Question

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

• A. 2
• B. 0
• C. 3
• D. 1

Answer 1

The correct option is A 2

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

$\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{ccc}3(x+2{)}^2& .& -3<x\le 1\\ \frac{2}{3}{x}^{-\frac{1}{3}}& ;& -1<x<2\end{array}$


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

lso $f^{\prime }\left(0\right)$ does not exist but $f^{\prime }(0-)<0$ and $f^{\prime }(0+)>0$ , it can only be inferred that $f\left(x\right)$ has a possibility of a minima at $x=0$.

Hence the given function has one local maxima at $x=–1$ and one local minima at $x=0$.