Question

E^(8x)+e^(-x) find the interval on which f is increasing

Answer 1

Assuming ya mean
$f(x)=e^{8x}+e^{-x}$
take derivitive
remember dy/dx e^x=e^x
also the chain rule

so
$f'(x)=8e^{8x}+-1e^{-x}$
find where it equals 0
$0=8e^{8x}+-e^{-x}$
$e^{-x}=8e^{8x}$
multiply both sides by e^x
$1=8e^{9x}$
divide both sides by 8
$\frac{1}{8}=e^{9x}$
take ln of both sides
$ln(\frac{1}{8})=ln(e^{9x})$
$ln(\frac{1}{8})=9x$
divide both sides by 9
$\frac{ln(\frac{1}{8})}{9}=x$

x≈-1/5
f'(-1)<0
f'(0)>0

so it is increasing from x to infinity

the interval it is increasing on is $(\frac{ln(\frac{1}{8})}{9}, \infty)$