Animated Solution for Mathematics - Differentiation: The total number of local maxima and local minima of the function $f(x)=\{\begin{array}{ll}(2+x{)}^3,& -3<x\le -1\\ x^{2/3},& -1<x<2\end{array}$ is

Imagine you are standing at the edge of a mathematical landscape. You have been given a function, $f(x)$, but it is not a simple, single-lane road. It is a piecewise function, a path that changes its nature at a specific junction.

In the world of JEE Advanced, these problems are not just about calculation; they are about navigation. You are the explorer, and your goal is to find the peaks and the valleys—the local maxima and minima.

Before we start differentiating, we must respect the geometry of the problem. We have two distinct paths: $f(x)=(2+x{)}^3$ for the region $x\in (-3,-1]$ and $f(x)=x^{2/3}$ for the region $x\in (-1,2)$.

The first thing any seasoned mathematician does is check the junction. We calculate the Left Hand Limit (LHL) as $x\to -1^{-}$ using the first branch:

Now, we check the Right Hand Limit (RHL) as $x\to -1^{+}$ using the second branch:

Since the limits match, the function is continuous. This is a massive relief, as it confirms we are dealing with a connected curve without jump discontinuities.

Now, let us look at the first branch: $f(x)=(2+x{)}^3$. We differentiate to find the slope:

Here is where the JEE examiners love to test your intuition. We set $f^{\prime }(x)=0$ and find $x=-2$.

Because of the square, $3(2+x{)}^2$ is always greater than or equal to zero. The function is monotonically increasing, and at $x=-2$, the slope is zero, but the function just pauses before continuing its climb. This is an inflection point, not an extremum.

We move to the second branch: $f(x)=x^{2/3}$. We differentiate this to get:

Now, look at $x=0$. The denominator becomes zero, meaning the derivative is undefined. In calculus, an undefined derivative at a continuous point often signifies a cusp—a sharp turn in the graph.

Let us test the slope around $x=0$: For $x<0$, $f^{\prime }(x)<0$ (the function is decreasing). For $x>0$, $f^{\prime }(x)>0$ (the function is increasing).

A change from decreasing to increasing is the definition of a local minimum. We have found our first true extremum at $x=0$.

Finally, we return to our junction at $x=-1$. We know the function is increasing to the left of $-1$ (from the first branch) and decreasing to the right of $-1$ (from the second branch).

A change from increasing to decreasing signifies a local maximum. We have successfully identified two extrema: a local maximum at $x=-1$ and a local minimum at $x=0$.

The total count of extrema is 2. By breaking the problem into logical phases and respecting the behavior of the derivative, we have conquered the function.