Q. 75. - 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.

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

Step-by-Step Solution

Answer

The value of the function is 1.

$$\begin{gathered} L.H.D.=2 \\ R.H.D.=2 \end{gathered}$$.

Step 1. Given Information.

The given function is 

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

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:

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

Step 3. Evaluation.

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 piecewisedefined function  to evaluate $f\left(1+h\right)$. Of course we still have , so we still use the second part of  to evaluate :

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

Since the left and right derivatives of  at $x=1$are equal to each other, the derivative

$f\left(1\right) at x=1$ is defined.

The value of $f\left(1\right)=1$.