Question

Let $f\left(x\right)$ be defined as follows:
$f\left(x\right)=\{\begin{array}{c}{x}^6,x^2>1\\ x^3,x^2\le 1\end{array}$.

Then $f\left(x\right)$ is?

• A. continuous everywhere
• B. differentiable everywhere
• C. discontinuous at $x=-1$
• D. not differentiable at $x=1$

Answer 1

Answer: (C), (D)

The correct options are
C not differentiable at $x=1$
D discontinuous at $x=-1$

$f\left(x\right)=\{\begin{array}{c}{x}^6,x^2>1\\ x^3,x^2\le 1\end{array}$
$f\left(x\right)=\{\begin{array}{c}{x}^6,x<-1\\ x^3,-1\le x\le 1\\ x^6,x>1\end{array}$
At $x=-1$, L.H.L $=1\ne$R.H.L $=-1\Rightarrow f\left(x\right)$ is not continuous at $x=-1$
At $x=1$, L.H.D $=3\ne$R.H.D $=6\Rightarrow f\left(x\right)$ is not differentiable at $x=1$.