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

Answer: (C)

$[\frac{ 3}{7} , \frac{1}{2} )$

Answer 2

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

We can express the numerator by completing the square as:
$f(x) =\frac{x^2+2x+3+1}{2x^2+4x+9}$

$f(x)=\frac{(x+1)^2+3}{2x^2+4x+9}$

Next, we aim to obtain $(x + 1)^2$ in the denominator too:
$f(x)=\frac{(x+1)^2+3}{2(x+1)^2+7}$

At this point, we can extract $(x+1)^2 + 3$ from both the numerator and the denominator and represent it as 'k'. Substituting $(x+1)^2 + 3$ with 'k', we have:
$f(x)=\frac{k}{2k+1}$
The smallest possible value for 'k' is 3, given that the squared component is consistently positive, and another positive number is introduced into the equation.
Therefore, the fraction reaches its minimum value when 'k' equals 3.
$f(x)=\frac{3}{2}(3)+1=\frac{3}{7}$

Furthermore, as 'k' approaches its maximum value, we can observe that the fraction will have $2k + 1$ in the denominator, making it consistently one more than twice the numerator.
Consequently, as 'k' increases, the addition of one becomes relatively less significant in the overall denominator. While the fraction approaches $\frac{1}{2}$, it never precisely reaches $\frac{1}{2}$.

Therefore, the range is defined as follows: $[\frac{3}{7},\frac{1}{2})$