December 14th, 2018, 08:19 AM  #1 
Newbie Joined: Nov 2018 From: Iran Posts: 8 Thanks: 0  Range
could you give some ways to define the function's range? f(x)=(x^2x1)/(x^2x+1) thanks 
December 14th, 2018, 09:55 AM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
December 14th, 2018, 10:26 AM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,887 Thanks: 1506 
If analysis using calculus is disallowed, you can find the domain of the function's inverse. Some algebra grunt work yields the inverse relation $x = \dfrac{(y1) \pm \sqrt{(1y)(5+3y)}}{2(y1)}$ domain of this inverse relation is $\left[\dfrac{5}{3}, 1 \right)$ which would be the range of the original function. 
December 14th, 2018, 10:35 AM  #4  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Quote:
$\displaystyle y=\frac{x^2x1}{x^2x+1}$ Arrange this as: $\displaystyle (y1)x^2(y1)x+(y+1)=0$ where $\displaystyle y\ne1$ Require the discriminant to be nonnegative: $\displaystyle ((y1))^24(y1)(y+1)\ge0$ $\displaystyle (y1)(3y+5)\le0$ 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: iPhone Posts: 486 Thanks: 75 
For this one the limit is allowed $\displaystyle y<\lim_{x\rightarrow \infty } y =1$ Write $\displaystyle x^2x=(x1/2)^2 +1/4$ The lower bound holds for $\displaystyle x=1/2$ $\displaystyle y(1/2)\leq y <1$ 

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