Graphing using derivatives

Question

Sketch the graph of the following equation. Show steps of finding out critical numbers, intervals of increase and decrease, absolute maximum and minimum values and concavity.

$$y= xe^{x^2}$$

I found the first derivative which is $y'=(2x^2 + 1)e^{x^2}$ and I know that in order to find min and max the zeroes for $y'$ must be found, but $y'$ doesn't have any real zeroes, and I'm confused about how to go on with solving the problem.

If someone could help me out, that would be appreciated. Thank you in advance.

Answer 1

If $f(x)=xe^{x^2}$, then $f'(x)=2x^2e^{x^2}+e^{x^2}=e^{x^2}(2x^2+1)$, as you have written.

Since the derivative never equals zero or is undefined across the domain, you can conclude there are no critical numbers and no relative extrema.

Can you show that the function is always increasing by showing that $f'(x)>0$ for all $x$?

To determine the concavity of the function, you must now find $f''(x)$. Can you proceed?

Answer 2

You have discovered that the function is increasing, because its derivative is positive, so the function has no maximum nor minimum.

Compute also $$ \lim_{x\to-\infty}xe^{x^2}=-\infty, \qquad \lim_{x\to\infty}xe^{x^2}=\infty, $$ and note there's no oblique asymptote.

For concavity and convexity, compute the second derivative $$ f''(x)=4xe^{x^2}+(2x^2+1)\cdot 2xe^{x^2}=2(2x^3+3x)e^{x^2} $$ The second derivative only vanishes at $0$.