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]
