+0

# can someone help me with this pls?

0
239
4
+9

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

Dec 17, 2018

#1
+770
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
#2
+770
0
PartialMathematician  Dec 17, 2018
#4
+770
-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
+111357
+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

Dec 17, 2018