The total number of local maxima and local minima of the function $f(x)=\left\{\begin{array}{lc}(2+x)^3 ; & -3

Question

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

Aicharya · Accepted Answer

Given that, $f(x)=\left\{\begin{array}{l}(2+x)^3 ;-3 < x \leq-1 \\ x^{2 / 3} ; \quad-1 < x < 2\end{array}\right.$ $ \Rightarrow \quad f^{\prime}(x)=\left\{\begin{array}{l} 3(2+x)^2 ;-3 < x \leq 1 \\ \frac{2}{3} x^{-\frac{1}{3}} ; \quad-1 < x < 2 \end{array}\right. $

Clearly, $f^{\prime}(x)$ changes its sign at $x=-1$ from $+$ ve to -ve and so $f(x)$ has local maxima at $x=-1$. Also, $f^{\prime}(0)$ does not exist but $f^{\prime}\left(0^{-}\right) < 0$ and $f^{\prime}\left(0^{+}\right) < 0$. It can only be inferred that $f(x)$ 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 $x=0$

The total number of local maxima and local minima of the function $f(x)=\left\{\begin{array}{lc}(2+x)^3 ; &…

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