# Find the domain of the function. Write your answer in both set builder and interval notation. the square root of 3x + 1

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### 2 Answers

Find the domain of `f(x)=sqrt(3x+1)` :

Restrictions on the domain typically involve division by zero, taking a logarithm of a nonnegative number, and taking even roots of negative numbers.

Here, the expression in the radicand cannot be negative if the expression is to be real.

3x+1<0 ==> `x<-1/3` ; so x cannot be less than `1/3` .

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The domain is `{x|x in RR, x>= -1/3 }` or `[-1/3,oo)`

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The domain is all the valid x values for the function.

Our function is:

`sqrt(3x+1)`

The only restriction for functions with square roots is that there can't be any negatives under the root (unless you have imagination! :) ).

So, to make sure nothing under the root is less that zero:

`3x+1 >= 0`

Solving this equation, we get:

`x>=-1/3`

In set builder notation:

`{x|x>=-1/3}`

In interval notation:

`[-1/3,oo)`