`f(x)=cos^(-1)(x^(2)/sqrt(1+x^(2)))`

Answer

To find the range of the function \( f(x) = \cos^{-1}\left(\frac{x^2}{\sqrt{1+x^2}}\right) \), we will follow these steps:

### Step 1: Identify the domain of the function inside the cosine inverse
The function \( \cos^{-1}(a) \) is defined for \( a \) in the interval \([-1, 1]\). Therefore, we need to ensure that:
\[ -1 \leq \frac{x^2}{\sqrt{1+x^2}} \leq 1 \]

### Step 2: Analyze the expression \( \frac{x^2}{\sqrt{1+x^2}} \)
Since \( x^2 \) is always non-negative for all real \( x \), we have:
\[ \frac{x^2}{\sqrt{1+x^2}} \geq 0 \]
This means the left part of the inequality \(-1 \leq \frac{x^2}{\sqrt{1+x^2}}\) is always satisfied.

### Step 3: Determine the upper bound
Next, we need to check the upper bound:
\[ \frac{x^2}{\sqrt{1+x^2}} \leq 1 \]
Squaring both sides (which is valid since both sides are non-negative):
\[ \frac{x^4}{1+x^2} \leq 1 \]
This simplifies to:
\[ x^4 \leq 1 + x^2 \]
Rearranging gives:
\[ x^4 - x^2 - 1 \leq 0 \]
Let \( y = x^2 \). The inequality becomes:
\[ y^2 - y - 1 \leq 0 \]

### Step 4: Solve the quadratic inequality
To solve \( y^2 - y - 1 = 0 \), we use the quadratic formula:
\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{5}}{2} \]
The roots are:
\[ y_1 = \frac{1 + \sqrt{5}}{2}, \quad y_2 = \frac{1 - \sqrt{5}}{2} \]
Since \( y \) (which is \( x^2 \)) must be non-negative, we only consider \( y_1 = \frac{1 + \sqrt{5}}{2} \).

### Step 5: Determine the valid range for \( y \)
The quadratic \( y^2 - y - 1 \) opens upwards (as the coefficient of \( y^2 \) is positive), and it is negative between its roots:
\[ y_2 \leq y \leq y_1 \]
Since \( y_2 = \frac{1 - \sqrt{5}}{2} \) is negative, we only consider \( 0 \leq y \leq \frac{1 + \sqrt{5}}{2} \).

### Step 6: Find the range of \( f(x) \)
Now we can find the range of \( f(x) \):
- When \( y = 0 \) (i.e., \( x = 0 \)), \( f(0) = \cos^{-1}(0) = \frac{\pi}{2} \).
- When \( y = \frac{1 + \sqrt{5}}{2} \), we calculate:
\[ f(x) = \cos^{-1}\left(1\right) = 0 \]

Thus, the range of \( f(x) \) is:
\[ f(x) \in [0, \frac{\pi}{2}] \]

### Final Result
The range of the function \( f(x) = \cos^{-1}\left(\frac{x^2}{\sqrt{1+x^2}}\right) \) is:
\[ [0, \frac{\pi}{2}] \]

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