Find f^-1 and the domain and range of f^-1?

Question

The graph pf the function in the problem is one-quarter of the graph of the circle the radius 3 and center (0,0).

Find f^-1 and the domain and range of f^-1

f(x) = √(9 - x^2) , 0 ≤ x ≤ 3

Answer 1

y = √(9 - x^2)

interchange variables and solve for y

x = √(9 - y^2)

x^2 = 9 - y^2

y^2 = 9 - x^2

y = √(9 - x^2)

.. . .

thus

f^(-1)(x) = √(9 - x^2)

domain: 0 ≤ x ≤ 3

range: 0 ≤ y ≤ 3

.. . .. .

Answer 2

Like "DC - ドン" said, but you can also observe that swapping x & y is the same as reflecting the graph in the line y=x.

Your function is symmetrical about this line, so it is its own inverse, without algebra.