Find the number of critical points of the function
f (x) = min (tan x , cos x) , `x in (0 , pi)`

Answer

To find the number of critical points of the function \( f(x) = \min(\tan x, \cos x) \) for \( x \in (0, \pi) \), we will follow these steps:

### Step 1: Understand the functions involved
We need to analyze the functions \( \tan x \) and \( \cos x \) over the interval \( (0, \pi) \).

### Step 2: Identify the behavior of \( \tan x \) and \( \cos x \)
- The function \( \tan x \) is defined and continuous in the interval \( (0, \pi) \) except at \( x = \frac{\pi}{2} \), where it has a vertical asymptote.
- The function \( \cos x \) is continuous and oscillates between 1 and -1 in the interval \( (0, \pi) \).

### Step 3: Find points where \( \tan x = \cos x \)
To find the critical points, we need to determine where \( \tan x \) and \( \cos x \) intersect. We set:
\[ \tan x = \cos x \]
This can be rewritten as:
\[ \frac{\sin x}{\cos x} = \cos x \implies \sin x = \cos^2 x \]
Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can substitute:
\[ \sin x = 1 - \sin^2 x \]
Let \( y = \sin x \). Then we have:
\[ y = 1 - y^2 \implies y^2 + y - 1 = 0 \]
Using the quadratic formula:
\[ y = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]
We take the positive root since \( y = \sin x \) must be between 0 and 1:
\[ y = \frac{-1 + \sqrt{5}}{2} \]

### Step 4: Find the corresponding \( x \) values
Now we need to find \( x \) such that:
\[ \sin x = \frac{-1 + \sqrt{5}}{2} \]
This will give us one critical point in the interval \( (0, \pi) \).

### Step 5: Analyze the behavior around \( x = \frac{\pi}{2} \)
At \( x = \frac{\pi}{2} \), \( \tan x \) is undefined, and we need to check the behavior of \( f(x) \) around this point:
- As \( x \) approaches \( \frac{\pi}{2} \) from the left, \( \tan x \to +\infty \) and \( \cos x \to 0 \).
- As \( x \) approaches \( \frac{\pi}{2} \) from the right, \( \tan x \to -\infty \) and \( \cos x \to 0 \).

Thus, \( f(x) \) switches from \( \tan x \) to \( \cos x \) at \( x = \frac{\pi}{2} \), which is a point of discontinuity.

### Step 6: Count the critical points
1. The intersection point we found gives us one critical point.
2. The point \( x = \frac{\pi}{2} \) is a point of discontinuity, which we also count as a critical point.

### Conclusion
Thus, the total number of critical points of the function \( f(x) = \min(\tan x, \cos x) \) in the interval \( (0, \pi) \) is **2**.

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