Get Answers to all your Questions

header-bg qa

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

Answers (1)



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


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.


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


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}}}.


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support