assuming ya mean
[tex]f(x)=e^{8x}+e^{-x}[/tex]
take derivitive
remember dy/dx e^x=e^x
also the chain rule
so
[tex]f'(x)=8e^{8x}+-1e^{-x}[/tex]
find where it equals 0
[tex]0=8e^{8x}+-e^{-x}[/tex]
[tex]e^{-x}=8e^{8x}[/tex]
multiply both sides by e^x
[tex]1=8e^{9x}[/tex]
divide both sides by 8
[tex]\frac{1}{8}=e^{9x}[/tex]
take ln of both sides
[tex]ln(\frac{1}{8})=ln(e^{9x})[/tex]
[tex]ln(\frac{1}{8})=9x[/tex]
divide both sides by 9
[tex]\frac{ln(\frac{1}{8})}{9}=x[/tex]
x≈-1/5
f'(-1)<0
f'(0)>0
so it is increasing from x to infinity
the interval it is increasing on is [tex](\frac{ln(\frac{1}{8})}{9}, \infty)[/tex]
