The one-to-one functions g and h are defined as follows. g= {(-6, -5), (-2,3), (0,5) (9,0)} h(x)= 9x-8 Find g^-1(0), h^-1(x) and (h open circle h^-1)(1)

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The one-to-one functions g and h are defined as follows. g= {(-6, -5), (-2,3), (0,5) (9,0)} h(x)= 9x-8 Find g^-1(0), h^-1(x) and (h open circle h^-1)(1)

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please help me!
for g, it is quite simple , just switch the coordinate and you get \(g^{-1}= \{(-5,-6), (3,-2), (5,0), (0,9)\}\) on the new set, you see (0,9) , right? that is \(g^{-1}(0) = 9\). Dat sit
thank you! do you know how to solve the others?

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That's all I know.
y=9x-8 is a linear equation it is one to one to find h's inverse solve y=9x-8 for x. Then interchange the x and y at the end (some people like to do this as first step and then solve for y but whatever it is same deal-o; there is a reason I like to which them at the end sometimes it helps me in figuring out the correct domain to restrict the original to but you don't have to worry about any restrictions here)

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