For the function \(f(x) = e^x + e^{-x}\)
(A) \(f'(x) = e^x - e^{-x}\)
(B) The critical point is \(x = 0\)
(C) The minimum value is 1
(D) \(x = 0\) is the point of local minimum.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
\(f(x) = e^x + e^{-x}\)
\(f'(x) = e^x - e^{-x}\)
\(f''(x) = e^x + e^{-x}\)
(A) \(f'(x) = e^x- e^{-x}\) — Correct. The derivative of \(e^{-x}\) is \(-e^{-x}\), so \(f'(x) = e^x - e^{-x}\).
(B) The critical point is \(x = 0\) — Correct. \(f'(x) = 0\) gives \(e^x = e^{-x}\), so \(e^{2x} = 1\) and \(x = 0\), the only critical point.
(C) The minimum value is 1 — Incorrect. The minimum value is \(f(0) = 1 + 1 = 2\).
(D) \(x = 0\) is the point of local minimum. — Correct. \(f''(0) = 2 \gt 0\), so by the second derivative test x = 0 is a point of local minimum.
Shortcut: By the AM–GM inequality, \(e^x + e^{-x} \ge 2\sqrt{e^x \cdot e^{-x}} = 2\), with equality at x = 0. So the minimum value is 2, not 1.