how do you find the domain of (x-3)^-1/4?

Question

campbell_st · Answer

well you are rally looking at 

$$\frac{1}{\sqrt[4]{x - 3}}$$

so x cannot equal 3 as this would result in a zero denominator... 

and the have a real number in the denominator 

x > 3 would be the domain.

tabithax3 · Answer

@campbell_st the problem is $$h(x)=\left( x-3 \right)^{-1/4}$$ so it means$$-\frac{ 1 }{ \sqrt[4]{x-3} }$$ right? and the domain is $$\left( 3,\infty \right)$$ or do i ignore the - in $$(x-3)^{-1/4}$$

tabithax3 · Answer

i hope that made sense...

campbell_st · Answer

if you have 

$$h(x) = (x - 3)^\frac{-1}{4}$$

then it can be written as 

$$h(x) = \frac{1}{\sqrt[4]{x - 3}}$$

so if you are dealing with real numbers then 

x cannot be less than 3 as it result in the 4th root of a negative... which means complex numbers... 

so you need to look at the set of numbers where x > 3

or 

$$(3, \infty)$$

this set of numbers will result in a real number answer. 

hope this helps

tabithax3 · Answer

got it, thank you for the help!