let y = f(x) to get:
y = sqrt(x-3)
let y = x and x = y to get:
x= sqrt(y-3)
square both sides of the equation to get:
x^2 = (y-3)
add 3 to both sides of the equation to get:
x^2 + 3 = y
that should be your inverse equation.
on the graph, both the equation and its inverse are shown as reflections of the line y = x.
the domain of the equation of y = sqrt(x-3) is all value of x >= 3.
this keeps the expression inside the square root sign either 0 or positive.
this is necessary since it can't be negative, because, if it is negative, then you don't get a real answer.
the range of the equation of y = sqrt(x-3) is all value of y greater than or equal to 0.
the inverse equation of y = x^2 + 3 has a range of y >= 3 and a domain of x >= 0.
this is because, the range of the inverse function is equal to the domain of the original function and the domain of the inverse function is equal to the range of the original function.
you can see on the graph that the domain of the inverse function is less than 0.
while that is true of the equation, it is not part of the inverse function.
the range of the inverse function must be equal to the domain of the original function and cannot be anything else.
therefore, the domain of the inverse function is restricted to the range of the original function.
the range of the original function is y >= 0.
the domain of the inverse function must be x >= 0 and nothing more.
the reason you see y = x^2 + 3, where x is negative, is because the graphing software can't restrict it.
if you were to draw the inverse function manually, or if you had software that could restrict the domain or the range, you would only draw the part that has x >= 0.
i can do the same thing by shading out the part of the graph where x is less than 0.
i can also chase out the part of the graph where y is less than 0.
i can also shade out the partof the graph where y <= -x + 3.
the leaves only the part of the graph where the original function and its inverse reside.
that part is not shaded.
the original and the inverse equation without the shading looks like this:
the originl and the inverse equation with the shading looks like this:
since the equations are inverses of each other, than the coordinate point (x,y) on the original equation is equal to the coordinate point of (y,x) on the inverse equation.
an example of this is the coordinate point of (3,0) on the original equation is equal to the coordinate point of (0,3) on the inverse equation.
this is because those points are reflections of each other about the line y = x.
reflections of each other about the line y = x means that, if you draw a line perpendicular to the line y = x that starts with the point (0,3) on the original equation, it will intersect with the point (3,0) on the inverse equation.
that happens with all reflected points, as shown on the graph for the other 2 points where a line perpendicular to the line y = x has been drawn.
another test for whether you have an inverse equation is that, if f(x) is the equation and g(x) is the inverse equation, then f(g(x)) = x and g(f(x)) = x.
i'll do both to show you.
f(x) = sqrt(x-3)
that's the original equation.
g(x) = x^2 + 3
that's the inverse equation.
f(g(x)) becomes sqrt((x^2+3) - 3)
simplify that to get:
f(g(x)) = sqrt(x^2 + 3 - 2) which becomes:
f(g(x)) = sqrt(x^2) which becomes:
f(g(x)) = x.
g(x) = x^2 + 3
f(x) = sqrt(x-3)
g(f(x)) = (sqrt(x-3))^2 + 3
that becomes:
g(f(x)) = x-3 + 3 which becomes:
g(f(x)) = x.
you were asked what is the domain of the inverse of f(x).
the domain of the inverse of f(x) is equal to the range of f(x).
since the domain of f(x) is all values of x >= 3, then the range of the inverse of f(x) is all values of y >= 3.
this can be seen on the graphs.
if you are using interval notation, then:
the domain of the inverse equation is y = [3,plus infinity)
that's the same as saying that y is greater than or equal to 3.
here's a reference on inverse equatons that you might find helpful.
https://www.purplemath.com/modules/invrsfcn.htm
if there's anything in the reference, or in what i just told you that you might still be struggling with, let me know and i'll do my best to clarify better for your understanding.