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

We have the function $f(x) = \frac{x^2+2x+4}{2x^2+4x+9}$

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Complete the square for both numerator and denominator.

For the numerator: $x^2 + 2x + 4$

Take half of the coefficient of $x$: $2 \div 2 = 1$, then square it: $1^2 = 1$

Add and subtract: $x^2 + 2x + 1 + 3 = (x+1)^2 + 3$

For the denominator: $2x^2 + 4x + 9$

Factor out $2$ from first two terms: $2(x^2 + 2x) + 9$

Complete the square inside: $2(x^2 + 2x + 1 - 1) + 9 = 2(x+1)^2 - 2 + 9 = 2(x+1)^2 + 7$

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

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Let $t = (x+1)^2$. Since this is a square, $t \geq 0$.

Now $f(x) = \frac{t + 3}{2t + 7}$ where $t \geq 0$

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Find the minimum value when $t = 0$:

$f = \frac{0 + 3}{2(0) + 7} = \frac{3}{7}$

As $t$ gets very large, the fraction approaches:

$\frac{t + 3}{2t + 7} \approx \frac{t}{2t} = \frac{1}{2}$

But it never actually equals $\frac{1}{2}$.

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The range is $\left[\frac{3}{7}, \frac{1}{2}\right)$

$[$ means $\frac{3}{7}$ is included, $)$ means $\frac{1}{2}$ is not included.