Range of this Function : $\frac{(x^2 + x +1) }{ (x - 4)}$

Question

I got answer with Wolfram Alpha. But I do not know how the answer was obtained?

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what's method should I use?

score 2 · Answer 1 · answered Nov 22 '13 at 13:51

2

Hint: suppose $\;a\in\Bbb R\;$ is sucht that

$$\frac{x^2+x+1}{x-4}=a\iff x^2+(1-a)x+(4a+1)=0$$

Thus, $\;a\;$ is in in te range of the function iff the above quadratic has at least one solution, ie.e iff its discriminat is non-negative:

$$\Delta=(1-a)^2-4(4a+1)\ge0\iff a^2-18a-3\ge0$$

In order to factor the last quadraitc in $\;a\;$ we use the roots formula:

$$a_{1,2}=\frac{18\pm\sqrt{336}}2=9\pm 2\sqrt{21}$$

and we get

$$a^2-18a-3=\left(a-(9-2\sqrt{21}\right)\left(a-(9+2\sqrt{21})\right)\ge0\iff$$

$$\iff\begin{cases}a\le9-2\sqrt{21}\\{}\\a\ge9+2\sqrt 21\end{cases}$$

answered Nov 22 '13 at 13:51

DonAntonio

1

Nice work! (+1 - oops, there it is now.) But a bit more than a hint! ;-) – amWhy Nov 22 '13 at 13:53
Yes, I know: I got carried away, yet at this level the mere justification of the quadratics via their discriminants...hmmm. Thanks. – DonAntonio Nov 22 '13 at 13:57
I've done the same thing! Intention: A hint. Once I get going...sometimes...I get carried away! – amWhy Nov 22 '13 at 13:58

score 2 · Answer 2 · answered Nov 22 '13 at 14:17

2

Note that $$y=\frac{x^2+x+1}{x-4}=\frac{(x-4)^2+9(x-4)+21}{x-4}=(x-4)+\frac{21}{x-4}+9$$

answered Nov 22 '13 at 14:17

Shuchang

2 Answers2