Question

Consider the function f(x)=x+9−−−−−√−3f(x)=x+9−3 on the domain [−9,[infinity])[−9,[infinity]). The inverse of this function is f−1(x)=x2+6xf−1(x)=x2+6x. A: What is the domain of f−1f−1? B: Why must the domain of f−1f−1 be restricted?

Answer 1

Answer with Step-by-step explanation:

We are given that a function $f(x)=\sqrt{x+9}-3$

Domain of f(x)=[-9,$\infty$)

The inverse of given function $f^{-1}(x)=x^2+6x$

a.We have to find the domain of $f^{-1}(x)$

We know that domain of f(x) is convert into range of $f^{-1}(x)$ and range of f(x) is convert into domain of $f^{-1}(x)$

If we substitute x=-9 in the given function then we get

$f(x)=\sqrt{-9+9}-3=-3$

Therefore, range of f(x) =[-3,$\infty$)

Domain of $f^{-1}(x)=[-3,[tex]\infty)$

b.Range  of f(x) is restricted .Therefore, domain of $f^{-1}(x)$ must be restricted because range of f(x) is converted into domain of $f^{-1}(x)$ and range of f(x) is restricted.