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

1 Answer
tom ยท Stefan V.
Jul 16, 2018

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.

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