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Solution

✅ Correct Option: 3

f′(x)=ex−e−x=0f'(x)=e^x-e^{-x}=0 gives e2x=1e^{2x}=1, so x=0x=0. Since f′′(x)=ex+e−x>0f''(x)=e^x+e^{-x}>0, this is a minimum.

The minimum value is f(0)=1+1=2f(0)=1+1=2. (Equivalently, AM-GM on exe^x and e−xe^{-x} gives the same bound.)

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