Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f:(-1,1)\to R$ be a function defined by $f(x)=\max \{-|x|,-\sqrt{1-x^2}\}$. If K be the set of all points at which f is not differentiable, then K has exactly :

Imagine you are standing on a coordinate plane, looking at two distinct paths. One is a sharp, inverted V-shape defined by $g(x)=-|x|$.

The other is a smooth, gentle curve, the lower half of a unit circle, defined by $h(x)=-\sqrt{1-x^2}$.

Our function $f(x)=\max \{-|x|,-\sqrt{1-x^2}\}$ is a traveler walking along the higher of these two paths. This "upper envelope" is the heart of our problem.

Where Paths Cross

To understand where our traveler might stumble—that is, where the function might not be differentiable—we must first find where these paths cross. We set the two components equal:

The negative signs cancel out, leaving us with $|x|=\sqrt{1-x^2}$. Squaring both sides is our key to unlocking the algebra:

This simplifies beautifully to $2x^2=1$, or $x^2=\frac{1}{2}$. Thus, our paths intersect at $x=\frac{1}{\sqrt{2}}$ and $x=-\frac{1}{\sqrt{2}}$.

These are the critical junctions where the function switches its identity from a line to a curve.

We now hunt for the points where the function fails to be smooth. A function is not differentiable if it has a sharp corner, a kink, or a discontinuity. We have three candidates for these "trouble spots":

1. The Origin ($x=0$): Look at the component $g(x)=-|x|$. At $x=0$, it forms a sharp vertex. The slope from the left is $1$, and the slope from the right is $-1$. Since $1 eq -1$, the function is not differentiable here.

2. The Intersections ($x=\pm \frac{1}{\sqrt{2}}$): At these points, the function switches from the line to the circle. Even if the graphs meet, they do so at different angles.

The slope of the line is constant, while the slope of the circle is changing. Because the slopes of the two functions do not match at the point of intersection, a sharp kink is created. This happens at both $x=\frac{1}{\sqrt{2}}$ and $x=-\frac{1}{\sqrt{2}}$.

We have identified three distinct points where the smoothness of our function breaks: $x=0$, $x=\frac{1}{\sqrt{2}}$, and $x=-\frac{1}{\sqrt{2}}$.

These are the elements of our set $K$. Counting them, we find exactly three elements.

This problem teaches us that calculus is not just about symbols; it is about visualizing the behavior of functions. When you see a "max" function, think of it as a path that always chooses the higher road, and watch out for the sharp turns where the roads meet!