Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f(x)=\min \{1,x^2,x^3\}$, then

Welcome, young scholars! Today we embark on a journey into the heart of calculus, where we unravel the mystery of the minimum function.

Imagine you are standing on a vast, undulating landscape defined by three distinct paths: a flat, infinite plain at $y=1$, a graceful, upward-opening parabola $y=x^2$, and a daring, twisting cubic curve $y=x^3$.

Your task is to walk along the lowest possible path at every point. This is the essence of the function $f(x)=\min \{1,x^2,x^3\}$. To master this, we must first visualize the terrain.

For $x<0$, the cubic curve $y=x^3$ dives deep into the negative abyss, far below the positive values of $x^2$ and $1$. Thus, the floor is $x^3$.

As we cross into the interval $0\le x<1$, we encounter a fascinating property of fractions: when you raise a fraction between $0$ and $1$ to a higher power, it shrinks! Thus, $x^3<x^2<1$, and our path remains the cubic curve.

Finally, at $x=1$, the landscape shifts. For $x\ge 1$, the constant $1$ becomes the most humble, staying below the rapidly growing $x^2$ and $x^3$.

We have constructed our piecewise function:

Continuity and Differentiability

Now, for the grand finale: continuity and differentiability. At $x=1$, the left-hand limit is:

The right-hand limit is:

Since the limits match, the function is continuous.

But what about the slope? The derivative of $x^3$ is $3x^2$, which at $x=1$ is $3$. The derivative of $1$ is $0$.

Since $3 eq 0$, we have a sharp corner! The function is continuous but not differentiable at $x=1$.

You have just conquered a classic JEE problem with the elegance of a mathematician. Keep exploring, keep questioning, and never lose your wonder for the beauty of calculus!