Math  /  Algebra

QuestionThe one-to-one functions gg and hh are defined as follows. g={(7,8),(1,8),(1,3),(3,3)}h(x)=9x+8\begin{array}{l} g=\{(-7,-8),(-1,8),(1,3),(3,-3)\} \\ h(x)=9 x+8 \end{array}
Find the following.
Q \square g1(3)=g^{-1}(3)= \square h1(x)=(h1h)(1)=\begin{array}{r} h^{-1}(x)= \\ \left(h^{-1} \circ h\right)(-1)= \end{array}

Studdy Solution
g1(3)=1g^{-1}(3) = 1 h1(x)=x89h^{-1}(x) = \frac{x - 8}{9} (h1h)(1)=1(h^{-1} \circ h)(-1) = -1

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