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)
[tex]\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}[/tex]
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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)
[tex]g\placeholder{⬚}^{-1}(5)=-9[/tex]
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Find the inverse of h
[tex]h(x)=\frac{-x-2}{7}[/tex]
- Substitute the h(x) by y:
[tex]y=\frac{-x-2}{7}[/tex]
Solve x:
[tex]\begin{gathered} 7y=-x-2 \\ 7y+2=-x \\ \\ x=-7y-2 \end{gathered}[/tex]
Substitute the x by h^-1 (x) and y by x:
[tex]h\placeholder{⬚}^{-1}(x)=-7x-2[/tex]
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Use the second property of one to one functions:
[tex](h\circ h\placeholder{⬚}^{-1})(-1)=-1[/tex]
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Final answers:
[tex]\begin{gathered} g\placeholder{⬚}^{-1}(5)=-9 \\ \\ h\placeholder{⬚}^{-1}(x)=-7x-2 \\ \\ (h\circ h\placeholder{⬚}^{-1})(-1)=-1 \end{gathered}[/tex]
