Find the inverse of f?

Replace F(x) with y to make this easier

y = 3+√(x+7)

Now swap the x and y

x = 3 + √(y+7)

Solve for y

x - 3 = √(y+7)

x^2 - 6x + 9 = y + 7

y = x^2 - 6x + 2

So, the domain of an inverse function is the range of your original function and the range of your inverse function is the domain of the original function

F(x) has a range of [3,infinity), so that is also the domain of your inverse function

y = 3 + sqrt(x + 7)
y - 3 = sqrt(x + 7)
(y - 3)^2 = x + 7
y^2 - 6y + 9 = x + 7
x = y^2 - 6y + 9 - 7
x = y^2 - 6y + 2
The inverse is:
y = x^2 - 6x + 2

The domain of the original function is:
x + 7 >= 0
x >= -7
The range of the original function is:
sqrt(x + 7) >= 0
3 + sqrt(x + 7) >= 3
y >= 3

Normally, y = x^2 - 6x + 2 would have a domain of all real numbers. But in this case it's the inverse of F(x), so the range of the original function becomes the domain of the inverse: x >= 3

Where did you get x^2 - 16?

f = 3 + √ ( x + 7 ) ---> ( f - 3 )² = x + 7 ---> x = inverse f = ( f - 3)² - 7 ╪ (f² - 16 )