# Range of this function

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

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. :)

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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?

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

HallsofIvy
Science Advisor
Homework Helper
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
Science Advisor
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].

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.