Provide solution for R.D.Sharma class 12 chapter 6 Maxima and Minima Excercise 17.3 Question 1 Sub Question 7

Point of local minima value is $-\frac{7}{4}$and it’s local minimum value is $-\frac{3}{(4)^{\frac{4}{3}}}$

Hint:

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If $f^{\prime \prime}\left(c_{1}\right)>0$ then $c_1$ is point of local minima.

If $f^{\prime \prime}\left(c_{2}\right)>0$ then $c_2$ is point of local maxima .

where $c_1$ &$c_2$ are critical points.

Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.

Given:

$f(x)=(x+1)(x+2)^{\frac{1}{3}}, x>=-2$

Explanation:

we have ,

\begin{aligned} &f(x)=(x+1)(x+2)^{-} \\ &f^{\prime}(x)=(x+2)^{\frac{1}{3}}+\frac{1}{3}(x+1)(x+2)^{\left(-\frac{2}{3}\right)} \ldots \text{ using } \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime \prime}(x)=\frac{2}{3(x+2)^{2 / 3}}-\frac{2}{9}(x+1)(x+2)^{\left(-\frac{5}{3}\right)} \ldots \text{ using }\frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \end{aligned}

For maxima & minima, f’(x)=0

\begin{aligned} &(x+2)^{\frac{1}{3}}+\frac{1}{3}(x+1)(x+2)^{-\frac{2}{3}}=0 \\ &\frac{1}{3}(x+1)=-(x+2)^{\frac{1}{3}}(x+2)^{\frac{2}{3}} \\ &\frac{1}{3}(x+1)=-(x+2) \\ &x+1=-3 x-6 \\ &x=-\frac{7}{4} \end{aligned}

At x=$-\frac{7}{4}$

\begin{aligned} f^{\prime \prime}\left(-\frac{7}{4}\right) &=\frac{2}{3}\left(-\frac{7}{4}+2\right)^{\frac{-2}{3}}-\frac{2}{9}\left(-\frac{7}{4}+1\right)\left(-\frac{7}{4}+2\right)^{\left(-\frac{5}{3}\right)} \\ &=\frac{2}{3}\left(\frac{1}{4}\right)^{-\frac{2}{3}}+\frac{1}{18}\left(\frac{1}{4}\right)^{-\frac{5}{3}}>0 \end{aligned}

So , $x=-\frac{7}{4}$  is a point of local minima

Now at  $x=-\frac{7}{4}$

\begin{aligned} f\left(-\frac{7}{4}\right) &=\left(-\frac{7}{4}+1\right)\left(-\frac{7}{4}+2\right)^{\frac{1}{3}} \\ &=\left(-\frac{3}{4}\right)\left(\frac{1}{4}\right)^{\frac{1}{3}} \\ &=-\frac{3}{4^{\frac{4}{3}}} \end{aligned}

Thus,  point of local minima is $-\frac{7}{4}$& its value is $-\frac{3}{4^{\frac{4}{3}}}$.