Answer:
B. [tex]k(x)=(-x+3)^{1/2}[/tex]
Step-by-step explanation:
When we're talking about a restricted domain, we're looking for a function that's undefined for certain values of x. A classic example is the function f(x) = 1/x. Division by 0 is undefined, so this restricts our domain to all real numbers except 0.
A key feature of our problem is in the exponents - are there any exponents that would limit the possible values of x? Let's strip away the rest of the details and just look at those powers for each option:
For A, let's look at the function [tex]f(x) = x^{1/5}[/tex]. Remember that fractional exponents are the same as roots, so we could also say [tex]f(x)=\sqrt[5]{x}[/tex]. Any odd root like this can take any value for x, positive and negative. This has to do with the fact that any odd power of a negative is always negative. So no restrictions on A.
For B, just looking at the exponent, let's see if there are restrictions on the function [tex]k(x) = x^{1/2}[/tex]. Again, we can rewrite the right side as a root: [tex]k(x) = \sqrt{x}[/tex].
Unline f(x), k(x) is restricted. The square root of a negative number has no real solutions, so k(x) comes with the restriction that x ≥ 0. We could look at C and D, but we've already found our solution at this point, so I'll stop here!
