Question

Discuss the continuity and differentiability of the following function $f\left(x\right)=\{\begin{array}{cc}{x}^2& x<-2\\ 4& -2\le x\le 2\\ x^2& x>2\end{array}$

• A. continuous everywhere except at $x=2$ and $-2$
• B. continuous everywhere
• C. differentiable everywhere
• D. differentiable everywhere except at $x=2$ and $-2$

Answer 1

Answer: (B), (D)

The correct options are
B continuous everywhere
D differentiable everywhere except at $x=2$ and $-2$

$f\left(x\right)=\{\begin{array}{cc}{x}^2,& x<-2\\ 4,& -2\le x\le 2\\ x^2& x>2\end{array}$
$f(-2-)=4$ and $f(-2+)=4$
$f(2-)=4$ and $f(2+)=4$
Therefore, f is continuous everywhere
$f^{\prime }\left(x\right)=\{\begin{array}{cc}2x,& x<-2\\ 0,& -2\le x\le 2\\ 2x& x>2\end{array}$
$f^{\prime }(-2-)=-4$ and $f^{\prime }(-2+)=0$
$f^{\prime }(2-)=0$ and $f^{\prime }(2+)=4$
Therefore, f is differentiable everywhere except at $x=\pm 2$