What is the range of the function f(x)=1-x/(√5+7x-x²)?

[yik-boh]

$$f(x) = \frac{1 - x}{\sqrt{5 + 7x - x^{2}}}$$

I know that the range of that function is $$(-\infty, 0) U (0, +\infty)$$

But how do I get it? I'm having difficulties in isolating x. Can you explain it to me? Thank you. :)

 

[HallsofIvy]

You don't get that range- it's wrong. It is easy to see that
$$f(1)= \frac{1- 1}{\sqrt{5+ 7- 1}}= \frac{0}{\sqrt{11}}= 0$$
so 0 certainly is in the range.

Since you have determined what x can be, what values of f do those values of x give?

 

[adriank]

Moreover, for sufficiently large |x|, the denominator is not real.

 

[HallsofIvy]

Right. $5+ 7x- x^2= 0$ when $x= (7\pm\sqrt{49+ 20})/2$ so that the domain of the function is only from $(7- \sqrt{69})/2$ to $7+\sqrt{69})/2$, or from between -1 and 0 to between 7 and 8. The function goes to infinity as x approaches the end points. The lower bound on the range will occur at the minimum of the function.

 

[uart]

Assuming we're talking about real functions then the domain is [itex](7-\sqrt{69})/2 < x < (7+\sqrt{69})/2[/tex] and the range is all real [itex]y[/tex].<br /> <br /> Halls, note that the function goes to +infinity at the lower end of it's domain and to -infinity at the upper end of it's domain and in continuous in between.[/itex][/itex]