# What is the domain of g(t)=root4(t+4)?

Oct 31, 2016

Let's look at the similarities between the something you're probably very familiar with, square roots, and this fourth root problem.

Consider the following example.

Determine the value of:

a) ${\left(- 2\right)}^{2}$

b) ${\left(- 2\right)}^{3}$

c) ${\left(- 2\right)}^{4}$

a) $\left(- 2\right) \left(- 2\right) = 4$

b) $\left(- 2\right) \left(- 2\right) \left(- 2\right) = - 8$

c) $\left(- 2\right) \left(- 2\right) \left(- 2\right) \left(- 2\right) = 16$

As you can see, there is a pattern here. Whenever the power is even, the answer will always be positive, while if it is odd it will be negative.

So, we can also conclude by this that $\sqrt[3]{- 64}$ is defined while $\sqrt[4]{- 16}$ isn't.

Hence, the function $g \left(t\right) = \sqrt[4]{t + 4}$ is only defined when the number under the $4$th root is equal to or greater than $0$.

$0 \le \sqrt[4]{t + 4}$

$0 \le t + 4$

$- 4 \le t$

So, the domain is $t \ge - 4$.

Hopefully this helps!