We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
150
4
avatar+9 

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

 Dec 17, 2018
 #1
avatar+701 
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+701 
-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+101798 
+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

5 Online Users