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?

ant101
Nov 28, 2018

#2**+2 **

To have a domain of all real numbers

1x^2 + 4x + c must not = 0

We have the form ax^2 + bx + c

We need to find the x root that makes this 0.... to do this,set the discriminant to 0

b^2 - 4ac = 0

4^2 - 4(1)(c) = 0

4^2 - 4c = 0

And when c = 4....we will have one root that makes the denominator = 0

So.... c = 5 is the smallest value that produces a non-zero denominator for all x values because the value makes the discriminant < 0 and we will have no real roots

Look at the graph here to confirm this : https://www.desmos.com/calculator/4qtwwfaidb

When c is an integer < 4, we have two zeroes

When c = 4, we have one zero

When c is an integer > 4, we have no real zeroes

CPhill
Nov 28, 2018