Chapter 3, Problem 57RE

To determine

To find: The given function is differentiable, continuous but not differentiable, and neither continuous nor differentiable.

Answer to Problem 57RE

Differentiable at $x=\left(-1,0\right)\cup \left(0,4\right)$

Continuous but not differentiable at $x=0$.

Neither continuous nor differentiable at nowhere.

Explanation of Solution

Given information: $\begin{array}{l}f\left(x\right)=\{\begin{array}{l}2x-3,-1\le x<0\\ x-3,0\le x\le 4\end{array}\end{array}$.

The given functions $y=2x-3$ and $x-3$ is continuous over its entire interval because it is linear.

Differentiating $y=2x-3$ ,

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left(2x-3\right)\\ \frac{dy}{dx}=2=m\end{array}$

Differentiating $x-3$ ,

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left(x-3\right)\\ \frac{dy}{dx}=1=m\end{array}$

  $f^{\prime }\left(x\right)=\{\begin{array}{l}2,-1\le x<0\\ 1,0\le x\le 4\end{array}$

Therefore the function is differentiable for $x=\left(-1,0\right)\cup \left(0,4\right)$.

The function is not differentiable at $x=0$.

Conclusion:

The given function is differentiable at $x=\left(-1,0\right)\cup \left(0,4\right)$

The function is continuous but not differentiable at $x=0$.

The function is neither continuous nor differentiable at nowhere.