Answer:
The domain of [tex]f(x)[/tex] is restricted to [tex]x\leq 0[/tex], and the domain of [tex]f^{-1}(x)[/tex] is restricted to [tex]x\geq 4[/tex].
Step-by-step explanation:
From Function Theory we know that domain of a function is the set of values such that an image exist. Let [tex]f(x) = x^{2}+4[/tex] and [tex]f^{-1}(x) = \sqrt{x-4}[/tex] 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 [tex]x-4 \geq 0[/tex], which means that [tex]x\geq 4[/tex].
Finally, we concluded that following answer offers the best approximation to our result:
The domain of [tex]f(x)[/tex] is restricted to [tex]x\leq 0[/tex], and the domain of [tex]f^{-1}(x)[/tex] is restricted to [tex]x\geq 4[/tex].
Answer:
C. The domain of f(x) is restricted to x ≤ 0, and the domain of f–1(x) is restricted to x ≥ 4.
Step-by-step explanation:
