Animated Solution for Mathematics - Differentiation: The function $f(x)=2x+3(x{)}^{\frac{2}{3}},x\in \mathbb{R}$, has

Imagine you are standing on a landscape defined by the function $f(x)=2x+3x^{\frac{2}{3}}$. At first glance, it looks like a simple combination of a linear term and a power function.

But as we dive into the calculus, we uncover a fascinating, sharp-edged reality. Our goal is to find the local maxima and minima, the peaks and valleys of this mathematical terrain.

The Derivative

To understand the slope of our landscape, we must differentiate. Using the power rule, we find the derivative:

This simplifies beautifully to:

This expression is the key to everything. It tells us exactly how the function is changing at any point $x$.

Many students fall into the trap of only looking for where $f^{\prime }(x)=0$. But in JEE Advanced, we must be more vigilant.

Critical points occur where the derivative is zero OR where it is undefined. Setting the numerator to zero, we get $x^{\frac{1}{3}}+1=0$, which leads us to $x=-1$. This is our first critical point.

Now, look at the denominator. If $x=0$, the derivative $f^{\prime }(x)$ becomes undefined. This is our second critical point.

This point is special; it is where the graph has a sharp cusp, a point where the slope suddenly jumps.

Now, let us test the behavior of the function in the intervals defined by our critical points: $x=-1$ and $x=0$.

1. For $x<-1$: Let us pick $x=-8$. The derivative $f^{\prime }(-8)=2(1+\frac{1}{-2})=2(0.5)=1$. Since $1>0$, the function is increasing.

2. For $-1<x<0$: Let us pick $x=-\frac{1}{8}$. The derivative $f^{\prime }(-\frac{1}{8})=2(1+\frac{1}{-0.5})=2(1-2)=-2$. Since $-2<0$, the function is decreasing.

3. For $x>0$: Let us pick $x=1$. The derivative $f^{\prime }(1)=2(1+\frac{1}{1})=4$. Since $4>0$, the function is increasing again.

At $x=-1$, the slope changes from positive to negative. The function climbs up and then starts to fall—this is a local maximum.

At $x=0$, the slope changes from negative to positive. The function falls into a valley and then climbs back up—this is a local minimum.

We have found exactly one point of local maxima and one point of local minima. The elegance of this result lies in how the derivative's behavior at the cusp ($x=0$) and the zero-slope point ($x=-1$) perfectly define the shape of the curve. You have successfully navigated the terrain!