Question

Let $f\left(x\right)=\{\begin{array}{cc}{x}^3,& x<0\\ x^2,& x\ge 0\end{array}$ then

• A. $f\left(x\right)$ is continuous at x = 0
• B. $f\left(x\right)$ is differentiable at x = 0
• C. $f^{\prime }\left(x\right)$ is continuous on R
• D. $f^{\prime \prime }\left(x\right)$ exists for all $xϵR$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f\left(x\right)$ is continuous at x = 0
B $f\left(x\right)$ is differentiable at x = 0
C $f^{\prime }\left(x\right)$ is continuous on R

$f\left(0^{+}\right)=f\left(0^{-}\right)=f\left(0\right)$
$f\left(0^{+}\right)=f\left(0^{-}\right)=f\left(0\right)=0$
So, $f\left(x\right)$ is continuous at $x=0$
$f^{\prime }\left(x\right)=\{\begin{array}{cc}3x^2,& x<0\\ 2x,& x\ge 0\end{array}$
$f^{\prime }\left(0^{+}\right)=f^{\prime }\left(0^{-}\right)=0,f^{\prime }\left(0\right)=0$
So, $f\left(x\right)$ is differentiable and continuous at $x=0$ and $f^{\prime }\left(x\right)$ is continuous but non differentiable at $x=0$. Hence, $f^{\prime \prime }\left(x\right)$ will not be defined at $x=0$
$f^{\prime \prime }\left(x\right)=\{\begin{array}{cc}6x,& x<0\\ 2,& x>0\end{array}$
$\begin{array}{cc}f"\left(0^{+}\right)=2& \\ f"\left(0^{-}\right)=0& \end{array}$
So, $f^{\prime \prime }\left(x\right)$ is not defined at $x=0$