How do you find the domain and range for #y =sqrt(4x-1)#?

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Answer 1 · 2015-05-25T23:05:10

Assuming we're working with real numbers, #sqrt(a)# is only defined for #a>=0#, but it is defined for all #a>=0#.

So we require #4x - 1 >= 0#.

Adding #1# to both sides, then dividing both sides by #4# we get:

#x >= 1/4#

So our domain is #{x in RR: x >= 1/4}# or #[1/4, oo)# in interval notation.

The range is #{y in RR: y >= 0}# or #[0, oo)# in interval notation.

To prove this:

Given any #y >= 0#, let #x = (y^2+1)/4#

Then #4x - 1 = y^2#

and since #y >= 0# we have #y = sqrt(4x-1)# as required.