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

1 Answer
Apr 10, 2016

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