To find the range of the function \( f(x) = \sqrt{C(2x^2 + 3, x^2 + 4x)} \), we need to analyze the binomial coefficient \( C(n, r) \) which is defined as:
\[
C(n, r) = \frac{n!}{r!(n-r)!}
\]
where \( n \) must be non-negative and \( r \) must be between \( 0 \) and \( n \) (inclusive).
### Step 1: Identify the parameters of the binomial coefficient
In our case, we have:
- \( n = 2x^2 + 3 \)
- \( r = x^2 + 4x \)
### Step 2: Determine the constraints for \( C(n, r) \)
For \( C(2x^2 + 3, x^2 + 4x) \) to be valid:
1. \( 2x^2 + 3 \geq 0 \) (always true for all real \( x \))
2. \( 0 \leq x^2 + 4x \leq 2x^2 + 3 \)
### Step 3: Solve the inequalities
First, let's solve the inequality \( 0 \leq x^2 + 4x \):
\[
x^2 + 4x \geq 0
\]
Factoring gives:
\[
x(x + 4) \geq 0
\]
The critical points are \( x = 0 \) and \( x = -4 \). Testing intervals, we find:
- For \( x < -4 \): both factors are negative, product is positive.
- For \( -4 < x < 0 \): one factor is negative, product is negative.
- For \( x > 0 \): both factors are positive, product is positive.
Thus, \( x^2 + 4x \geq 0 \) for \( x \leq -4 \) or \( x \geq 0 \).
Next, we solve the inequality \( x^2 + 4x \leq 2x^2 + 3 \):
\[
0 \leq 2x^2 - x^2 + 3 - 4x
\]
This simplifies to:
\[
0 \leq x^2 - 4x + 3
\]
Factoring gives:
\[
0 \leq (x - 1)(x - 3)
\]
The critical points are \( x = 1 \) and \( x = 3 \). Testing intervals, we find:
- For \( x < 1 \): both factors are negative, product is positive.
- For \( 1 < x < 3 \): one factor is negative, product is negative.
- For \( x > 3 \): both factors are positive, product is positive.
Thus, \( (x - 1)(x - 3) \geq 0 \) for \( x \leq 1 \) or \( x \geq 3 \).
### Step 4: Combine the intervals
Combining the intervals from both inequalities:
- From \( x^2 + 4x \geq 0 \): \( x \leq -4 \) or \( x \geq 0 \)
- From \( (x - 1)(x - 3) \geq 0 \): \( x \leq 1 \) or \( x \geq 3 \)
The common intervals are:
1. \( x \leq -4 \)
2. \( 0 \leq x \leq 1 \)
3. \( x \geq 3 \)
### Step 5: Evaluate \( f(x) \) at critical points
Now, we evaluate \( f(x) \) at the critical points \( x = 1 \) and \( x = 3 \):
- For \( x = 1 \):
\[
f(1) = \sqrt{C(5, 5)} = \sqrt{1} = 1
\]
- For \( x = 3 \):
\[
f(3) = \sqrt{C(21, 9)} = \sqrt{1} = 1
\]
### Step 6: Determine the range
Since \( f(x) \) can only take values from \( 0 \) to \( 1 \) based on the properties of the binomial coefficient and the evaluations, the range of the function is:
\[
\text{Range of } f(x) = \{ -1, 1 \}
\]
