# What are the critical points of f(x) =x^(e^x)?

Jan 7, 2017

There are none.

#### Explanation:

A point on $f$ at $x = a$ is a critical point if $f \left(a\right)$ is defined and either $f ' \left(a\right) = 0$ or $f ' \left(a\right)$ is undefined.

So, we first need to find $f ' \left(x\right)$ then set it equal to $0$ and see if it is ever undefined.

$f \left(x\right) = {x}^{{e}^{x}}$

We will use logarithmic differentiation, which is helpful when finding a derivative such as this one, where we have a function whose power is another function. To use this method, begin by taking the natural logarithm of both sides of the equation.

$\ln \left(f \left(x\right)\right) = \ln \left({x}^{{e}^{x}}\right)$

The right side can be simplified by bringing the power of ${e}^{x}$ out of the logarithm using the rule $\ln \left({a}^{b}\right) = b \ln \left(a\right)$:

$\ln \left(f \left(x\right)\right) = {e}^{x} \ln \left(x\right)$

Now take the derivative of both sides of the equation. The left side will need the chain rule, since we're dealing with a composite function (a function within a function). On the right side, we'll use the product rule.

$\frac{1}{f} \left(x\right) f ' \left(x\right) = \left(\frac{d}{\mathrm{dx}} {e}^{x}\right) \ln \left(x\right) + {e}^{x} \left(\frac{d}{\mathrm{dx}} \ln \left(x\right)\right)$

$\frac{1}{f} \left(x\right) f ' \left(x\right) = {e}^{x} \ln \left(x\right) + {e}^{x} \left(\frac{1}{x}\right)$

$\frac{1}{f} \left(x\right) f ' \left(x\right) = \frac{x {e}^{x} \ln \left(x\right) + {e}^{x}}{x}$

$\frac{1}{f} \left(x\right) f ' \left(x\right) = \frac{{e}^{x} \left(x \ln \left(x\right) + 1\right)}{x}$

Now solving for the derivative by multiplying both sides by $f \left(x\right)$, which is equal to ${x}^{{e}^{x}}$:

$f ' \left(x\right) = \frac{{x}^{{e}^{x}} {e}^{x} \left(x \ln \left(x\right) + 1\right)}{x}$

Rewriting the following: ${x}^{{e}^{x}} / x = {x}^{{e}^{x}} / {x}^{1} = {x}^{{e}^{x} - 1}$

$f ' \left(x\right) = {x}^{{e}^{x} - 1} {e}^{x} \left(x \ln \left(x\right) + 1\right)$

We can try to solve for critical points. But first, realize that $f \left(x\right) = {x}^{{e}^{x}}$ is only defined on $x > = 0$. This excludes $x = 0$ from being a critical point, since it's an endpoint of the function.

Letting the derivative equal $0$:

$0 = {x}^{{e}^{x} - 1} {e}^{x} \left(x \ln \left(x\right) + 1\right)$

We have three parts:

${x}^{{e}^{x} - 1} = 0 \text{ "" or "" "e^x=0" " or """ } x \ln \left(x\right) + 1 = 0$

The first is only true if ${e}^{x} - 1 = 0$, or ${e}^{x} = 1$ so $x = 0$, but this cannot be a critical point.

The second is never true.

The third can be graphed to show that $x \ln \left(x\right) + 1 = 0$ is never true.

So, we have found no critical points.

A graph of $f$ confirms this:

graph{x^(e^x) [-.2, 3, -20, 100]}