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

Jul 16, 2018

$t = \ln 2$

#### 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

${e}^{t} {\left({e}^{t} - 2\right)}^{3} \cdot {\left({e}^{t} + 7\right)}^{4} \cdot \left\{5 \left({e}^{t} - 2\right) + 4 \left({e}^{t} + 7\right)\right\}$

Now the last 'term' simplifies to $9 {e}^{t} - 18$ which conveniently simplifies to $9 \left({e}^{t} - 2\right)$.
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 $t = \ln 2$.