Question

The minimum value of \(e^x + e^{-x}\) is

• A. \(-1\)
• B. 0
• C. 1
• D. 2

Hint:

Use AM-GM: \(e^x + e^{-x} \geq 2\sqrt{e^x e^{-x}} = 2\).

Answer 1

The Correct Option is D

Step 1: Set up the function:
Let \(f(x) = e^x + e^{-x}\). It is defined and smooth for all real \(x\).

Step 2: Find the critical point:
\(f'(x) = e^x - e^{-x}\). Setting it to 0 gives \(e^x = e^{-x}\), so \(e^{2x} = 1\) and \(x = 0\).

Step 3: Check that it is a minimum:
\(f''(x) = e^x + e^{-x}\), and \(f''(0) = 2 > 0\). So \(x = 0\) gives a minimum.

Step 4: Find the minimum value:
\(f(0) = e^0 + e^0 = 2\).

Step 5: Check the options:
\(e^x\) is always positive, so the sum can never be \(-1\) or 0. The sum is never below 2, so 1 is not reached either.

Final Answer:
The minimum value is 2, option 4. \[ \boxed{2} \]