For the function `f(x) = cos^(-1)x + cos ^(-1) x^(2)` which of the following statements is/are true

Options

A. f(x) is always decreasing
B. f(x) is decreasing in [0,1] and increasing in [-1, 0]
C. f(x) has only one local maxima
D. f(x) has only one local minima

Answer

To analyze the function \( f(x) = \cos^{-1}(x) + \cos^{-1}(x^2) \) and determine the intervals where it is increasing or decreasing, we will follow these steps:

### Step 1: Find the derivative of the function

To determine where the function is increasing or decreasing, we first need to find its derivative \( f'(x) \).

The derivative of \( f(x) \) is given by:

\[ f'(x) = \frac{d}{dx}(\cos^{-1}(x)) + \frac{d}{dx}(\cos^{-1}(x^2)) \]

Using the derivative of \( \cos^{-1}(x) \):

\[ \frac{d}{dx}(\cos^{-1}(x)) = -\frac{1}{\sqrt{1 - x^2}} \]

For \( \cos^{-1}(x^2) \), we apply the chain rule:

\[ \frac{d}{dx}(\cos^{-1}(x^2)) = -\frac{2x}{\sqrt{1 - (x^2)^2}} = -\frac{2x}{\sqrt{1 - x^4}} \]

Thus, the derivative can be combined as follows:

\[ f'(x) = -\frac{1}{\sqrt{1 - x^2}} - \frac{2x}{\sqrt{1 - x^4}} \]

### Step 2: Set the derivative to zero

To find the critical points, we set \( f'(x) = 0 \):

\[ -\frac{1}{\sqrt{1 - x^2}} - \frac{2x}{\sqrt{1 - x^4}} = 0 \]

This simplifies to:

\[ \frac{1}{\sqrt{1 - x^2}} = -\frac{2x}{\sqrt{1 - x^4}} \]

### Step 3: Analyze the critical points

Next, we need to find the values of \( x \) where this equation holds true. Squaring both sides gives:

\[ \frac{1}{1 - x^2} = \frac{4x^2}{1 - x^4} \]

Cross-multiplying leads to:

\[ 1 - x^4 = 4x^2(1 - x^2) \]

Rearranging gives:

\[ 1 - x^4 = 4x^2 - 4x^4 \]

Combining like terms results in:

\[ 3x^4 - 4x^2 + 1 = 0 \]

Let \( y = x^2 \). The equation becomes:

\[ 3y^2 - 4y + 1 = 0 \]

### Step 4: Solve the quadratic equation

Using the quadratic formula:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6} \]

This gives:

\[ y = 1 \quad \text{or} \quad y = \frac{1}{3} \]

Converting back to \( x \):

\[ x^2 = 1 \Rightarrow x = 1 \quad \text{or} \quad x = -1 \]
\[ x^2 = \frac{1}{3} \Rightarrow x = \frac{1}{\sqrt{3}} \quad \text{or} \quad x = -\frac{1}{\sqrt{3}} \]

### Step 5: Determine intervals of increase and decrease

We analyze the sign of \( f'(x) \) in the intervals determined by the critical points \( -1, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, 1 \).

1. **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \)
- \( f'(-2) < 0 \) (decreasing)

2. **Interval \( (-1, -\frac{1}{\sqrt{3}}) \)**: Choose \( x = -0.5 \)
- \( f'(-0.5) > 0 \) (increasing)

3. **Interval \( (-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) \)**: Choose \( x = 0 \)
- \( f'(0) < 0 \) (decreasing)

4. **Interval \( (\frac{1}{\sqrt{3}}, 1) \)**: Choose \( x = 0.9 \)
- \( f'(0.9) < 0 \) (decreasing)

5. **Interval \( (1, \infty) \)**: Choose \( x = 2 \)
- \( f'(2) < 0 \) (decreasing)

### Conclusion

The function \( f(x) \) is:
- Increasing on \( (-1, -\frac{1}{\sqrt{3}}) \)
- Decreasing on \( (-\infty, -1) \), \( (-\frac{1}{\sqrt{3}}, 1) \), and \( (1, \infty) \)

### Summary of Statements

1. The function is always decreasing: **False**
2. The function is decreasing in \( (0, 1) \) and increasing in \( (-1, 0) \): **False**
3. The function has only one local maxima and one local minima: **True**
4. The function has only one local minima at \( x = -\frac{1}{\sqrt{3}} \): **True**