prove that the equation is concave up and find minimum value of y

Question

Prove that the graph of $y = 4e^{2x} + e^{-x}$ is concave up, and find the minimum value of y and where it occurs.

I have gotten as far as finding the 1st and 2nd derivative:

$y' = 8e^{2x} - e^{-x}$

$y'' = 16e^{2x} + e^{-x}$

However I am unsure how to prove the concavity of the graph at this point..

you have to state the function is greater or less than 0 then solve and evaluate?

plz help

Answer 1

Recall that $e^x > 0$ for each $x$. Thus, $y'' > 0$ everywhere, and so the function $y = 4e^{2x} + e^{-x}$ is concave up everywhere. To find the minimum, we set the derivative equal to zero and solve the resulting equation. A change of variables should help you to solve this equation.