For the function \( f(x)=(x+1)^2(x-2)^2 \), which of the following statements are TRUE?
(A) \( f(x) \) is decreasing on \( (-\infty,-1) \)
(B) \( f(x) \) is increasing on \( \left(-1,\displaystyle\frac{1}{2}\right) \)
(C) \( f(x) \) is decreasing on \( \left(\displaystyle\frac{1}{2},2\right) \)
(D) \( f(x) \) is increasing on \( (2,\infty) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B), (C) and (D)
By the product rule,
\( \displaystyle f'(x)=2(x+1)(x-2)^2+2(x+1)^2(x-2) \)
\( \displaystyle =2(x+1)(x-2)\left[(x-2)+(x+1)\right] \)
\( \displaystyle =2(x+1)(x-2)(2x-1) \)
So \( f'(x)=0 \) at \( x=-1,\ \frac{1}{2},\ 2 \). One test point in each interval gives the sign of \( f'(x) \):
\( \displaystyle f'(-2)=2(-1)(-4)(-5)=-40 \)
\( \displaystyle f'(0)=2(1)(-2)(-1)=4 \)
\( \displaystyle f'(1)=2(2)(-1)(1)=-4 \)
\( \displaystyle f'(3)=2(4)(1)(5)=40 \)
So \( f'(x)\lt 0 \) on \( (-\infty,-1) \) and \( \left(\frac{1}{2},2\right) \), and \( f'(x)\gt 0 \) on \( \left(-1,\frac{1}{2}\right) \) and \( (2,\infty) \).
Checking the statements
(A) \( f(x) \) is decreasing on \( (-\infty,-1) \) — Correct. \( f'(-2)=-40\lt 0 \).
(B) \( f(x) \) is increasing on \( \left(-1,\frac{1}{2}\right) \) — Correct. \( f'(0)=4\gt 0 \).
(C) \( f(x) \) is decreasing on \( \left(\frac{1}{2},2\right) \) — Correct. \( f'(1)=-4\lt 0 \).
(D) \( f(x) \) is increasing on \( (2,\infty) \) — Correct. \( f'(3)=40\gt 0 \).
Shortcut: \( f(x)=\left[(x+1)(x-2)\right]^2 \) is the square of a parabola that is zero at x = −1 and x = 2, so f has minimum value 0 at both zeros and a local maximum midway, at \( x=\frac{1}{2} \). The graph falls, rises, falls, rises, which is exactly (A), (B), (C) and (D).