Great Question!
The problem we have at hand is known to be a function, which maps elements from one set of objects, ( the domain ) onto another, the range. If we were to consider an ordered pair, say ( x, y ), then the function would map x onto y. The inverse function is simply the reverse. Take the ordered pair (-4,0). Function g would map - 4 onto 0, such that [tex]g( - 4 ) = 0[/tex]. Therefore, the inverse function would map 0 onto - 4, resulting in [tex]g^{-1}( 0 ) = - 4[/tex]. And there you have it! Our first part is answered!
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This second bit here is interesting. Let [tex]y = h( x )[/tex] -
[tex]y = 4x + 3[/tex] - Switch x and y,
[tex]x = 4y + 3[/tex] - And now solve this equation for y,
[tex]x - 4y = 3,\\- 4y = - x + 3,\\y = 1 / 4x - 3 / 4[/tex]
As you can see, we have taken the inverse of h( x ). As y = h( x ), we can thus conclude the following -
[tex]h^{-1}(x) = 1 / 4x - 3 / 4[/tex]
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The composition (h^-1 o h)(-5) is, in other words, h^-1(h(-5)). We can therefore calculate h(-5) and then take it's inverse -
[tex]h(-5) = 4(-5) + 3,\\h(-5) = - 20 + 3,\\h(-5) = - 17[/tex]
Now we can take it's inverse -
[tex]h^{-1}(-17) = 1 / 4( - 17 ) - 3 / 4,\\- 17 / 4 - 3 / 4,\\= - 5![/tex]
Our solution for this last bit is - 5. And, if you don't feel like reading through this entire explanation just take a look at the " summed up " answer below,
[tex]g^{-1}( 0 ) = - 4,\\\\h^{-1}( x ) = 1 / 4x - 3 / 4,\\\\( h * h^{-1} )( - 5 ) = - 5[/tex] I do hope that helps you!
