# What are the inflection points of f(x) = (x^2)(e^18x)?

Sep 26, 2015

You can almost tell that the ${e}^{18}$ is a distractor. It doesn't matter what the exponent is because any ${e}^{C}$ is just a constant.

The first derivative gives you the slope. The second gives you the concavity.

Since that is the case, when the second derivative at some $x$ value is positive, it's concave up and vice versa. A function with an inflection point is either concave up, concave down, or neither. Therefore, if there's somewhere that the second derivative is equal to $0$, then it must be neither concave up nor down, and so it must be an inflection point.

$\frac{d}{\mathrm{dx}} \left[f \left(x\right)\right] = \frac{d}{\mathrm{dx}} \left[{e}^{18} {x}^{3}\right] = {e}^{18} \left(3 {x}^{2}\right)$

Simply from writing out $f \left(x\right)$, you can tell that you have a basic cubic function with no vertical or horizontal shifts. So, I would expect only one inflection point, and it's probably at $x = 0$.

$\frac{d}{\mathrm{dx}} \left[{e}^{18} \left(3 {x}^{2}\right)\right] = {e}^{18} \left(6 x\right)$

Now let's just set this to $0$:

${e}^{18} \left(6 x\right) = 0$

$\textcolor{b l u e}{x = 0}$

...Yep, that's really the only one.