`F(x) = 4 tan x-tan^(2)x+tan^(3)x,xnenpi+(pi)/(2)`

Options

A. is monotonically increasing
B. is monotonically decreasing
C. has a point of maxima
D. has a point of minima

Explanation

To solve the problem, we need to analyze the function \( F(x) = 4 \tan x - \tan^2 x + \tan^3 x \) and determine its monotonicity and whether it has any maxima or minima at the point \( x = n\pi + \frac{\pi}{2} \).

### Step 1: Differentiate the Function
First, we differentiate the function \( F(x) \) with respect to \( x \).

\[ F'(x) = \frac{d}{dx}(4 \tan x - \tan^2 x + \tan^3 x) \]

Using the derivative of \( \tan x \), which is \( \sec^2 x \), we get:

\[ F'(x) = 4 \sec^2 x - 2 \tan x \sec^2 x + 3 \tan^2 x \sec^2 x \]

### Step 2: Factor the Derivative
Next, we can factor out \( \sec^2 x \):

\[ F'(x) = \sec^2 x (4 - 2 \tan x + 3 \tan^2 x) \]

### Step 3: Analyze the Quadratic Expression
Now we need to analyze the quadratic expression \( 4 - 2 \tan x + 3 \tan^2 x \).

Let \( t = \tan x \). The expression becomes:

\[ 3t^2 - 2t + 4 \]

### Step 4: Determine the Nature of the Quadratic
To determine if this quadratic has real roots, we calculate the discriminant:

\[ D = b^2 - 4ac = (-2)^2 - 4 \cdot 3 \cdot 4 = 4 - 48 = -44 \]

Since the discriminant is negative, the quadratic \( 3t^2 - 2t + 4 \) has no real roots and is always positive.

### Step 5: Conclusion about Monotonicity
Since \( \sec^2 x \) is always positive for \( x \neq n\pi + \frac{\pi}{2} \) (where \( \tan x \) is undefined), and the quadratic \( 4 - 2 \tan x + 3 \tan^2 x \) is always positive, we conclude that:

\[ F'(x) > 0 \quad \text{for all } x \text{ in the domain} \]

Thus, the function \( F(x) \) is increasing for all \( x \) in its domain.

### Step 6: Check for Maxima or Minima
At the points \( x = n\pi + \frac{\pi}{2} \), \( F'(x) \) is undefined, indicating vertical asymptotes. Therefore, there are no local maxima or minima at these points.

### Final Conclusion
The function \( F(x) = 4 \tan x - \tan^2 x + \tan^3 x \) is increasing for all \( x \) in its domain, and there are no points of maxima or minima at \( x = n\pi + \frac{\pi}{2} \).

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