For the function \( f(x)=x^4-\displaystyle\frac{x^3}{3} \) which of the following statements are correct?
(A) \( f(x) \) is increasing on \( (-\infty,0) \)
(B) \( f(x) \) is decreasing on \( \left(\frac{1}{4},\infty\right) \)
(C) \( f(x) \) is increasing on \( \left(\frac{1}{4},\infty\right) \)
(D) \( f(x) \) is decreasing on \( (-\infty,0)\cup\left(0,\frac{1}{4}\right) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C) and (D) only
The sign of the derivative decides where f increases or decreases.
\( \displaystyle f'(x)=4x^3-x^2 \)
\( \displaystyle =x^2(4x-1) \)
Since \( x^2\gt 0 \) for x ≠ 0, the sign of f′(x) is the sign of the factor (4x − 1):
\( \displaystyle f'(x)\lt 0\ \text{for}\ x\lt\frac{1}{4},\ x\ne 0 \)
\( \displaystyle f'(x)\gt 0\ \text{for}\ x\gt\frac{1}{4} \)
So f is decreasing on \( (-\infty,0) \) and on \( \left(0,\frac{1}{4}\right) \), and increasing on \( \left(\frac{1}{4},\infty\right) \).
Checking the statements
(A) \( f(x) \) is increasing on \( (-\infty,0) \) — Incorrect. For x < 0, 4x − 1 < 0, so f′(x) < 0 and f is decreasing.
(B) \( f(x) \) is decreasing on \( \left(\frac{1}{4},\infty\right) \) — Incorrect. f′(x) > 0 for x > 1/4.
(C) \( f(x) \) is increasing on \( \left(\frac{1}{4},\infty\right) \) — Correct. f′(x) > 0 for x > 1/4.
(D) \( f(x) \) is decreasing on \( (-\infty,0)\cup\left(0,\frac{1}{4}\right) \) — Correct. f′(x) < 0 at every point of both intervals.