What is the domain and range of #F(x) = sqrt(x-3)#?

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Answer 1 · 2018-07-07T06:15:57

#x >= 3# or

in interval notation #[3, oo)#

Given: #F(x) = sqrt(x - 3)#

A function starts out having a domain of all Reals #(-oo, oo)#

A square root limits the function because you can't have negative numbers under the square root (they are called imaginary numbers).

This means #" "x - 3 >= 0#

Simplifying: #" "x >= 3#

Answer 2 · 2018-07-07T06:18:22

The domain is # x in [3, +oo)#. The range is #y in [0, +oo)#

Let #y=sqrt(x-3)#

What's under the #sqrt# sign must be #>=0#

Therefore,

#x-3>=0#

#=>#, #x>=3#

The domain is # x in [3, +oo)#

When #x=3#, #y=sqrt(3-3)=0#

And

#lim_(x->+oo)y=lim_(x->+oo)sqrt(x-3)=+oo#

Therefore,

The range is #y in [0, +oo)#

graph{sqrt(x-3) [-12.77, 27.77, -9.9, 10.38]}