# For the function g(x) = e^{-x^{2}}, a) find the critical numbers. b) find the intervals of...

## Question:

For the function {eq}\displaystyle\; g(x) = e^{-x^{2}} {/eq},

a) find the critical numbers.

b) find the intervals of increase and decrease.

c) find all inflection points.

## Qualitative Features of a Function

To extract qualitative features of a graph of a given function, {eq}\displaystyle f(x) {/eq} we will look for the intervals where the function is increasing or decreasing,

for critical points or intervals where the function is concave up or down.

To see where a function {eq}\displaystyle y=f(x) {/eq} is increasing or decreasing, we look at the sign of its derivative: {eq}\displaystyle \text{ if } f'(x)>0 \implies f(x) \text{ is increasing} \text{ and } \displaystyle \text{ if } f'(x)<0 \implies f(x) \text{ is decreasing}. {/eq}

The concavity of a function, when the graph is not given, is determined by the second derivative:

{eq}\displaystyle \text{ if } f''(x)>0 \implies f(x) \text{ is concave up}\\ \displaystyle \text{ if } f''(x)<0 \implies f(x) \text{ is concave down}. {/eq}

The points where a functions changes concavity are called inflection points.

## Answer and Explanation:

a) The critical points of the function {eq}\displaystyle g(x)=e^{-x^2} {/eq}

are given by the zeros of the first derivative or the points where the first derivative is not defined.

The derivative function is calculated using Chain rule, as below

{eq}\displaystyle \begin{align} g'(x)&=\frac{d}{dx}\left(e^{-x^{2}}\right)=-2xe^{-x^{2}}. \end{align} {/eq}

Because the derivative is well-defined everywhere, the critical points are obtained only from the zeros of the first derivative.

So, the critical points are the solutions of

{eq}\displaystyle \begin{align} g'(x)&=0 \iff -2xe^{-x^{2}}=0\implies \boxed{x=0 \text{ is the only critical point }}. \end{align} {/eq}

b. The intervals of increasing/decreasing are the intervals where {eq}\displaystyle g'(x)>0 \text{ or }g'(x)<0. {/eq}

So, the function is increasing when {eq}\displaystyle g'(x)>0\iff -2xe^{-x^2}>0\implies x<0 (\text{ because the exponential term is always positive })\iff \boxed{\text{ on }(-\infty, 0) \text{ the function is increasing}}. {/eq}

and the function is decreasing when {eq}\displaystyle g'(x)<0\iff -2xe^{-x^2}<0\implies x>0 \iff \boxed{\text{ on }(0,\infty) \text{ the function is decreasing}}. {/eq}

c. The inflection points, are the points where the second derivative changes signs or the graph changes concavity.

The second derivative is calculated with product rule and chain rule.

{eq}\displaystyle g''(x)=\frac{d}{dx}\left(-2xe^{-x^2}\right)=\left(-2+4x^2\right)e^{-x^2}. {/eq}

The second derivative changes signs when the quadratic function {eq}\displaystyle -2+4x^2 {/eq} changes signs, because the exponential term is positive for all values.

So, the quadratic function changes signs when {eq}\displaystyle 4x^2-2=0\iff x=\pm \frac{\sqrt{2}}{2}. {/eq}

Therefore, {eq}\displaystyle \boxed{\text{ the inflection points are } -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}}. {/eq}