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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
 #1
avatar+3198 
+2

got mangled somehow and CPhil answered anyway.

Rom  Nov 28, 2018
edited by Rom  Nov 28, 2018
 #2
avatar+92814 
+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

 

 

cool cool cool

CPhill  Nov 28, 2018
 #3
avatar+742 
+1

Exactly! Thanks, CPhill! 

ant101  Nov 28, 2018

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