Q. 11 - Calculus

Question

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

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

about from the left or right?

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

Step-by-Step Solution

Answer

The function $f$ is continuous but not differentiable at $x=1$.

The possible graph can be as follow:

Step 1. Given information

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

Step 2. Check continuity and differentiability at x = 1

Since $f\left(1\right)=3$, $\underset{x\to 1^{-}}{\lim }f\left(x\right)=3$, $\underset{x\to 1^{+}}{\lim }f\left(x\right)=3$, therefore the function is continuous at $x=1$.

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

Left hand derivatives and Right hand derivatives are not equal, So the function is not differentiable at $x=1$.

Step 3. Draw the possible graph of f .

The possible function will be

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

So, the graph will be