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

1 Answer
Jul 16, 2018

Answer:

#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#.