2
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $y\in (-5,{3\over 4})$.

$\endgroup$
  • $\begingroup$ So it is right? $\endgroup$ – Lifeforbetter Nov 8 '19 at 17:27
  • $\begingroup$ It seems so.... $\endgroup$ – Aqua Nov 8 '19 at 17:28
0
$\begingroup$

Your first inequality is fullfiled for $$x<-2$$ or $$\frac{1}{3}<x<8$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.