Answer:
all real nunbers
Step-by-step explanation:
when we exchange the place of x and y and solve for y we get the equation y=x^2 + 2
any number inserted in x will give a value y
The range of the inverse of the given function f(x) is all real numbers.
"The inverse function of a function is a function that undoes the operation of the function. The inverse of a function exists if and only if the given function is bijective, and if it exists."
The given function is
[tex]f(x)= \sqrt{x-2}[/tex]
⇒ [tex]y = \sqrt{x-2}[/tex]
In order to calculate the inverse of the given function, we have to write:
[tex]x = \sqrt{y-2}[/tex]
⇒ [tex]x^{2}=y-2[/tex]
⇒ [tex]y = x^{2} +2[/tex]
Therefore, the inverse of the given function f(x) is
[tex]f^{-1}(x) = (x^{2}+2)[/tex]
Here, for any value of 'x', there will be a value of [tex]f^{-1}(x)[/tex].
Therefore, the range of [tex]f^{-1}(x)[/tex] is all real numbers.
Learn more about the inverse of a function here: https://brainly.com/question/12063552
#SPJ2
