The function f(x)=x^3+2x^2-1 is(A) increasing in ≤ft((-4)/3,0) (B) de…

The correct answer is Option (3) → (B), (C) and (D) only

    \( \displaystyle f'(x)=3x^2+4x \)

    \( \displaystyle f'(x)=x(3x+4) \)

\( f'(x)=0 \) at x = 0 and x = −4/3. The sign of f′(x) from the factors x and (3x + 4):

On \( \left(-\infty,-\frac{4}{3}\right) \): (−)(−), so \( f'(x)\gt 0 \)
On \( \left(-\frac{4}{3},0\right) \): (−)(+), so \( f'(x)\lt 0 \)
On \( (0,\infty) \): (+)(+), so \( f'(x)\gt 0 \)

So f is increasing on \( \left(-\infty,-\frac{4}{3}\right) \) and on \( (0,\infty) \), and decreasing on \( \left(-\frac{4}{3},0\right) \).

Checking the statements

(A) increasing in \( \left(\displaystyle\frac{-4}{3},0\right) \) — Incorrect. \( f'(x)\lt 0 \) there, so f is decreasing.

(B) decreasing in \( \left(\displaystyle\frac{-4}{3},0\right) \) — Correct. \( f'(x)\lt 0 \) there.

(C) increasing in \( \left(-\infty,\displaystyle\frac{-4}{3}\right) \) — Correct. \( f'(x)\gt 0 \) there.

(D) increasing in \( (0,\infty) \) — Correct. \( f'(x)\gt 0 \) there.