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December 14th, 2018, 08:19 AM   #1
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Post Range

could you give some ways to define the function's range?

f(x)=(x^2-x-1)/(x^2-x+1)
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December 14th, 2018, 09:55 AM   #2
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Originally Posted by shadow dancer View Post
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
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December 14th, 2018, 10:26 AM   #3
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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.
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December 14th, 2018, 10:35 AM   #4
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Originally Posted by shadow dancer View Post
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:

$\displaystyle -\frac{5}{3}\le f(x)<1$
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December 14th, 2018, 11:02 AM   #5
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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 $\displaystyle x=1/2$
$\displaystyle y(1/2)\leq y <1$
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