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 December 14th, 2018, 08:19 AM #1 Newbie   Joined: Nov 2018 From: Iran Posts: 12 Thanks: 0 Math Focus: calculus Range could you give some ways to define the function's range? f(x)=(x^2-x-1)/(x^2-x+1) thanks
December 14th, 2018, 09:55 AM   #2
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Quote:
 Originally Posted by shadow dancer could you give some ways to define the function's range? f(x)=(x^2-x-1)/(x^2-x+1) thanks
Normally you could mess around with critical points and first and second derivatives to sketch it, but it'd be faster to graph it. The only other way is to try out a bunch of points and see what happens.

-Dan

 December 14th, 2018, 10:26 AM #3 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 If analysis using calculus is disallowed, you can find the domain of the function's inverse. Some algebra grunt work yields the inverse relation $x = \dfrac{(y-1) \pm \sqrt{(1-y)(5+3y)}}{2(y-1)}$ domain of this inverse relation is $\left[-\dfrac{5}{3}, 1 \right)$ which would be the range of the original function. Thanks from topsquark
December 14th, 2018, 10:35 AM   #4
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Quote:
 Originally Posted by shadow dancer could you give some ways to define the function's range? f(x)=(x^2-x-1)/(x^2-x+1) thanks
I would write:

$\displaystyle y=\frac{x^2-x-1}{x^2-x+1}$

Arrange this as:

$\displaystyle (y-1)x^2-(y-1)x+(y+1)=0$ where $\displaystyle y\ne1$

Require the discriminant to be non-negative:

$\displaystyle (-(y-1))^2-4(y-1)(y+1)\ge0$

$\displaystyle (y-1)(3y+5)\le0$

From this we conclude:

$\displaystyle -\frac{5}{3}\le f(x)<1$

 December 14th, 2018, 11:02 AM #5 Senior Member   Joined: Dec 2015 From: somewhere Posts: 600 Thanks: 87 For this one the limit is allowed $\displaystyle y<\lim_{x\rightarrow \infty } y =1$ Write $\displaystyle x^2-x=(x-1/2)^2 +1/4$ The lower bound holds for $\displaystyle x=1/2$ $\displaystyle y(1/2)\leq y <1$ Thanks from topsquark

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