can anyone help me out on this problem?

Question

http://i181.photobucket.com/albums/x85/ctti_3/untitled-37.jpg

Answer 1

1) Draw a picture (or use the grapher). the graph for x from 0 to 3 looks like an arc, symmetric about identity, or line y=x.

This is an important clue.

2) do the usual:

y = sqrt(9-x^2), interchange x and y

x = sqrt(9-y^2), solve for y

x^2 = 9-y^2, so

y^2 = 9-x^2 or
y = sqrt(9-x^2).

No accident, this function is its own inverse, since it is symmetric about the identity. And yes, x is still from 0 to 3.

Answer 2

Given:
f(x) = ±√(9 - x^2), 0 ≤ x ≤ 3, ( - 3 ≤ y ≤ 3 )
Using y for f(x):
y = ±√(9 - x^2), 0 ≤ x ≤ 3, - 3 ≤ y ≤ 3
Swap x and y:
x = ±√(9 - y^2), 0 ≤ y ≤ 3, - 3 ≤ x ≤ 3
Expand:
x^2 = 9 - y^2, 0 ≤ y ≤ 3, - 3 ≤ x ≤ 3
y^2 = 9 - x^2, 0 ≤ y ≤ 3, - 3 ≤ x ≤ 3
Taking the root,
y = + √(9 - x^2), because 0 ≤ y ≤ 3, - 3 ≤ x ≤ 3
so
f^-1(x) = + √(9 - x^2), - 3 ≤ x ≤ 3

Answer 3

f(x)=square root of 9-x^2
lets f(x)=y

( 9-x^2)^1/2 = y
(9-x^2) = y^2
x^2 = 9-y^2
x= (9-y^2)^1/2

therefore, inverse of f(x) = (9-x^2)^1/2
answer: 1

Answer 4

f^-1 (x) is given by

x = sqrt(9-y^2)
x^2 = 9 - y^2
y^2 = 9 - x^2
y= +/- sqrt(9 - x^2)
but dom f = [0,3]
and range f = [0,3]

so dom f^-1 = [0,3], range f^-1 = [0,3],
so, f^-1(x) = sqrt(9 - x^2)

Answer 5

I can help you if you tell me the problem