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

1 Answer
Apr 10, 2016

Answer:

#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#