+0

-1
989
1

What is the smallest integer value of $c$ such that the function $f(x)=\frac{2x^2+x+5}{x^2+4x+c}$ has a domain of all real numbers?

Jul 12, 2018

#1
+1

The only part of the function that has the ability to restrict the domain is the denominator; the numerator plays no role in this. For this reason, we should only focus on the denominator and ignore the numerator.

Our job, in this case, is to determine the smallest value of c that will forever prevent a divisor by zero. Let's try and do that.

 $$x^2+4x+c=0$$ By setting the denominator equal to zero, we will see what solutions result in a real input that equals zero. Let's use the quadratic formula. $$a=1,b=4,c=c;\\ x_{1,2}=\frac{-4\pm\sqrt{\textcolor{red}{4^2-4*1*c}}}{2*1}$$ We want to focus on the discriminant because that determines whether or not there are real solutions for a given quadratic. $$16-4c<0$$ When the discriminant is less than zero, there are no real solutions. This means that there will no input that causes a domain restriction. $$16<4c$$ $$4 < c\text{ or }c > 4$$ The immediate integer after 4 is 5, so this is the smallest integer that will result in no domain restriction. $$c=5$$
Jul 12, 2018
edited by TheXSquaredFactor  Jul 12, 2018

#1
+1
 $$x^2+4x+c=0$$ By setting the denominator equal to zero, we will see what solutions result in a real input that equals zero. Let's use the quadratic formula. $$a=1,b=4,c=c;\\ x_{1,2}=\frac{-4\pm\sqrt{\textcolor{red}{4^2-4*1*c}}}{2*1}$$ We want to focus on the discriminant because that determines whether or not there are real solutions for a given quadratic. $$16-4c<0$$ When the discriminant is less than zero, there are no real solutions. This means that there will no input that causes a domain restriction. $$16<4c$$ $$4 < c\text{ or }c > 4$$ The immediate integer after 4 is 5, so this is the smallest integer that will result in no domain restriction. $$c=5$$