Application Of Derivatives Ques 71

71. The total number of local maxima and local minima of the function $f(x)=\begin{array}{cc}(2+x)^{3}, & -3<x \leq-1 \\ x^{\frac{2}{3}}, & -1<x<2\end{array}$ is

(2008, 3M)

0

1

2

3

Show Answer

Answer:

Correct Answer: 71.(d)

Solution:

Formula:

Maxima and Minima of functions of one variable :

  1. Given, $\quad f(x)=\begin{array}{cc}(2+x)^{3}, & \text { if }-3<x \leq-1 \\ x^{2 / 3} & \text { if }-1<x<2\end{array}$

$\Rightarrow \quad f^{\prime}(x)=\begin{array}{ll}3(x+2)^{2}, & \text { if }-3<x \leq-1 \\ \frac{2}{3} x^{-\frac{1}{3}} & , \text { if }-1<x<2\end{array}$

Clearly, $f^{\prime}(x)$ changes its sign at $x=-1$ from $+\mathrm{ve}$ to $-\mathrm{ve}$ and so $f(x)$ has local maxima at $x=-1$.

Also, $f^{\prime}(0)$ does not exist but $f^{\prime}(0^{+})<0$. It can only be inferred that $f(x)$ has a possibility of a $\operatorname{minima}$ at $x=0$. Hence, the given function has one local maxima at $x=-1$ and one local minima at $x=0$.



sathee Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on

Please select your preferred language
कृपया अपनी पसंदीदा भाषा चुनें