Question

Suppose $f\left(0\right)=1$, $\underset{x\to 0^{+}}{\lim }f\left(x\right)=1$, $\underset{x\to 0^{-}}{\lim }f\left(x\right)=1$, $\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x}=3$, $\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x}=-2$

(a) Is $f$continuous and/or differentiable at $x=0$? What

about from the left or right?

(b) Sketch a possible graph of $f$.

Short answer

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

The possible graph can be as follow:

Step-by-step solution

Step 1. Given information

It is given that $f\left(0\right)=1$,$\underset{x\to 0^{+}}{\lim }f\left(x\right)=1$,$\underset{x\to 0^{-}}{\lim }f\left(x\right)=1$,$\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x}=3$,$\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x}=-2$

Step 2. Check continuity and differentiability at x=0

Since $f\left(0\right)=1$, $\underset{x\to 0^{+}}{\lim }f\left(x\right)=1$, $\underset{x\to 0^{-}}{\lim }f\left(x\right)=1$, therefore the function is continuous at $x=0$.

$\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x}=3$and $\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x}=-2$

Left hand derivatives and Right hand derivatives are not equal, So the function is not differentiable at$x=0$.

Step 3. Draw the possible graph of f.

The possible function will be

$f\left(x\right)=\left\{\begin{array}{l}3x+1ifx\ge 0\\ -2x+1ifx<0\end{array}\right.$

So, the graph will be