Question

What is the domain and range of `f(x)=sqrt(4x+2)`?

Answer 1

`x in [-1/2, +oo)`

Explanation:

The function is a Square Root Function

To easily determine the domain and range, we should first convert the equation to General Form:

`y=a*sqrt(x-b)+c`
Where the point `(b,c)` is the endpoint of the function (essentially the place at which the graph begins).

Let's now convert the given function to General Form:
`y=sqrt(4(x+1/2))`
We can now simplify this by taking the square root of 4 outside:
`y=2*sqrt(x+1/2)`

Therefore, from general form, we can now see that the endpoint of the graph is present at the point `(-1/2,0)` due to the fact that `b=-1/2` and `c=0`.

Additionally from General Form we can see that neither `a` is negative, nor is `x` negative, therefore no reflections about the `x` or `y` axis are present. This implies that the function originates from the point `(-1/2,0)` and continues to positive infinity.

For reference, the graph of the function `(y=sqrt(4x+2))` is below:
graph{sqrt(4x+2) [-10, 10, -5, 5]}

Therefore, the domain of the function can be expressed as:
1. Domain: `x in [-1/2, +oo)`
2. Domain: `x >=-1/2`
3. Domain: `-1/2<=x< +oo`