Question

For all real values of x, the range of the function \(f(x) =\frac{ x^2+2x+4}{2x^2+4x+9}\) is

• A. \([\frac{ 3}{7} ,\frac{ 8}{9} )\)
• B. \([ \frac{4}{9} ,\frac{ 8}{9} ]\)
• C. \([\frac{ 3}{7} , \frac{1}{2} )\)
• D. \(( \frac{3}{7} , \frac{1}{2} )\)

Answer 1

The Correct Option is C

Finding the Range of the Function 

We're given the function:

\( f(x) = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} \)

Step 1: Completing the Square

Let's simplify the numerator by completing the square:

\( x^2 + 2x + 4 = (x+1)^2 + 3 \)

Now handle the denominator:

\( 2x^2 + 4x + 9 = 2(x^2 + 2x) + 9 = 2((x+1)^2 + 1) + 7 = 2(x+1)^2 + 7 \)

Step 2: Rewriting with k-substitution

So, the function becomes:

\( f(x) = \frac{(x+1)^2 + 3}{2(x+1)^2 + 7} \)

Let’s define: \( k = (x+1)^2 + 3 \).
Then: \( (x+1)^2 = k - 3 \)

Substitute back into the denominator: 
\( 2(x+1)^2 + 7 = 2(k - 3) + 7 = 2k - 6 + 7 = 2k + 1 \)

So the function becomes: 
\( f(x) = \frac{k}{2k + 1} \)

Step 3: Analyzing the Expression

We're now studying 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.

Find minimum of \( f(x) \): 
\( f_{\text{min}} = \frac{3}{2(3) + 1} = \frac{3}{7} \)

As \( k \to \infty \), the function becomes: 
\( \frac{k}{2k + 1} \to \frac{1}{2} \) 
But it never actually reaches \( \frac{1}{2} \), because of the "+1" in the denominator.

Final Answer

Therefore, the range of the function is: 
\[ \left[ \frac{3}{7}, \frac{1}{2} \right) \]