Chapter 2.2, Problem 35E

To determine

To show: The set of points where the points is differentiable.

Answer to Problem 35E

The function is differentiable at all real values except at $x=3$ .

Explanation of Solution

Given information:

The function is. $g\left(x\right)=\left\{\begin{array}{c}{\left(x+1\right)}^2,x\le 0\\ 2x+1,0<x<3\\ {\left(4-x\right)}^2,x\ge 3\end{array}\right\}$

Concept used:

The right hand derivative is $\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ and the left hand derivative is $\underset{h\to 0^{-}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ .

Calculation:

The function is $g\left(x\right)=\left\{\begin{array}{c}{\left(x+1\right)}^2,x\le 0\\ 2x+1,0<x<3\\ {\left(4-x\right)}^2,x\ge 3\end{array}\right\}$ .

Right hand derivative at $x=0$ is as shown below.

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{\left[2\left(0+h\right)+1\right]-\left[{\left(0+1\right)}^2\right]}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{2h+1-1}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{2h}{h}\\ =2\end{array}$

Left hand derivative at $x=0$ is as shown below.

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{{\left(0-h+1\right)}^2-{\left(0+1\right)}^2}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{h^2-2h+1-1}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{h\left(h+2\right)}{h}\\ =2\end{array}$

Since the left hand derivative and right hand derivative are equal so the function is differentiable at $x=0$ .

Right hand derivative at $x=3$ is as shown below.

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{f\left(3+h\right)-f\left(3\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{\left[4-{\left(3+h\right)}^2\right]-\left[4-{\left(3\right)}^2\right]}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{4-9-h^2-6h-4+9}{h}\\ =\underset{h\to 0^{+}}{\lim }-\frac{h\left(h+6\right)}{h}\\ =-6\end{array}$

Left hand derivative at $x=3$ is as shown below.

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(3+h\right)-f\left(3\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{\left[2\left(0+h\right)+1\right]-\left[4-{\left(0\right)}^2\right]}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{2h+2-4}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{2h}{h}-\underset{h\to 0^{-}}{\lim }\frac{2}{h}\\ =\infty \end{array}$

Since the left hand derivative and right hand derivative are not equal so the function is not differentiable at $x=3$ .

Conclusion: Thus, the function is differentiable at all real values except at $x=3$ .