Find all values of c such that the domain of f(x)=(x+1)/(x^2+2cx+4) where the domain is all real numbers?

I've been stuck on this problem for over twenty minutes, I am not even sure if there is a value of C where the domain is all real numbers because I see that there will always be an asymptote. Thanks in advance.

Find all values of c such that the domain of f(x)=(x+1)/(x^2+2cx+4) where the domain is all real numbers?
You want values of c that will not make the bottom of the fraction zero.





Thus x^2 + 2cx + 4 must not be equal to zero for those values of c.





This means that x^2 + 2cx + 4 must not have any real roots.





This occurs when the discriminant %26lt; zero





Discriminant = 4c^2 - 16 %26lt; 0





c^2 - 4 %26lt; 0





(c-2)(c+2) %26lt; 0





This occurs when -2 %26lt; c %26lt; 2
Reply:what about c=0????





whit x^2 + 4 in the denominator, there are no asymptotes.





You want the denominator to be different of 0.


Let's solve x^2 + 2cx + 4 = 0





this equation has no solution if 4c^2 - 16%26lt;0





which means:





c^2%26lt;4





%26lt;=%26gt; c is in (-2,2)





therefor, the domain of f(x)=(x+1)/(x^2+2cx+4) is all real numbers if c is between -2,2

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