The difference of absolute maximum (largest) and absolute minimum (smallest) values of the function \( f(x)=12x^{\frac{4}{3}}-6x^{\frac{1}{3}},\ x\in[-1,1] \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{81}{4} \)
On a closed interval, the absolute extreme values occur at critical points or at the end points.
\( \displaystyle f'(x)=16x^{\frac{1}{3}}-2x^{-\frac{2}{3}} \)
\( \displaystyle =\frac{2(8x-1)}{x^{\frac{2}{3}}} \)
f′(x) = 0 at \( x=\frac{1}{8} \), and f′(x) does not exist at x = 0. So the values at x = −1, 0, 1/8 and 1 are compared:
\( \displaystyle f(-1)=12(1)-6(-1)=18 \)
\( \displaystyle f(0)=0 \)
\( \displaystyle f\left(\frac{1}{8}\right)=12\cdot\frac{1}{16}-6\cdot\frac{1}{2} \)
\( \displaystyle =\frac{3}{4}-3 \)
\( \displaystyle =-\frac{9}{4} \)
\( \displaystyle f(1)=12-6=6 \)
Absolute maximum = 18 (at x = −1) and absolute minimum = −9/4 (at x = 1/8).
\( \displaystyle \text{Difference}=18-\left(-\frac{9}{4}\right)=\frac{81}{4} \)