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Find the range of the function (x^2 +14x+9)/(x^2 + 1)
as x varies over all real numbers.

 Apr 17, 2022
 #1
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Here's a solution without using calculus.

 

Suppose that \(k \in \operatorname{range}\left(\dfrac{x^2 + 14x + 9}{x^2 + 1}\right)\).

 

Then the equation \(\dfrac{x^2 + 14x + 9}{x^2 + 1} = k\) has solutions.

We rewrite the equation as follows to transform it into a quadratic equation.

\(\begin{array}{rcl} \dfrac{x^2 + 14x + 9}{x^2 + 1} &=& k\\ x^2 + 14x +9 &=& k(x^2 + 1)\\ (1 -k )x^2 +14x+(9-k)&=&0 \end{array}\)

 

Consider the discriminant of this quadratic equation. Since it has solutions, \(\Delta \geq 0\).

Then,

\(14^2 - 4(1 - k)(9- k) \geq 0\\ 14^2- 36 +40k-4k^2 \geq 0\\ 4k^2 -40k-160\leq 0\\ k^2 - 10k - 40 \leq 0\\ \dfrac{10 - \sqrt{10^2 - 4(1)(-40)}}2\leq k\leq \dfrac{10 + \sqrt{10^2 - 4(1)(-40)}}2\\ 5-\sqrt{65} \leq k \leq 5 + \sqrt{65}\)

 

Therefore, we conclude that if k is a number in the range of this function, then it must be between \(5\pm \sqrt{65}\).

Conversely, it can be obviously seen (using the derivation above) that if k is a number between \(5\pm \sqrt{65}\), then k must be in the range.

 

Therefore, we have \(\operatorname{range}\left(\dfrac{x^2 + 14x + 9}{x^2 + 1}\right)=[5-\sqrt{65}, 5 + \sqrt{65}]\)

 Apr 17, 2022

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