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# help algebra

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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

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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