Consider the equation yx²+(y-1)x+(y-1)=0
1. Find the set values of y for which this equation has real roots.
2.Hence determine the range of the function

f(x)=(x+1)/(x²+x+1)

3. Explain why f(x) has no inverse.

Question

Consider the equation yx²+(y-1)x+(y-1)=0
1. Find the set values of y for which this equation has real roots.
2.Hence determine the range of the function 

f(x)=(x+1)/(x²+x+1)

3. Explain why f(x) has no inverse.

deoxna · Answer

I get that for the first question you can use the quadratic formula to get y in terms of x. Then solving for it I got y=1. According to wolframalpha there's also an "integer root" (not sure what that means) at y=0 for both cases. What does this mean? I got question 2, but I'm not sure about 3 (since my only explanation relies on the fact that the equation from (1) has two roots). Thanks for the help!

deoxna · Answer

This is the WolframAlpha page for the first question: http://www.wolframalpha.com/input/?i=yx%C2%B2%2B%28y-1%29x%2B%28y-1%29%3D0+in+terms+of+x

anonymous · Answer

Use the Discriminant:
$$b^2-4ac \ge 0$$

anonymous · Answer

$$(y-1)^2-4y(y-1) \ge 0$$

deoxna · Answer

For question 2? Yeah I already got that, the range of the function is$$-\frac{ 1 }{ 3 } \le y \le 1$$
But my question is what does the "integer square root" mean (wikipedia wasn't very helpful) and whether its relevant to question 1 and 3. Also, how to solve question 3.