Q. 76. - Calculus

Question

In Exercises 69–80, determine whether or not f is continuous and/or differentiable at the given value of x. If not, determine any left or right continuity or differentiability. For the last four functions, use graphs instead of the definition of the derivative.

$f\left(x\right)=\left\{\begin{array}{l}{x}^2 , if x\le 1\\ 2x+4 , if x>1,\end{array}\right.x=1$.

Step-by-Step Solution

Answer

The L.H.D. $=\infty .$

$R.H.D.=2$

Step 1. Given Information.

The given function is:

$f\left(x\right)=\left\{\begin{array}{l}{x}^2 , if x\le 1\\ 2x+4 , if x>1,\end{array}\right.x=1$.

Step 2. Differentiability.

We start by finding the right derivative of $f at x=1$ . In this case we examine $h\to 0^{+}$, which means that $h>0$, and thus $1+h>1$. Therefore we will use the second part of the piecewise-defined function $f$to evaluate $f\left(1+h\right)$ in this case:

$\begin{array}{rcl}f{\text{'}}_{+}\left(1\right)& =& \underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ & =& \underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ & =& \underset{h\to 0^{+}}{\lim }\frac{2\left(1+h\right)+4-1}{h}\\ & =& \underset{h\to 0^{+}}{\lim }\frac{2+2h+3}{h}\\ & =& \underset{h\to 0^{+}}{\lim }\frac{2h+5}{h}\\ & =& 2+\infty \\ & =& \infty .\end{array}$

Step 3. Calculation.

In contrast, when we calculate the left derivative of $f at x=1$ we will have $h\to 0^{-}$, and thus $h<0$. This means that $1+h<1$, so we will use the first part of the piecewise defined function $f$ to evaluate $f\left(1+h\right)$. Of course we still have $1\ge 1$, so we still use the second part of  $f$ to evaluate $f\left(1\right)$:

$\begin{array}{rcl}f{\text{'}}_{-}\left(1\right)& =& \underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ & =& \underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ & =& \underset{h\to 0^{-}}{\lim }\frac{{\left(1+h\right)}^2-1}{h}\\ & =& \underset{h\to 0^{-}}{\lim }\frac{1+2h+h^2-1}{h}\\ & =& \underset{h\to 0^{-}}{\lim }\frac{2h+h^2}{h}\\ & =& 2.\end{array}$

Since the left and right derivatives of $f$at $x=1$are not equal to each other, the derivative$f\text{'}\left(1\right) of f at x=1$ is undefined. Note that $f$ is left differentiable and right differentiable at $x=1$, but not differentiable at $x=1$.