Question

Consider the function and its inverse.

and


When comparing the functions using the equations, which conclusion can be made?

f(x) = x^2 +4 and f^-1 (x) = square root of x-4


The domain of f(x) is restricted to x ≥ 0, and the domain of f–1(x) is restricted to x ≥ 0.

The domain of f(x) is restricted to x ≥ 0, and the domain of f–1(x) is restricted to x ≤ 0.

The domain of f(x) is restricted to x ≤ 0, and the domain of f–1(x) is restricted to x ≥ 4.

The domain of f(x) is restricted to x ≤ 0, and the domain of f–1(x) is restricted to x ≤ 4.

Answer 1

C. The domain of f(x) is restricted to x ≤ 0, and the domain of f–1(x) is restricted to x ≥ 4.

Explanation:

Answer 2

The domain of
`f(x)` is restricted to
`x\leq 0`, and the domain of
`f^(-1)(x)` is restricted to
`x\geq 4`.

Explanation:

From Function Theory we know that domain of a function is the set of values such that an image exist. Let
`f(x) = x^(2)+4` and
`f^(-1)(x) = √(x-4)` the function and its inverse, respectively.

At first glance we notice that function is a second order polynomial and every polinomial is a continuous function and therefore, there exists an image for every element of domain.

But domain of its inverse is restricted to every value of x so that
`x-4 \geq 0`, which means that
`x\geq 4`.

Finally, we concluded that following answer offers the best approximation to our result:

The domain of
`f(x)` is restricted to
`x\leq 0`, and the domain of
`f^(-1)(x)` is restricted to
`x\geq 4`.