Draw the graph of `y=e^(x)+e^(-2x)`

Explanation

We have `y=f(x)=e^(x)+e^(-2x)`
Clearly, the doamin of the function is R.
`f(0) =1+1=2`
Also `f(x) gt 0, AA x in R`
`f^(')(x) = e^(x)-2e^(-x)= e^(-2x)e^(3x-2)`
`f^(')(x)=0 therefore e^(3x)=2` or `x=1/3log2`
`f(1/3log2) e^(1/3log2) + e^(-2/3)log2=2^(1//3) + 2^(-2//3) = 3/2^(2/3)`
`f(x)` decreases in `(-infty, 1/3 log2)` and increases in `(1/3log2, infty)`
Hence the graph of the function is as shown in the following figure.