Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $S$ be the set of all points in $(-\pi ,\pi )$ at which the function, $f(x)=\min \{\sin x,\cos x\}$ is not differentiable. Then $S$ is a subset of which of the following?

Imagine you are an ant walking along a terrain defined by two intersecting paths: the sine wave and the cosine wave. Your rule is simple: at any point $x$, you must choose the path that is lower.

This is the essence of the function $f(x)=\min \{\sin x,\cos x\}$. In this article, we will explore why this simple rule creates a landscape that challenges the very definition of differentiability.

We are working in the interval $(-\pi ,\pi )$. If you sketch $y=\sin x$ and $y=\cos x$ on the same axes, you see a beautiful dance of waves.

The sine wave starts at $0$, peaks at $\frac{\pi }{2}$, and dips to $-1$ at $-\frac{\pi }{2}$. The cosine wave starts at $1$ at $x=0$ and dips to $-1$ at $\pm \pi$.

When we define $f(x)=\min \{\sin x,\cos x\}$, we are essentially tracing the 'lower envelope' of these two graphs. For most of the interval, the function is smooth, following either the sine or cosine curve. But at the points where the two curves cross, the ant must jump from one path to the other.

In calculus, differentiability is synonymous with smoothness. If you can zoom in on a point and the graph looks like a straight line, it is differentiable.

However, at the intersection points, the graph of $f(x)$ forms a sharp corner—a cusp. At these points, the slope abruptly changes from the slope of the sine curve to the slope of the cosine curve (or vice versa).

Because the left-hand derivative and the right-hand derivative do not agree, the derivative does not exist. Thus, the points of non-differentiability are precisely the points of intersection.

To find these points, we set the two functions equal:

Dividing by $\cos x$ (assuming $\cos x eq 0$), we get:

We need to solve this within $(-\pi ,\pi )$. We know that $\tan x=1$ at $x=\frac{\pi }{4}$ (in the first quadrant).

Since the tangent function has a period of $\pi$, the next solution is:

These are our two critical points:

We have found that the function $f(x)$ is non-differentiable at exactly two points. When we look at the options provided, we simply need to find the set that contains both $-\frac{3\pi }{4}$ and $\frac{\pi }{4}$.

Option 1, $\left\{-\frac{3\pi }{4},-\frac{\pi }{4},\frac{\pi }{4},\frac{3\pi }{4}\right\}$, contains our set $S$. By visualizing the function as a physical path, we transformed a potentially abstract calculus problem into a clear, geometric reality.