Question

The function $f\left(x\right)=1–e^{–\frac{x^2}{2}}$

• A. Decreasing for all $x$
• B. Increasing for all $x$
• C. Decreasing for $x<0$ and increasing for $x>0$
• D. Increasing for $x<0$ and decreasing for $x>0$

Answer 1

The correct option is C

Decreasing for $x<0$ and increasing for $x>0$

Explanation for correct option

Given function, $f\left(x\right)=1–e^{–\frac{x^2}{2}}$

So, $f\text{'}\left(x\right)=\frac{d\left(1-e^{\frac{–x^2}{2}}\right)}{dx}$

$=–e^{\frac{–x^2}{2}}\frac{d\frac{–x^2}{2}}{dx}$[$\mathbf{}\mathbf{}\mathbf{}\mathbf{\because }\frac{\mathbf{d}{\mathbf{e}}^{\mathbf{x}}}{\mathbf{d}\mathbf{x}}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\mathbf{}$ and applying chain rule]

$=x{e}^{\frac{–x^2}{2}}$

The critical point of $f\left(x\right)$ are the values of $x$ for which $f\text{'}\left(x\right)=0$.
Thus, $f\text{'}\left(x\right)=0$ when $x=0$

So, $x=0$ is the only critical point.

Also, $f\text{'}\left(x\right)=x{e}^{\frac{–x^2}{2}}<0$ when $x<0$. Thus, $f\left(x\right)$ decreases in this interval.

And, $f\text{'}\left(x\right)=x{e}^{\frac{–x^2}{2}}>0$ when $x>0$. Thus, $f\left(x\right)$ increases in this interval.

Hence, the correct option is $\left(\mathrm{C}\right)$.