Chapter 4, Problem 58RE

a.

To determine

Determine where the function is differentiable.

Answer to Problem 58RE

Differentiable for[-3,3].

Explanation of Solution

Given information: Given function is $\begin{array}{l}f(x)=\sin 2x\text{if}-3\le x<0.\\ f(x)=x^2+2x,\text{if}0\le x<3\end{array}$

Calculation:

  $f(x)=\sin 2x$ is continuous for real values of xand $x^2+2x$ is continuous for all x . The only possible location of a discontinuity is then x =0 since $f(x)$ switches from $\sin 2x\text{to}{x}^2+2x$ at this value. To determine if $f(x)$ is continuous at x =0, we must show $f(0)\text{exists}\underset{x\to 0}{\lim }f(x)\text{exists}\text{and}\underset{x\to 0}{\lim }f(x)=f(0).$

When $\begin{array}{l}x=0,f(0)=0^2+2(0)=0\\ f(0)=0.\end{array}$

  $\begin{array}{l}\underset{x\to 0^{-}}{\lim }f(x)=\underset{x\to 0^{-}}{\lim }\sin 2x=\sin 0=0.\\ \underset{x\to 0^{+}}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }(x^2+2x)=0^2+2(0)=0\end{array}$

The left and right hand limits both equal 0 so $\underset{x\to 0}{\lim }f(x)=0$.Therefore $\underset{x\to 0}{\lim }f(x)=0=f(0).$

So, $f(x)$ is continuous at x =0.

The function is then continuous for (-3,3) since the two pieces of the domain were (-3,0) and (0,3) and it was continuous at x =0.

b.

To determine

Determine where the function is continuous but not differentiable.

Answer to Problem 58RE

Continuous but not differentiable,nowhere..

Explanation of Solution

Given information: Given function is $\begin{array}{l}f(x)=\sin 2x\text{if}-3\le x<0.\\ f(x)=x^2+2x,\text{if}0\le x<3\end{array}$

Calculation:

  $f(x)$ is continuous on its entire domain of (-3,3) so to determine if there are any values of x where $f(x)$ is not differentiable we must determine if there are any values of x that makes $f\text{'}(x)$ discontinuous .

  $\begin{array}{l}f(x)=\sin 2x\text{if}-3\le x<0.\\ f(x)=x^2+2x,\text{if}0\le x<3\\ f\text{'}(x)=\cos 2x.2,\text{if}-3\le x<0.\\ f\text{'}(x)=2x+2,\text{if}0<x<3.\\ f\text{'}(x)=2\cos 2x,\text{if}-3\le x<0.\\ f\text{'}(x)=2x+2,\text{if}0<x\le 3.\\ \underset{x\to 0^{-}}{\lim }f\text{'}(x)=\underset{x\to 0^{-}}{\lim }2\cos 2x.\\ \underset{x\to 0^{-}}{\lim }f\text{'}(0)=\underset{x\to 0^{-}}{\lim }2\cos 0=2\times 1=2\\ \underset{x\to 0+}{\lim }f\text{'}(x)=\underset{x\to 0^{+}}{\lim }(2x+2)\\ \underset{x\to 0+}{\lim }f\text{'}(0)=\underset{x\to 0^{+}}{\lim }(2(0)+2)=2.\\ \end{array}$

The left and right hand limits are equal so $f\text{'}(x)$ is continuous at x =0.Therfore $f(x)$ is n differentiable at x =0.

c.

To determine

Determine where the function neither continuous nor differentiable.

Answer to Problem 58RE

Neither continuous nor differentiable,nowhere.

Explanation of Solution

Given information: Given function is $\begin{array}{l}f(x)=\sin 2x\text{if}-3\le x<0.\\ f(x)=x^2+2x,\text{if}0\le x<3\end{array}$

Calculation:

In summary $f(x)$ neither continuous nor differentiable no where.