The minimum value of e^x+e^-x is

The correct answer is Option (4) → 2

Let \( f(x)=e^x+e^{-x} \). At a critical point, f′(x) = 0.

    \( \displaystyle f'(x)=e^x-e^{-x} \)

    \( \displaystyle e^x=e^{-x} \)

    \( \displaystyle e^{2x}=1\ \Rightarrow\ x=0 \)

    \( \displaystyle f''(x)=e^x+e^{-x}\gt 0\ \text{for all}\ x \)

So f has a local minimum at x = 0, and since it is the only critical point, this is the least value.

    \( \displaystyle f(0)=e^0+e^0 \)

    \( \displaystyle =1+1=2 \)

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

    \( \displaystyle \frac{e^x+e^{-x}}{2}\ge\sqrt{e^x\cdot e^{-x}}=1 \)

So \( e^x+e^{-x}\ge 2 \), with equality when \( e^x=e^{-x} \), that is, x = 0.