We have been given that
[tex] f(x)=e^{9x}+e^{-x} [/tex]
On differentiating we get
[tex] f'(x)=9e^{9x}-e^{-x} [/tex]
For the function to be increasing, the first derivative must be positive
f'(x)>0
[tex] 9e^{9x}-e^{-x}>0 [/tex]
[tex] \left (9e^{-9x}-e^{-x} \right )>0
9e^{9x}-\frac{1}{e^x}>0
\frac{9e^{10x}-1}{e^x}>0
9e^{10x}-1>0
e^{10x} >\frac{1}{9}
[/tex]
[tex] x>\frac{1}{10}ln(9^{-1})
[/tex]
[tex] x>-\frac{1}{5}ln3
[/tex]
Thus, the required interval is given by
[tex] x\in \left ( -\frac{ln3}{5},\infty \right ) [/tex]
