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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Find the absolute maximum and minimum values of a function f given by
$\text{f}(\text{x})=12\text{x}^\frac{4}{3}-6\text{x}^\frac{1}{3},\text{x}\in[-1,1]$
✓

Answer

$\text{f}(\text{x})=12\text{x}^\frac{4}{3}-6\text{x}^\frac{1}{3}$
$\therefore\ \text{f}'(\text{x})=16\text{x}^\frac{1}{3}-\frac{2}{\text{x}^{\frac{2}{3}}}=\frac{2(8\text{x}-1)}{\text{x}^{\frac{2}{3}}}$
Thus, f'(x) = 0
$\Rightarrow\ \text{x}=\frac{1}{8}$
Further note that f'(x) is not defined at x = 0.
So, the critical points are x = 0 and $\text{x}=\frac{1}{8}.$
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.
$\text{f}(-1)=12(-1)^{\frac{4}{3}}-6(-1)^{-\frac{1}{3}}=18$
$\text{f}(0)=12(0)-6(0)=0$
$\text{f}\Big(\frac{1}{8}\Big)=12(\frac{1}{8})^{\frac{4}{3}}-6(\frac{1}{8})^{\frac{1}{3}}=\frac{-9}{4}$
$\text{f}(1)=12(1)^{\frac{4}{3}}-6(1)^{\frac{1}{3}}=16$
Hence, we can clude thet absolute maximum value of f is 18 at x = -1 and absolute minimum value f is $\frac{-9}{4}\text{ at x}=\frac{1}{8}.$

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