right-hand limit is
lim_{x \to 2^+} \frac{f(x) - f(2)}{x - 2} = square,
and the left-hand limit is
lim_{x \to 2^-} \frac{f(x) - f(2)}{x - 2} = square.

since these limits are not equal, ( f(2) = lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} ) does not exist and ( f ) is not differentiable at 2.

find a formula for ( f ) and sketch its graph.
f(x) = \begin{cases} square & \text{if } x > 2 \\ square & \text{if } x < 2 end{cases}
graph lay...

Explanation:

Step1: Rewrite the absolute value function

For \(|x - 2|\), when \(x\geq2\), \(x - 2\geq0\), so \(|x - 2|=x - 2\). When \(x\lt2\), \(x - 2\lt0\), so \(|x - 2|=-(x - 2)=2 - x\). Thus, \(f(x)=

$$\begin{cases}x - 2, & x\geq2\\2 - x, & x\lt2\end{cases}$$

\)

Step2: Calculate the right - hand limit of the difference quotient

First, find \(f(2)\). Substitute \(x = 2\) into \(f(x)\), when \(x = 2\), \(f(2)=|2 - 2| = 0\). For the right - hand limit as \(x
ightarrow2^{+}\), \(x\gt2\), so \(f(x)=x - 2\). Then the difference quotient is \(\frac{f(x)-f(2)}{x - 2}=\frac{(x - 2)-0}{x - 2}\). Simplify this expression: \(\lim_{x
ightarrow2^{+}}\frac{f(x)-f(2)}{x - 2}=\lim_{x
ightarrow2^{+}}\frac{x - 2}{x - 2}=\lim_{x
ightarrow2^{+}}1 = 1\)

Step3: Calculate the left - hand limit of the difference quotient

For the left - hand limit as \(x
ightarrow2^{-}\), \(x\lt2\), so \(f(x)=2 - x\). The difference quotient is \(\frac{f(x)-f(2)}{x - 2}=\frac{(2 - x)-0}{x - 2}=\frac{-(x - 2)}{x - 2}\). Simplify this expression: \(\lim_{x
ightarrow2^{-}}\frac{f(x)-f(2)}{x - 2}=\lim_{x
ightarrow2^{-}}\frac{-(x - 2)}{x - 2}=\lim_{x
ightarrow2^{-}}(-1)=-1\)

Step4: Find the derivative formula

For \(x\gt2\), \(f(x)=x - 2\), and the derivative of \(y=x - 2\) with respect to \(x\) is \(f^{\prime}(x)=1\) (using the power rule \((x^n)^\prime=nx^{n - 1}\), here \(n = 1\) for \(x\) and the derivative of a constant \(-2\) is \(0\)). For \(x\lt2\), \(f(x)=2 - x\), and the derivative of \(y = 2-x\) with respect to \(x\) is \(f^{\prime}(x)=-1\) (the derivative of \(2\) is \(0\) and the derivative of \(-x\) is \(-1\)). So \(f^{\prime}(x)=

$$\begin{cases}1, & x\gt2\\-1, & x\lt2\end{cases}$$

\)

Filling in the blanks:

• \(f(x)=|x - 2|=

$$\begin{cases}\boldsymbol{x - 2} & \text{if }x\geq2\\\boldsymbol{2 - x} & \text{if }x\lt2\end{cases}$$

\)

• The right - hand limit \(\lim_{x

ightarrow2^{+}}\frac{f(x)-f(2)}{x - 2}=\boldsymbol{1}\)

• The left - hand limit \(\lim_{x

ightarrow2^{-}}\frac{f(x)-f(2)}{x - 2}=\boldsymbol{-1}\)

• \(f^{\prime}(x)=

$$\begin{cases}\boldsymbol{1} & \text{if }x\gt2\\\boldsymbol{-1} & \text{if }x\lt2\end{cases}$$

\)

Answer:

• Piecewise function for \(f(x)\): \(f(x)=

$$\begin{cases}x - 2, & x\geq2\\2 - x, & x\lt2\end{cases}$$

\)

• Right - hand limit: \(1\)
• Left - hand limit: \(-1\)
• Derivative \(f^{\prime}(x)\): \(f^{\prime}(x)=

$$\begin{cases}1, & x\gt2\\-1, & x\lt2\end{cases}$$

\)