Animated Solution for Mathematics - Differentiation: Let $f:R\to R$ and $g:R\to R$ be respectively given by $f(x)=|x|+1$ and $g(x)=x^2+1$. Define $h:R\to R$ by $h(x)=\{\begin{array}{ll}\max \{f(x),g(x)\}& \text{if }x\le 0,\\ \min \{f(x),g(x)\}& \text{if }x>0.\end{array}$ The number of points at which $h(x)$ is not differentiable is .........

Welcome, future engineer! Today, we are going to dissect a problem that perfectly captures the elegance of calculus. We are dealing with a piecewise function $h(x)$ constructed from two familiar friends: the absolute value function $f(x)=|x|+1$ and the quadratic function $g(x)=x^2+1$.

Before we dive into the algebra, I want you to close your eyes and visualize these graphs. $f(x)$ is a sharp V-shape, and $g(x)$ is a smooth, upward-opening parabola. Both of them share the exact same vertex at $(0,1)$.

They are essentially dancing around each other, crossing paths at specific points. Our goal is to trace the 'boundary' of these two shapes based on the rules of 'max' and 'min' and then hunt down the points where this new path becomes too sharp to be differentiable.

Where Paths Cross

To understand the behavior of $h(x)$, we first need to know exactly where these two functions meet. We set $f(x)=g(x)$, which gives us the equation:

Subtracting $1$ from both sides, we are left with the beautiful simplicity of $|x|=x^2$. Now, let's be careful here. We know that $x^2$ is the same as $|x{|}^2$.

So, the equation becomes $|x|=|x{|}^2$, or $|x|(|x|-1)=0$. This gives us two cases: $|x|=0$ (which means $x=0$) or $|x|=1$ (which means $x=1$ or $x=-1$).

So, our intersection points are $x=-1,0,1$. These are the critical junctions where our function $h(x)$ will switch its behavior.

Now, let's build $h(x)$ piece by piece. For $x\le 0$, we are told $h(x)=\max \{f(x),g(x)\}$. This means we are tracing the upper boundary of the two graphs.

If you look at the region $x\in (-\infty ,-1]$, the parabola $g(x)=x^2+1$ is actually higher than the line $f(x)=-x+1$. So, $h(x)=x^2+1$. But as we move into the interval $(-1,0]$, the line $f(x)=-x+1$ rises above the parabola, so $h(x)=-x+1$.

For $x>0$, the rule changes to $h(x)=\min \{f(x),g(x)\}$. Now we are tracing the lower boundary.

In the interval $(0,1]$, the parabola $g(x)=x^2+1$ dips below the line $f(x)=x+1$, so $h(x)=x^2+1$. Finally, for $x>1$, the line $f(x)=x+1$ stays below the rapidly climbing parabola, so $h(x)=x+1$. We have successfully constructed our piecewise function $h(x)$ across four distinct intervals.

Now, the moment of truth: differentiability. A function is differentiable only if it is smooth. If there is a 'kink' or a 'sharp corner,' the derivative does not exist because the slope from the left doesn't match the slope from the right.

We check our transition points: $x=-1,0,1$.

At $x=-1$, the Left Hand Derivative (LHD) is the derivative of $x^2+1$, which is $2x$. Plugging in $-1$, we get $-2$. The Right Hand Derivative (RHD) is the derivative of $-x+1$, which is $-1$. Since $-2 eq -1$, we have a sharp corner!

At $x=0$, the LHD is the derivative of $-x+1$, which is $-1$. The RHD is the derivative of $x^2+1$, which is $2x$. Plugging in $0$, we get $0$. Since $-1 eq 0$, we have another sharp corner!

At $x=1$, the LHD is the derivative of $x^2+1$, which is $2x$. Plugging in $1$, we get $2$. The RHD is the derivative of $x+1$, which is $1$. Since $2 eq 1$, we have our third sharp corner!

By carefully analyzing the slopes at these transition points, we have found that $h(x)$ fails to be differentiable at $x=-1,0,1$. That gives us a total of 3 points of non-differentiability.

It is a beautiful result, isn't it? Calculus allows us to take a complex, piecewise definition and break it down into simple, manageable pieces to reveal the underlying geometry. Keep practicing this visualization, and you will find that no function is too intimidating to solve!