If \( f(x)=x^4-\displaystyle\frac{2x^3}{3} \), then \( f(x) \) is
A. increasing in \( \left(\displaystyle\frac{1}{2},\infty\right) \)
B. decreasing in \( \left(\displaystyle\frac{1}{2},\infty\right) \)
C. increasing in \( (-\infty,0) \)
D. decreasing in \( (-\infty,0)\cup\left(0,\displaystyle\frac{1}{2}\right) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A and D only
\( f'(x)=4x^3-2x^2 \)
\( f'(x)=2x^2(2x-1) \)
The factor \( 2x^2 \) is positive for every x ≠ 0, so the sign of \( f'(x) \) is the sign of \( (2x-1) \).
So \( f'(x)\gt 0 \) for x > 1/2, and \( f'(x)\lt 0 \) for x < 1/2 (x ≠ 0).
Checking the statements
(A) increasing in \( \left(\displaystyle\frac{1}{2},\infty\right) \) — Correct. \( f'(x)\gt 0 \) for x > 1/2.
(B) decreasing in \( \left(\displaystyle\frac{1}{2},\infty\right) \) — Incorrect. \( f'(x)\gt 0 \) for x > 1/2, so f is increasing there.
(C) increasing in \( (-\infty,0) \) — Incorrect. \( f'(x)\lt 0 \) for x < 0, so f is decreasing there.
(D) decreasing in \( (-\infty,0)\cup\left(0,\displaystyle\frac{1}{2}\right) \) — Correct. \( f'(x)\lt 0 \) on both intervals.