Question

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

Answer 1

There are none.

Explanation:

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

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

`f(x)=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(f(x))=ln(x^(e^x))`

The right side can be simplified by bringing the power of `e^x` out of the logarithm using the rule `ln(a^b)=bln(a)`:

`ln(f(x))=e^xln(x)`

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.

`1/f(x)f'(x)=(d/dxe^x)ln(x)+e^x(d/dxln(x))`

`1/f(x)f'(x)=e^xln(x)+e^x(1/x)`

`1/f(x)f'(x)=(xe^xln(x)+e^x)/x`

`1/f(x)f'(x)=(e^x(xln(x)+1))/x`

Now solving for the derivative by multiplying both sides by `f(x)`, which is equal to `x^(e^x)`:

`f'(x)=(x^(e^x)e^x(xln(x)+1))/x`

Rewriting the following: `x^(e^x)/x=x^(e^x)/x^1=x^(e^x-1)`

`f'(x)=x^(e^x-1)e^x(xln(x)+1)`

We can try to solve for critical points. But first, realize that `f(x)=x^(e^x)` is only defined on `xgt=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(xln(x)+1)`

We have three parts:

`x^(e^x-1)=0" "" or "" "e^x=0" " or """ "xln(x)+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 `xln(x)+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]}