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)

Question

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)

faf3 · Answer

please help me!

loser66 · Answer

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

faf3 · Answer

thank you! do you know how to solve the others?

loser66 · Answer

That's all I know.

freckles · Answer

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)