A function f(x) is given by, f(x)=12x^4/3-6x^1/3\ ,x∈[-1,1] then whic…

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

    \( \displaystyle f'(x)=16x^{1/3}-2x^{-2/3} \)

    \( \displaystyle =\frac{2(8x-1)}{x^{2/3}} \)

So \( f'(x)=0 \) at \( x=\frac{1}{8} \), and \( f' \) does not exist at x = 0. The absolute extreme values are among the values at these points and at the end points:

    f(−1) = 12 + 6 = 18
    f(0) = 0
    \( f\left(\frac{1}{8}\right)=12\times\frac{1}{16}-6\times\frac{1}{2}=-\frac{9}{4} \)
    f(1) = 12 − 6 = 6

So \( x=\frac{1}{8} \) is a critical point, the absolute maximum value is 18 (at x = −1) and the absolute minimum value is \( -\frac{9}{4} \) (at \( x=\frac{1}{8} \)).

Checking the statements

(A) \( f(x) \) has a critical point \( x=\displaystyle\frac{1}{8} \) — Correct. \( f'\left(\frac{1}{8}\right)=0 \).

(B) Absolute maximum value of \( f(x) \) is 18. — Correct. f(−1) = 18 is the largest of the four values.

(C) Absolute maximum value of \( f(x) \) is 6. — Incorrect. 6 is f(1), but f(−1) = 18 is larger.

(D) Absolute minimum value of \( f(x) \) is \( -\displaystyle\frac{9}{4} \). — Correct. \( f\left(\frac{1}{8}\right)=-\frac{9}{4} \) is the smallest of the four values.