The function \( f(x)=24x^4-32x^3-96x^2+7 \) on the interval \( \left[-\displaystyle\frac{3}{2},4\right] \) is :
(A) increasing in (−1, 0)
(B) decreasing in (0, 4)
(C) increasing in (2, 4)
(D) decreasing in \( \left(-\displaystyle\frac{3}{2},-1\right) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C) and (D) only
Differentiating,
\( \displaystyle f'(x)=96x^3-96x^2-192x \)
\( \displaystyle =96x(x^2-x-2) \)
\( \displaystyle =96x(x+1)(x-2) \)
The critical points are x = −1, 0, 2. The sign of f′(x) in each part of \( \left[-\frac{3}{2},4\right] \):
\( \displaystyle -\frac{3}{2}\lt x\lt -1:\ x\lt 0,\ x+1\lt 0,\ x-2\lt 0,\ f'(x)\lt 0 \)
\( \displaystyle -1\lt x\lt 0:\ x\lt 0,\ x+1\gt 0,\ x-2\lt 0,\ f'(x)\gt 0 \)
\( \displaystyle 0\lt x\lt 2:\ x\gt 0,\ x+1\gt 0,\ x-2\lt 0,\ f'(x)\lt 0 \)
\( \displaystyle 2\lt x\lt 4:\ x\gt 0,\ x+1\gt 0,\ x-2\gt 0,\ f'(x)\gt 0 \)
So f is decreasing on \( \left(-\frac{3}{2},-1\right) \), increasing on (−1, 0), decreasing on (0, 2) and increasing on (2, 4).
Checking the statements
(A) increasing in (−1, 0) — Correct. f′(x) > 0 on (−1, 0).
(B) decreasing in (0, 4) — Incorrect. f decreases on (0, 2) but increases on (2, 4), so it is not decreasing on the whole of (0, 4).
(C) increasing in (2, 4) — Correct. f′(x) > 0 on (2, 4).
(D) decreasing in \( \left(-\frac{3}{2},-1\right) \) — Correct. f′(x) < 0 on \( \left(-\frac{3}{2},-1\right) \).