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

$t = \setminus \ln 2 , \setminus \ln 38$

#### Explanation:

Given function

$s \left(t\right) = \setminus \frac{{\left({e}^{t} - 2\right)}^{4}}{{\left({e}^{t} + 7\right)}^{5}}$

$\setminus \frac{d}{\mathrm{dt}} \left(s \left(t\right)\right) = \setminus \frac{{\left({e}^{t} + 7\right)}^{5} \setminus \frac{d}{\mathrm{dt}} {\left({e}^{t} - 2\right)}^{4} - {\left({e}^{t} - 2\right)}^{4} \setminus \frac{d}{\mathrm{dt}} {\left({e}^{t} + 7\right)}^{5}}{{\left({\left({e}^{t} + 7\right)}^{5}\right)}^{2}}$

$= \setminus \frac{{\left({e}^{t} + 7\right)}^{5} 4 {\left({e}^{t} - 2\right)}^{3} {e}^{t} - {\left({e}^{t} - 2\right)}^{4} 5 {\left({e}^{t} + 7\right)}^{4} {e}^{t}}{{\left({e}^{t} + 7\right)}^{10}}$

$= \setminus \frac{{e}^{t} {\left({e}^{t} + 7\right)}^{4} {\left({e}^{t} - 2\right)}^{3} \left(4 \left({e}^{t} + 7\right) - 5 \left({e}^{t} - 2\right)\right)}{{\left({e}^{t} + 7\right)}^{10}}$

$= \setminus \frac{{e}^{t} {\left({e}^{t} - 2\right)}^{3} \left(38 - {e}^{t}\right)}{{\left({e}^{t} + 7\right)}^{6}}$

For critical points, we have

$\setminus \frac{d}{\mathrm{dt}} \left(s \left(t\right)\right) = 0$

$\setminus \frac{{e}^{t} {\left({e}^{t} - 2\right)}^{3} \left(38 - {e}^{t}\right)}{{\left({e}^{t} + 7\right)}^{6}} = 0$

${\left({e}^{t} - 2\right)}^{3} \left(38 - {e}^{t}\right) = 0 \setminus \quad \left(\setminus \because \setminus \setminus {e}^{t} > 0\right)$

${e}^{t} - 2 = 0 \setminus \mathmr{and} \setminus 38 - {e}^{t} = 0$

${e}^{t} = 2 \setminus \mathmr{and} \setminus {e}^{t} = 38$

$t = \setminus \ln 2 \setminus \mathmr{and} \setminus t = \setminus \ln 38$

$t = \setminus \ln 2 , \setminus \ln 38$