Question

If $f\left(x\right)=|x|e^x$, then at $x=0$

• A. $f$ is continuous
• B. $f$ is continuous bot not differentiable
• C. $f$ is differentiable
• D. the derivative is $1$

Answer 1

The correct option is B $f$ is continuous bot not differentiable

The function $f\left(x\right)$ is continuous at $x=0$ but not differentiable at $x=0$.

As,

$Lf\left(0\right)=\underset{x\to 0-}{lim}f\left(x\right)$$=\underset{x\to 0-}{lim}(-x{e}^x)=0$

$Rf\left(0\right)=\underset{x\to 0+}{lim}f\left(x\right)$$=\underset{x\to 0+}{lim}\left(x{e}^x\right)=0$

Again $f\left(0\right)=0$.

So $Rf\left(0\right)=Lf\left(0\right)=f\left(0\right)$.

So the function $f\left(x\right)$ is continuous at $x=0$.

But,

$R{f}^{\prime }\left(0\right)=\underset{x\to 0+}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}$

or, $R{f}^{\prime }\left(0\right)=\underset{x\to 0+}{lim}\frac{x{e}^x-0}{x-0}=1$

And,

$L{f}^{\prime }\left(0\right)=\underset{x\to 0-}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}$

or, $L{f}^{\prime }\left(0\right)=\underset{x\to 0-}{lim}\frac{-x{e}^x-0}{x-0}=-1$/

Since $R{f}^{\prime }\left(0\right)\ne L{f}^{\prime }\left(0\right)$.

So the function $f\left(x\right)$ is not differentiable at $x=0$.