Question

Find the inverse of the function, no solutions doesn't work and things like sqrt(9-x^2) doesn't either.

Answer 1

The range of
`f(x)` is
`f(x)\in[0,3]`

`f^(-1)(x)=-√(-x^2+9)` with a domain
`x\in[0,3]`

Explanation:

Domain: set of input values (x-values)

Range: set of output values (y-values)

Given function:

`f(x)=√(9-x^2)\quad(x\in[-3,0])`

Therefore, as the original function has a restricted domain, its range is also restricted:

`f(-3)=√(9-(-3)^2)=0`

`f(0)=√(9-0)=3`

`\therefore\textsf{Range}\:f(x)\in[0,3]`

To determine the inverse of a function

Change
`x` to
`y`:

`\implies x=√(9-y^2)`

Square both sides:

`\implies x^2=9-y^2`

Switch sides:

`\implies 9-y^2=x^2`

Subtract 9 from both sides:

`\implies -y^2=x^2-9`

Divide both sides by -1:

`\implies y^2=-x^2+9`

Therefore:

`\implies y=√(-x^2+9)\textsf{ and }y=-√(-x^2+9)`

As the range of the inverse function is the same as the domain of the original function:

`\implies f^(-1)(x)=-√(-x^2+9)` only as the range is [-3, 0]

The domain of the inverse function is the same as the range of the original function.

`\therefore\textsf{Domain of}\:f^(-1)(x):x \in [0,3]`

The inverse of a function is ordinarily the reflection of the original function in the line
`y=x`.

**Please see attached graph**