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# domain and range!

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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?

Nov 28, 2018

#1
+5172
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got mangled somehow and CPhil answered anyway.

Nov 28, 2018
edited by Rom  Nov 28, 2018
#2
+101369
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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

Nov 28, 2018
#3
+866
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Exactly! Thanks, CPhill!

ant101  Nov 28, 2018