Consider the following statements about real functions:
(A) A function that is continuous on a closed interval [a, b] has an absolute maximum value and an absolute minimum value on [a, b].
(B) If a function f, differentiable on an open interval I, has a local maximum at a point c of I, then \(f'(c) = 0\).
(C) If \(f'(c) = 0\), then f has either a local maximum or a local minimum at x = c.
(D) A local minimum value of a function is always less than each of its local maximum values.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
A continuous function on a closed interval always attains an absolute maximum and an absolute minimum. At a local extremum of a differentiable function the derivative is zero, but the converse fails:
\( f(x) = x^{3} \): \( f'(0) = 0 \), yet f is increasing through 0
A local minimum value can exceed a local maximum value:
\( \displaystyle g(x) = x + \frac{1}{x} \): local maximum g(−1) = −2, local minimum g(1) = 2
So (A) and (B) are true results, while x3 at 0 and x + 1/x show that (C) and (D) fail.
Checking the statements
(A) A function that is continuous on a closed interval [a, b] has an absolute maximum value and an absolute minimum value on [a, b]. — Correct. Continuity on a closed interval guarantees both.
(B) If a function f, differentiable on an open interval I, has a local maximum at a point c of I, then \(f'(c) = 0\). — Correct. f rises up to c and falls after it, so f′(c) = 0.
(C) If \(f'(c) = 0\), then f has either a local maximum or a local minimum at x = c. — Incorrect. For x3, f′(0) = 0 but 0 is neither.
(D) A local minimum value of a function is always less than each of its local maximum values. — Incorrect. For x + 1/x the local minimum 2 exceeds the local maximum −2.