How do you find the domain and range of #Y=(2x-1)^(1/2)#?

Question

Kennedy Adams · Accepted Answer

The domain is # x in [1/2, +oo)#
The range is #y in [0,+oo)# Our role is #y=f(x)=sqrt(2x-1)# What's under the square root sign is #>=0# Consequently, #2x-1>=0# #x>=1/2# The domain is # x in [1/2, +oo)# The minimum value of #y=0# and the maximum value is #+oo# The range is #y in [0,+oo)# graph{sqrt(2x-1) [-3.12, 3.123, 6.243, -6.244]}

Nathan Axel · Answer

undefined To find the domain of the function $ y = \sqrt{2x - 1} $, set the expression under the square root, $ 2x - 1 $, greater than or equal to zero and solve for $ x $. The range of the function depends on the domain of $ 2x - 1 $. Since the square root of a real number is always non-negative, the range of $ y $ is also non-negative.

**Domain:** Solve $ 2x - 1 \geq 0 $ for $ x $:
$$ 2x \geq 1 $$
$$ x \geq \frac{1}{2} $$

**Range:** Since the square root of a real number is always non-negative, the range of $ y $ is all real numbers greater than or equal to zero: $ y \geq 0 $.