Is anyone good with calculus and exponential functions?

Question

Anyone good with basic calc? I'm stuck on these two types of problems.

Find the extrema and points of inflection of the function:
f(x) = (e^x + e^(-x))/(2)

I got to the first derivative, giving me (2(e^x - e^(-x)))/(4) = 0... and I'm stuck trying to solve for x.

Then there's:
int( (e^(1/x^2))/(x^3))

I just have no idea where to go from there. I used u-substitution, but I can't find du.

Answer 1

f(x) = (e^x + e^(-x))/(2)
First, distribute the division by 2:
f(x) = 1/2 e^x + 1/2 e^(-x)

Derivative: if e^u = e^u du/dx. So the derivative of e^-x = -e^-x.
So, here are our 2 derivatives you need to find your extrema:
f'(x) = 1/2 e^x - 1/2 e^(-x)
f''(x) = 1/2 e^x + 1/2 e^(-x)

2.) int: ((e^(1/x^2))/(x^3))
The trick here is to find something to get rid of the 1/x^2 and 1/x^3.

Let's rewrite the formula to:
int: e^(x^-2) * x^-3 dx

Let u = x^-2 (or 1/x^2)
du = -2x^-3 dx
-1/2 du = x^-3 dx

That gives us an integral of -1/2 e^u du

Integrate to get -1/2 e^u = -1/2 e^x-2 + c or -1/2 e^(1/x^2) + c (solution1)

Answer 2

Try graphing these functions to get some intuition.

To solve for x, multiply both sides by 4/2, and you end up
with e^x=e^-x. If the answer isn't obvious try graphing
the left-hand and right-hand side of the equation.