Question

Explain how the domain of the function
g(x)=√x compares to the domain of g(x-k), where k≥ 0.

Answer 1

The function g(x) = √x has a domain of x ≥ 0, meaning that x must be greater than or equal to zero in order for the function to be defined.

When we consider the function g(x-k), where k ≥ 0, we are introducing a shift or translation in the x-direction by the value of k. This means that the graph of the function g(x) is shifted to the right by a distance of k units.

The effect of this shift on the domain of g(x) is that it remains the same. The shift does not affect the restrictions on the domain, which still require x ≥ 0. Regardless of the value of k, the function g(x-k) will still have a domain of x ≥ 0.

To illustrate this, let's consider an example. If we have the function g(x) = √x, the domain of g(x) is x ≥ 0. If we introduce a shift of 2 units to the right, g(x-2), the domain will still remain x ≥ 0. The function is simply shifted horizontally, but the range of allowable x-values remains unchanged.

In summary, the domain of the function g(x)=√x, which is x ≥ 0, remains the same when we consider g(x-k), where k ≥ 0. The shift or translation of the function does not affect the restrictions on the domain.

Explanation:

Answer 2

The function g(x) = √x represents the square root of x. Its domain consists of all non-negative real numbers because you can't take the square root of a negative number in the real number system.

Now, let's consider the function g(x - k), where k ≥ 0. When you subtract a non-negative constant (k) from x, it shifts the graph of the function horizontally to the right.

So, the domain of g(x - k) will still consist of all non-negative real numbers. In other words, shifting the function to the right by a non-negative value (k) does not change the domain. It remains the same as g(x) = √x, which is all non-negative real numbers or [0, ∞).