Question

The total number of points of local maxima and local minima of the function $f\left(x\right)=\{\begin{array}{c}(2+x{)}^3,-3<x\le -1\\ x^{\frac{2}{3}},-1<x<2\end{array}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$



$f\left(x\right)=\{\begin{array}{c}(2+x{)}^3,-3<x\le -1\\ x^{\frac{2}{3}},-1<x<2\end{array}{f}^{\prime }\left(x\right)=\{\begin{array}{c}3(2+x{)}^2,-3<x\le -1\\ \frac{2}{3}{x}^{\frac{-1}{3}},-1<x<2\end{array}$

From $f^{\prime }\left(x\right)=0$ we get $x=-2$
and at $x=0,f^{\prime }\left(x\right)$ becomes undefined,
and $f\left(x\right)$ is not differentiable at $x=-1$
So, here we get three critical points i.e. $x=-2,-1,0$
But there is no sign change for $f^{\prime }\left(x\right)$ at $x=-2$.
So, at $x=-2,f\left(x\right)$ has neither minimum nor maximum

Since, $f^{\prime }\left(x\right)$ changes sign from negative to positive as $x$ crosses $0$ from left to right, therefore $x=0$ is a point of local minima.

From the above graph clearly, $x=-1$ is a point of local maxima because $f(x\to -1^{-})<f(-1)>f(x\to -1^{+})$

The total number of local maximum or minimum $=2$