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Maxima and Minima — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMaxima and Minima5 Marks
Question
Find the points of local maxima or local minima and corresponding local maximum and local minimum values of the following functions. Also, find the points of inflection,
$\text{f}(\text{x})=(\text{x}+1)(\text{x}+2)^\frac{1}{3}$
✓

Answer

Given, $\text{f}(\text{x})=(\text{x}+1)(\text{x}+2)^\frac{1}{3}$
$\Rightarrow\text{f}'(\text{x})=(\text{x}+2)^\frac{1}{3}+\frac{1}{3}(\text{x}+1)(\text{x}+2)^\frac{-2}{3}$
For the local maxima or minima, We must have f"(x) = 0
$\Rightarrow(\text{x}+2)^\frac{1}{3}+\frac{1}{3}(\text{x}+1)(\text{x}+2)^\frac{-2}{3}=0$
$\Rightarrow\frac{1}{3}(\text{x}+1)=-(\text{x}+2)^\frac{1}{3}\times(\text{x}+2)^\frac{2}{3}$
$\Rightarrow\frac{1}{3}(\text{x}+1)=-(\text{x}+2)$
$\Rightarrow\text{x}+1=-3\text{x}-6$
$\Rightarrow\text{x}=\frac{-7}{4}$
Thus, $\text{x}=\frac{-7}{2}$ and the possible points of local maxima or a local minima.
Now, $\text{f}"\Big(\frac{-7}{4}\Big)=\frac{2}{3}(\text{x}+2)^\frac{-2}{3}-\frac{2}{9}(\text{x}+1)(\text{x}+2)^\frac{-5}{3}$
At $\text{x}=\frac{-7}{4}$
$\text{f}''\Big(\frac{-7}{4}\Big)=\frac{2}{3}\Big(\frac{-7}{4}+2\Big)^\frac{-2}{3}-\frac{2}{9}\Big(\frac{-7}{2}+1\Big)\Big(\frac{-7}{2}+2\Big)^\frac{-5}{3}$
$=\frac{2}{3}\Big(\frac{1}{4}\Big)^\frac{-2}{3}+\frac{1}{18}\Big(\frac{1}{4}\Big)^\frac{-5}{2}>0$
So, $\text{x}=\frac{-7}{2}$ is point of local minimum.
The local minimum value is given by 
$\text{f}''\Big(\frac{-7}{4}\Big)=\Big(\frac{-7}{4}+1\Big)\Big(\frac{-7}{4}+2\Big)^\frac{1}{3}=\frac{-3}{4}\Big(\frac{1}{4}\Big)^\frac{1}{3}=\frac{-3}{4^{\frac{4}{3}}}$

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