$\frac{-3x+1}{x^2-6x-16}>0$ and $y=-2/x +1$ find the interval of y

The known info $$\frac{-3x+1}{x^2-6x-16}>0$$ so, i find that : x not 8 nor -2. And $$x \geq 1/3$$

$$y=-2/x +1$$ For y i find that $$y > -5$$ and y not 3/4 or 2.

Based on that. So interval for $$y = -5 < y < 3/4$$

is it right? What is the interval of y?

It is obviously that for $$x\to \infty$$ we have a negative sign for the function $$f(x) =\frac{-3x+1}{x^2-6x-16}$$ which change the sign at every critical point, that is at $$8,1/3$$ and $$-2$$, since each if has an odd degree. So $$x\in(-\infty,-2)\cup ({1\over 3},8)$$.

Now since $$x={2\over 1-y}$$ we have to solve:

$${2\over 1-y}<-2\;\;\;{\rm and}\;\;\;{1\over 3}<{2\over 1-y}<8$$

From the first one we get $$1>y-1$$ so $$\boxed{y<2}$$ and from the second $$1-y<6$$ so $$\boxed{y>-5}$$ and $$1<4-4y$$ so $$\boxed{ y<{3\over 4}}$$.

Thus the final result is $$y\in (-5,{3\over 4})$$.

• So it is right? – Lifeforbetter Nov 8 '19 at 17:27
• It seems so.... – Aqua Nov 8 '19 at 17:28

Your first inequality is fullfiled for $$x<-2$$ or $$\frac{1}{3}