Question

The range of the real valued function \[ f(x) = \sqrt{\frac{x^2 + 2x + 8}{x^2 + 2x + 4}} \] is

• A. $\left[\sqrt{\frac{7}{3}}, \infty \right)$
• B. $(0, \infty)$
• C. $(1, \infty)$
• D. $\left(1, \sqrt{\frac{7}{3}}\right]$

Hint:

To find range of such expressions, try rewriting in a simpler form and analyze the denominator's minimum/maximum.

Answer 1

The Correct Option is D

Rewrite the function as \[ f(x) = \sqrt{1 + \frac{4}{x^2 + 2x + 4}} \] Since the denominator $x^2 + 2x + 4$ is always positive and minimized at $x=-1$ giving value $3$, the maximum value of $\frac{4}{x^2 + 2x + 4}$ is $\frac{4}{3}$.
Thus, \[ 1<f(x)^2 \leq 1 + \frac{4}{3} = \frac{7}{3} \] Taking square roots, \[ 1<f(x) \leq \sqrt{\frac{7}{3}} \] Hence the range is \[ \left(1, \sqrt{\frac{7}{3}}\right] \]