\([\frac{ 3}{7} ,\frac{ 8}{9} )\)
\([ \frac{4}{9} ,\frac{ 8}{9} ]\)
\([\frac{ 3}{7} , \frac{1}{2} )\)
\(( \frac{3}{7} , \frac{1}{2} )\)
The given function is:
\( f(x) = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} \)
Complete the square for the numerator:
\( x^2 + 2x + 4 = (x+1)^2 + 3 \)
Complete the square for the denominator:
\( 2x^2 + 4x + 9 = 2(x^2 + 2x) + 9 = 2((x+1)^2 + 1) + 7 = 2(x+1)^2 + 7 \)
The function can be rewritten as:
\( f(x) = \frac{(x+1)^2 + 3}{2(x+1)^2 + 7} \)
Let \( k = (x+1)^2 + 3 \). This implies \( (x+1)^2 = k - 3 \).
Substitute \( (x+1)^2 \) into the denominator expression:
\( 2(x+1)^2 + 7 = 2(k - 3) + 7 = 2k - 6 + 7 = 2k + 1 \)
The function now simplifies to:
\( f(x) = \frac{k}{2k + 1} \)
Analyze the behavior of \( \frac{k}{2k + 1} \).
Minimum Value of k: Since \( (x+1)^2 \geq 0 \), the minimum value of \( k = (x+1)^2 + 3 \) is 3.
Calculate the minimum value of \( f(x) \):
\( f_{\text{min}} = \frac{3}{2(3) + 1} = \frac{3}{7} \)
As \( k \) approaches infinity, the function's behavior is:
\( \frac{k}{2k + 1} \to \frac{1}{2} \)
Note that the function never reaches \( \frac{1}{2} \) due to the '+1' in the denominator.
The range of the function is:
\[ \left[ \frac{3}{7}, \frac{1}{2} \right) \]