Question

What are the critical points of `s(t)=(e^t-2)^4(e^t+7)^5`?

Answer 1

`t=ln2`

Explanation:

This is one of those crazy product rule problems where you have to patiently apply the product rule and then factor out all of the common terms: you should end up with

`e^t(e^t-2)^3 * (e^t+7)^4 * {5(e^t-2) +4(e^t+7)}`

Now the last 'term' simplifies to `9e^t-18` which conveniently simplifies to `9(e^t-2)`.
Soooooo...the only time the equation can be zero is when `e^t-2=0`. The answer is `ln 2`.

Notes: `e^t +7` can never be zero.

If you graph this monstrosity with a very high y max, you will see the function has a minimum at `t=ln2`.