Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $h(x)=\min \{x,x^2\}$, for every real number of $x$, Then

The function $h(x)=\min \{x,x^2\}$ acts as a decision-maker, selecting the smaller value between the linear path $y=x$ and the parabolic path $y=x^2$ at every point $x$ on the real number line.

To identify the transition points, we solve for the intersection of the two curves:

The junction points occur at $x=0$ and $x=1$. These are the critical values where the function switches its definition.

By testing intervals, we can define $h(x)$ explicitly:

For $x<0$, the line $y=x$ is negative, while the parabola $y=x^2$ is positive. Thus, $h(x)=x$.

For $0\le x\le 1$, the parabola $y=x^2$ lies below the line $y=x$. Thus, $h(x)=x^2$.

For $x>1$, the line $y=x$ is smaller than the parabola $y=x^2$. Thus, $h(x)=x$.

In summary, the function is defined as:

Continuity requires that the pieces meet at the transition points. At $x=0$, both $x$ and $x^2$ equal $0$. At $x=1$, both $x^2$ and $x$ equal $1$. Therefore, the function is continuous everywhere.

Differentiability requires the slopes to match at these junctions. We examine the derivative $h^{\prime }(x)$:

For $x<0$, $h^{\prime }(x)=1$. For $0<x<1$, $h^{\prime }(x)=2x$. For $x>1$, $h^{\prime }(x)=1$.

At $x=0$: The left-hand derivative is $1$, while the right-hand derivative is $2(0)=0$. Since $1 eq 0$, the function has a sharp corner and is not differentiable at $x=0$.

At $x=1$: The left-hand derivative is $2(1)=2$, while the right-hand derivative is $1$. Since $2 eq 1$, the function has another sharp corner and is not differentiable at $x=1$.

We have demonstrated that while $h(x)$ is continuous across its entire domain, it fails to be differentiable at the points $x=0$ and $x=1$. This confirms that continuity is a necessary but insufficient condition for differentiability, and that the $\min$ function often introduces non-differentiable "kinks" at its switching points.