# The one-to-one functions g and h are defined as follows. g={(-6,-5),(1,5),(4,8),(5,-3)} h(x)=3x+4 Find the following. g^(-1)(5)=? h^(-1)(x)=? (h * h^(-1))(7)

The one-to-one functions g and h are defined as follows.
$g=\left\{\left(-6,-5\right),\left(1,5\right),\left(4,8\right),\left(5,-3\right)\right\}$
h(x)=3x+4
Find the following.
${g}^{-1}\left(5\right)=?\phantom{\rule{0ex}{0ex}}{h}^{-1}\left(x\right)=?\phantom{\rule{0ex}{0ex}}\left(h\cdot {h}^{-1}\right)\left(7\right)$
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Makenna Lin
$g=\left\{\left(-6,-5\right),\left(1,5\right),\left(4,8\right),\left(5,-3\right)\right\}\phantom{\rule{0ex}{0ex}}h\left(x\right)=3x+4\phantom{\rule{0ex}{0ex}}{g}^{-1}=\left\{\left(-5,-6\right),\left(5,1\right),\left(8,4\right),\left(-3,5\right)\right\}\phantom{\rule{0ex}{0ex}}{g}^{-1}\left(5\right)=\left\{1\right\}$
$h\left(x\right)=3x+4$
Let
$\left(h\cdot {h}^{-1}\right)\left(7\right)=h\left[{h}^{-1}\left(7\right)\right]\phantom{\rule{0ex}{0ex}}=h\left[\frac{7-4}{3}\right]\phantom{\rule{0ex}{0ex}}=h\left[\frac{3}{3}\right]\phantom{\rule{0ex}{0ex}}=h\left(1\right)\phantom{\rule{0ex}{0ex}}=3\left(1\right)+4\phantom{\rule{0ex}{0ex}}=3+4\phantom{\rule{0ex}{0ex}}\left(h\cdot {h}^{-1}\right)\left(7\right)=7$