Question

Let $f\left(x\right)=\{\begin{array}{cc}\stackrel{x}{\underset{0}{\int }}\{1+|1-t|\}dt& ifx>2\\ 5x-7& ifx\le 2\end{array}$ then

• A. $f$ is not continuous at $x=2$
• B. $f$ is continuous but not differentiable at $x=2$
• C. $f$ is differentiable everywhere
• D. $f^{\prime }(2+)$ doesn't exist

Answer 1

Answer: (B)

$f$ is continuous but not differentiable at $x=2$

Let $f\left(x\right)=\left\{\begin{array}{cc}{\int }_0^x\{1+|1-t|\}dt& x>2\\ 5x-7& x\le 2\end{array}\right\}$

Let $y={\int }_0^x\{1+|1-t|\}dt$ as $x>2$

$y={\int }_0^1(1+1-t)dt+{\int }_1^{x}tdt$

$y=2-\frac{1}{2}+\frac{{(x}^2-1)}{2}$

$y=1+\frac{x^2}{2}$

$f\left(x\right)=\left\{\begin{array}{cc}1+\frac{x^2}{2},& x>2\\ 5x-7,& x\le 2\end{array}\right\}$

$f\left(2^{+}\right)=1+\frac{4}{2}=3$

$f\left(2^{-}\right)=10-7=3$

Differentiating with respect to $x$ we get

$f^1\left(x\right)=\left\{\begin{array}{cc}x;& x>2\\ 5;& x\le 2\end{array}\right\}$

We can see that it is not differentiable at $x=2$ but continuous at $x=2$