Question

Need help with this assignment if you can help that would be great

Answer 1

Properties of a one to one function:

1) The domain of f equals the range of f^-1 and the range of f equals the domain of f^-1

2)

`\begin{gathered} f\placeholder{⬚}^(-1)(f(x))=(f\placeholder{⬚}^(-1)\circ f)(x)=x \\ \\ f(f\placeholder{⬚}^(-1)(x))=(f\circ f\placeholder{⬚}^(-1))(x)=x \end{gathered}`

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As g is a one to one function, the domain of g^-1 is equal to the range of g

In g: (x,y) (domain,range)

g^-1(5) (find the range corresponding to the domain 5 in g^-1):

Use the point in g (-9,5), then in g^-1 is (5,-9)

`g\placeholder{⬚}^(-1)(5)=-9`

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Find the inverse of h

`h(x)=(-x-2)/(7)`

- Substitute the h(x) by y:

`y=(-x-2)/(7)`

Solve x:

`\begin{gathered} 7y=-x-2 \\ 7y+2=-x \\ \\ x=-7y-2 \end{gathered}`

Substitute the x by h^-1 (x) and y by x:

`h\placeholder{⬚}^(-1)(x)=-7x-2`

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Use the second property of one to one functions:

`(h\circ h\placeholder{⬚}^(-1))(-1)=-1`

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Final answers:

`\begin{gathered} g\placeholder{⬚}^(-1)(5)=-9 \\ \\ h\placeholder{⬚}^(-1)(x)=-7x-2 \\ \\ (h\circ h\placeholder{⬚}^(-1))(-1)=-1 \end{gathered}`