#### Explain solution RD Sharma class 12 chapter Maxima and Minima exercise 17.4 question 4 maths

Absolute Maximum $=18 \text { at } x=-1$

Absolute minimum $=-\frac{9}{4} \mathrm{at} \; \; \mathrm{x}=\frac{1}{8}$

Hint:

check the value of f(x) at end points & where f ‘ (x) = 0

Given:

$f(x)=12 x^{4 / 3}-6 x^{1 / 3}, x \in(-1,1)$

Explanation:

$f^{\prime}(x)=12(4 / 3) x^{\frac{4}{3}-1}-6(1 / 3) x^{\frac{1}{3}-1}$

\begin{aligned} &=16 x^{1 / 3}-2 x^{-2 / 3} \\ &f^{\prime}(x)=0 \\ &16 x^{1 / 3}-2 x^{-2 / 3}=0 \\ &8 x^{1 / 3}=x^{-2 / 3} \\ &x^{1 / 3} / x^{-2 / 3}=\frac{1}{8} \\ &x=\frac{1}{8} \end{aligned}

Check the value of $f(x) \text { at } x=-1, \frac{1}{8}, 1$

\begin{aligned} &f(-1)=12(-1)^{\frac{4}{3}}-6(-1)^{\frac{1}{3}} \\\\ &=12+6=18 \\\\ &f\left(\frac{1}{8}\right)=12\left(\frac{1}{8}\right)^{\frac{4}{3}}-6\left(\frac{1}{8}\right)^{\frac{1}{3}} \end{aligned}

\begin{aligned} &=12\left(\left(\frac{1}{2}\right)^{3}\right)^{\frac{4}{3}}-6\left(\left(\frac{1}{2}\right)^{3}\right)^{\frac{1}{3}} \\\\ &=\frac{12}{16}-3=\frac{3}{4}-3=\frac{-9}{4} \\\\ &f(1)=12(1)^{\frac{4}{3}}-6(1)^{\frac{1}{3}} \\\\ &=12-6=6 \end{aligned}

Hence,

Absolute Maximum $=18 \text { at } x=-1$

Absolute minimum $=-\frac{9}{4} \mathrm{at} \; \; \mathrm{x}=\frac{1}{8}$