Question

f (x) = -3 + 7. Find the inverse of f(x) and its domain.
O A. f-1(x) 277 - 3, where x=-7
O B. f-1(x) = +7 +3, where x# 7
c. f-1(x) = 11 – 3, where x# 3
O D. f-1(x) = + 3, where x# 3
S

Answer 1

Note: Your function is missing some information. It seems your function is

$f\left(x\right)=\frac{-3}{x}+7$.

So, I am solving the question based on the function $f\left(x\right)=\frac{-3}{x}+7$,  because it would still solve your query.

Answer:

Please see the explanation.

Step-by-step explanation:

Given the function

$f\left(x\right)=\frac{-3}{x}+7$

A function g is the inverse of a function f for y = f(x), x=g(y)

$y=\frac{-3}{x}+7$

Replace x with y

$x=\frac{-3}{y}+7$

solve for y

$y=-\frac{3}{x-7}$

so the inverse of $f\left(x\right)=\frac{-3}{x}+7$ will be:

$f^{-1}\:\left(x\right)=-\frac{3}{x-7}$

Finding the domain of $f^{-1}\:\left(x\right)=-\frac{3}{x-7}$

As we know that domain is the set of the possible input values where the function is defined.

so the function domain must be: $x<7\quad \mathrm{or}\quad \:x>7$

Therefore,

$\mathrm{Domain\:of\:}\:-\frac{3}{x-7}\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x<7\quad \mathrm{or}\quad \:x>7\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:7\right)\cup \left(7,\:\infty \:\right)\end{bmatrix}$