The minimum value of f(x)=e^x+e^-x is

The correct answer is Option (3) → 2

At a critical point the first derivative is zero:

    \( f'(x)=e^{x}-e^{-x}=0 \)

    \( e^{2x}=1 \)

    \( x=0 \)

    \( f''(x)=e^{x}+e^{-x}\gt 0 \)

So x = 0 is a point of local minimum, and since it is the only critical point, it gives the least value:

    \( f(0)=e^{0}+e^{0}=2 \)

Shortcut: By the AM–GM inequality, for the positive numbers \( e^{x} \) and \( e^{-x} \):

    \( \displaystyle e^{x}+e^{-x}\ge 2\sqrt{e^{x}\cdot e^{-x}} \)

    \( \displaystyle =2 \)

with equality at x = 0.