The minimum value of \(f(x) = e^{x} + e^{-x}\), \(x \in R\), is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
Differentiating:
\(f'(x) = e^{x} - e^{-x}\)
Setting f’(x) = 0 gives \(e^{x} = e^{-x}\), that is \(e^{2x} = 1\), so the only critical point is x = 0.
\(f''(x) = e^{x} + e^{-x} \gt 0\)
So x = 0 gives the minimum (f decreases before 0 and increases after it), and the minimum value is
\(f(0) = 1 + 1 = 2\)
Why the other options are incorrect
Option (1) 0 — Incorrect. This is the critical point x = 0, not the value of f there; f is always positive.
Option (2) 1 — Incorrect. This counts only one of the two terms at x = 0.
Option (4) \(e + \displaystyle\frac{1}{e}\) — Incorrect. This is f(1), which is larger than f(0) = 2.