Range of a function

[Mullandowski]

I have question, how do I ALGEBRAICALLY find the RANGE of this function y= (x^2+x)/ sqrt (x^3+x^2) ? I know how to do this graphically, but I can't algebraically. Please help.

 

[Mullandowski]

I have question, how do I ALGEBRAICALLY find the RANGE of this function y= (x^2+x)/ s

the equation is:
Wout Pout SOLVE FOR Win
Win = Pin


The correct answer is this

Win = Wout Pout
Pin
But i dont know the step on how to get this answer please help, all opinions are welcomed

[FONT=&quot]I have question, how do I ALGEBRAICALLY find the RANGE of this function y= (x^2+x)/ sqrt (x^3+x^2) ? I know how to do this graphically, but I can't algebraically. Please help.[/FONT]

 

[HallsofIvy]

Mulandowski: please do not "hijack" another persons thread to ask a separate question. That is extremely rude. There is a "Post New Thread" button at the top of each category. Use that.

Finding the domain of a function is much easier than finding the range. So I would recommend converting "range" to "domain" by inverting the function.

Here, \(\displaystyle y= \frac{x^2+ x}{\sqrt{x^3+ x^2}}\). Multiply by that denominator to get \(\displaystyle y\sqrt{x^3+ x^2}= x^2+ x\). Square both sides to get rid of the square root: \(\displaystyle y^2(x^3+ x^2)= x^4+ 2x^3+ x^2\). We can divide both sides by \(\displaystyle x^2\): \(\displaystyle y^2(x+ 1)= yx^2+ y= x^2+ 2x+ 1\) and that reduces to the quadratic equation. \(\displaystyle (y- 1)x^2- 2x+ (y- 1)= 0\). Use the quadratic formula to solve for x as a function of y. The domain of that function is the range of the original function.