Chapter 3, Problem 54RE

a.

To determine

To find: The given function $f\left(x\right)=\{\begin{array}{l}2\sin x,x\le 0\\ mx,x>0\end{array}$ is continuous at $x=0$.

Answer to Problem 54RE

Yes, the given function $f\left(x\right)=\{\begin{array}{l}2\sin x,x\le 0\\ mx,x>0\end{array}$is continuous at $x=0$.

Explanation of Solution

Given:

  $f\left(x\right)=\{\begin{array}{l}2\sin x,x\le 0\\ mx,x>0\end{array}$

The given function $f$ approaches $0$ from both right side and left. Therefore it is continuous at $x=0$.

Conclusion:

Thus we can conclude that the given function $f\left(x\right)=\{\begin{array}{l}x,0\le x\le 1\\ 2-x,1<x\le 2\end{array}$is continuous at $x=0$

.

b.

To determine

To find: The given function $f\left(x\right)=\{\begin{array}{l}2\sin x,x\le 0\\ mx,x>0\end{array}$ is differentiable at $x=0$.

Answer to Problem 54RE

Yes, the given function $f$ is differentiable at $x=0$ .

Explanation of Solution

Given:

  $f\left(x\right)=\{\begin{array}{l}2\sin x,x\le 0\\ mx,x>0\end{array}$

Concept Used:

  $f$ at $x=0$ have the limit points $\underset{x\to 0^{-}}{\lim }{f}^{\prime }\left(x\right)=\underset{x\to 0^{+}}{\lim }{f}^{\prime }\left(x\right)$ is differentiable for any value of $m$.

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}2\sin x\\ \frac{dy}{dx}=2\frac{d}{dx}\sin x\\ \frac{dy}{dx}=2\cos x\end{array}$

When $x=0$ ,

  $\begin{array}{c}\frac{dy}{dx}=2\cos 0\\ \frac{dy}{dx}=2\left(1\right)\\ \frac{dy}{dx}=2\\ m=2\end{array}$

Conclusion:

Thus we can conclude that the function is differentiable at $x=0$.