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can i have the domain of definition of: f(x)=6x/x^3+4

 Dec 17, 2018
 #1
avatar+812 
0

Make sure that x^3 + 4 is not 0. Dividing by 0 is undefined. This means that \(x^3 \ne -4 \Rightarrow x \ne ^3\sqrt{-4} \Rightarrow x \ne 2i\).

This value is not real, so we do not need to consider it. The domain of \(f(x) = \dfrac{6x}{x^3+4}\) is \(x \in (-\infty, \infty)\).

 

- PM

 Dec 17, 2018
edited by PartialMathematician  Dec 17, 2018
 #4
avatar+812 
+1

Actually, I think \(^3\sqrt-4\) might actually be a real number, so the domain is \(x \in (-\infty, ^3\sqrt-4) \cup (^3\sqrt-4, \infty)\)

PartialMathematician  Dec 17, 2018
 #3
avatar+94275 
+4

y =  6x / [ x^3 + 4]

 

The domain will only be affected by any x that makes the denominator = 0

 

So....set the denominator to 0 to find this value

 

x^3 + 4  = 0

 

x^3 =  -4     take the cube root of both sides

 

x =  ∛-4  =  - ∛4

 

So....the domain is

 

(-infinity,  -∛ 4 ) U  ( - ∛4, infinity )    

 

Here's a graph : https://www.desmos.com/calculator/qih0ucpatr

 

cool cool cool

 Dec 17, 2018

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