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the one to one function g and h are defined as follows g={(-1,8),(1,3),(6,4),(8,5),(9,-9)}h(x)=x-8/7Find the following. g^-1 (8) = h^-1 (x) =(h • h^-1)(o) =

the one to one function g and h are defined as follows g={(-1,8),(1,3),(6,4),(8,5),(9,-9)}h-example-1
User Mitar
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1 Answer

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One to one function : if no two elements in the domain of f correspond to the same element in the range of f .

The function g and h are one to one function and defined as;

g={(-1,8),(1,3),(6,4),(8,5),(9,-9)}


h(x)=(x-8)/(7)

1)


\begin{gathered} g^(-1)(8) \\ \text{From the defination of function g;} \\ g=\text{ (-1,8)} \\ i\mathrm{}e\text{. it defined as g(-1) = 8} \\ \text{thus, -1=g}^(-1)(8) \\ g^(-1)(8)=-1 \end{gathered}

2)


\begin{gathered} h^(-1)(x) \\ \text{The function h(x) defined as;} \\ h(x)=(x-8)/(7) \\ \text{Let h(x)=y} \\ y=(x-8)/(7) \\ \text{Solve for x;} \\ 7y=x-8 \\ x=7y+8 \\ h^(-1)(x)=7x+8 \end{gathered}

3)


\begin{gathered} (h\circ h^(-1))(0) \\ (h\circ h^(-1))(x)=h(h^(-1)(x)) \\ (h\circ h^(-1))(x)=h(7x+8) \\ (h\circ h^(-1))(x)=((7x+8)-8)/(7) \\ (h\circ h^(-1))(x)=(7x+8-8)/(7) \\ (h\circ h^(-1))(x)=(7x+0)/(7) \\ (h\circ h^(-1))(x)=x \\ \text{Substitute x = 0} \\ (h\circ h^(-1))(0)=0 \end{gathered}

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User Magirtopcu
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