Range of a rational function with radicals

Question

Find the range of the function

$$\frac{6}{5\sqrt{x^2-10x+29} - 2}$$

I tried using inverses, but the equation got super messy and I dont think its a good method for this problem.

$\frac{6}{5\sqrt{x^2-10x+29} - 2} = y$

getting the inverse,

$\frac{6}{5\sqrt{y^2-10y+29} - 2} = x$

$\frac{4x^2+24x+36}{25x^2}= y^2-10y +29$

Then it would be a quadratic function in y, but the discriminant becomes really big

$100- 116(\frac{4x^2+24x+36}{25x^2})$

What do you mean by "I tried using inverses"? Could you show your work? — A. Pongrácz, Aug 19 '18 at 09:43
A graph will show you some things that will help. It isn't hard to see that it can't be negative and a little calculus will help you find the single maximum. — Mandelbrot, Aug 19 '18 at 09:49
You can also write $ y^2-10y +29=\left ( {1 \over 5}\left ( {6\over x}+2\right ) \right )^2$ — N74, Aug 19 '18 at 17:05

score 7 · Answer 1 · answered Aug 19 '18 at 09:53

7

Hint

If you rewrite $x^2-10x+29=(x-5)^2+4$, it's easier to see that: $$(x-5)^2+4 \in [4,+\infty)$$ $$\sqrt{(x-5)^2+4} \in [2,+\infty)$$ $$5\sqrt{(x-5)^2+4}-2 \in [8,+\infty)$$ Does that help?

answered Aug 19 '18 at 09:53

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score 2 · Accepted Answer · answered Aug 19 '18 at 09:54

$\frac{6}{5\sqrt(x^2-10x+29)-2}=\frac{6}{5\sqrt((x-5)^2+4)-2}$

so you can see that there will be a maximum at $x=5$ and as $x$ tends to either positive or negative infinity the graph goes to zero.

So just plug in $x=5$ to get $\frac{3}{4}$ and you have a range $]0,\frac{3}{4}]$

Range of a rational function with radicals

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