Question

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

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

Hint:

For range questions involving quadratic expressions, complete the square and substitute \(t=(x+a)^2\), where \(t\geq 0\). This makes the expression easier to analyze.

Answer 1

The Correct Option is D

Step 1: Rewrite the quadratic expressions.
Given function is
\[ f(x)=\sqrt{\dfrac{x^2+2x+8}{x^2+2x+4}} \]
Now, write the quadratic terms by completing the square:
\[ x^2+2x+8=(x+1)^2+7 \]
and
\[ x^2+2x+4=(x+1)^2+3 \]

Step 2: Substitute a non-negative variable.
Let
\[ t=(x+1)^2 \]
Since square of a real number is always non-negative,
\[ t\geq 0 \]
Therefore,
\[ f(x)=\sqrt{\dfrac{t+7}{t+3}} \]

Step 3: Find the range of the expression inside the square root.
Let
\[ y=\dfrac{t+7}{t+3} \]
Then,
\[ y=\dfrac{t+3+4}{t+3} \]
\[ y=1+\dfrac{4}{t+3} \]
Since \(t\geq 0\), we have
\[ t+3\geq 3 \]
So,
\[ 0\lt \dfrac{4}{t+3}\leq \dfrac{4}{3} \]
Hence,
\[ 1\lt y\leq 1+\dfrac{4}{3} \]
\[ 1\lt y\leq \dfrac{7}{3} \]

Step 4: Apply square root to get the range of \(f(x)\).
Since
\[ f(x)=\sqrt{y} \]
and
\[ 1\lt y\leq \dfrac{7}{3} \]
we get
\[ 1\lt f(x)\leq \sqrt{\dfrac{7}{3}} \]

Step 5: Final conclusion.
Therefore, the range of the given function is
\[ \left(1,\sqrt{\dfrac{7}{3}}\right] \]
Hence,
\[ \boxed{\left(1,\sqrt{\dfrac{7}{3}}\right]} \]