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The function f(x)=24x4−32x3−96x2+7f(x) = 24x^4 - 32x^3 - 96x^2 + 7 on the interval [−32,4]\left[-\dfrac{3}{2}, 4\right] is :

A. increasing in (−1,0)(-1,0)

B. decreasing in (0,4)(0,4)

C. increasing in (2,4)(2,4)

D. decreasing in (−32,−1)\left(-\dfrac{3}{2},-1\right)

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 4

f′(x)=96x(x−2)(x+1)f'(x) = 96x(x-2)(x+1), with sign changes at −1-1, 00 and 22. So ff decreases on (−32,−1)\left(-\frac{3}{2},-1\right), increases on (−1,0)(-1,0), decreases on (0,2)(0,2) and increases on (2,4)(2,4). Hence A, C, D hold, while B fails on (2,4)(2,4).

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