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

1 Answer
Jan 7, 2017

Answer:

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]}