Application of Derivatives — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSApplication of Derivatives1 Mark
Question
Find absolute maximum and minimum values of a function f given by $f(x)=12 x^{\frac{4}{3}}-6 x^{\frac{1}{3}}, x \in[-1,1]$
✓
Answer
We have $f(x)=12 x^{\frac{4}{3}}-6 x^{\frac{1}{3}}$ or f'($x$) = $16 x^{\frac{1}{3}}-\frac{2}{x^{\frac{2}{3}}}=\frac{2(8 x-1)}{x^{\frac{2}{3}}}$ Thus, f ′($x$) = 0 gives x = $\frac{1}{8}$. Further note that f ′(x) is not defined at x = 0. So the critical points are x = 0 and x = $\frac{1}{8}$. Now evaluating the value of f at critical points x = 0, $\frac{1}{8}$ and at end points of the interval x = -1 and x = 1, we have $f (–1) = 12(-1)^\frac43 -6(-1)^\frac13 = 18$ $f (0) = 12 (0) – 6(0) = 0$ $f\left(\frac{1}{8}\right)=12\left(\frac{1}{8}\right)^{\frac{4}{3}}-6\left(\frac{1}{8}\right)^{\frac{1}{3}}=\frac{-9}{4}$ $f(1)=12(1)^{\frac{4}{3}}-6(1)^{\frac{1}{3}}=6$ Hence, we conclude that absolute maximum value of f is 18 that occurs at x = –1 and absolute minimum value of f is $\frac{-9}{4}$ that occurs at x = $\frac{1}{8}$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.