The minimum value of the function `f(x)=(tanx)/(3+2tanx), AA x in [0, (pi)/(2))` is

Options

A. 0
B. `(1)/(2)`
C. `(1)/(3)`
D. `(1)/(6)`

Answer

To find the minimum value of the function \( f(x) = \frac{\tan x}{3 + 2\tan x} \) for \( x \in [0, \frac{\pi}{2}) \), we can follow these steps:

### Step 1: Substitute \( t = \tan x \)

Since \( \tan x \) varies from \( 0 \) to \( +\infty \) as \( x \) goes from \( 0 \) to \( \frac{\pi}{2} \), we can rewrite the function in terms of \( t \):

\[ f(t) = \frac{t}{3 + 2t} \]

where \( t \in [0, +\infty) \).

**Hint:** Substituting \( t = \tan x \) simplifies the analysis of the function.

### Step 2: Find the derivative of \( f(t) \)

To find the minimum value, we need to differentiate \( f(t) \):

\[ f'(t) = \frac{(3 + 2t)(1) - t(2)}{(3 + 2t)^2} \]

This simplifies to:

\[ f'(t) = \frac{3 + 2t - 2t}{(3 + 2t)^2} = \frac{3}{(3 + 2t)^2} \]

**Hint:** Use the quotient rule for differentiation to find \( f'(t) \).

### Step 3: Analyze the derivative

The derivative \( f'(t) = \frac{3}{(3 + 2t)^2} \) is always positive for \( t \geq 0 \). This indicates that \( f(t) \) is an increasing function on the interval \( [0, +\infty) \).

**Hint:** A positive derivative means the function is increasing.

### Step 4: Evaluate the function at the endpoints

Since \( f(t) \) is increasing, the minimum value will occur at the left endpoint \( t = 0 \):

\[ f(0) = \frac{0}{3 + 2 \cdot 0} = 0 \]

**Hint:** Check the value of the function at the endpoints of the interval to find the minimum.

### Step 5: Conclusion

Thus, the minimum value of the function \( f(x) \) for \( x \in [0, \frac{\pi}{2}) \) is:

\[ \boxed{0} \]