if h(x)=2-x/2, find (h o h^-1)(2) A)X B)1/2 C)2 D)0 THANK YOU SO MUCH!!

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if h(x)=2-x/2, find (h o h^-1)(2) A)X B)1/2 C)2 D)0 THANK YOU SO MUCH!!

Calculus1
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is the function this? \[\Large h(x) = 2 - \frac{x}{2}\] OR is the function this? \[\Large h(x) = \frac{2-x}{2}\]
No, it is..
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ok so the first one I wrote
So yes, the first one you stated. Thank you.
think of it as \[\Large y = 2 - \frac{x}{2}\]
then swap x and y \[\Large x = 2 - \frac{y}{2}\] then solve for y
How
what about the h o h ^-1(2)
I would first multiply both sides by 2 that clears out the fraction
so you'll have 2x = 4 - y isolate y
-y=2x-4
then what happens with the h o h ^1 (2). That is my primary concern.
-y=2x-4 will turn into y = -2x+4
so the inverse is \[\Large h^{-1}(x) = -2x+4\]
\[\Large (h \circ h)^{-1}(2)\] is the same as \[\Large h(h^{-1}(2))\] the first task is to compute \(\Large h^{-1}(2)\)
Okay
sorry I meant to say \(\Large (h \circ h^{-1})(2)\)
anyways, what output do you get when you plug x = 2 into the inverse
0?
yep
Oh I see the -1
then that 0 is plugged into the h(x) function
\[\Large \Large h(\color{red}{h^{-1}(2)}) = h(\color{red}{0})\]
Thank you so much for the walkthrough, it really helped me. :)
so you found what h(0) is?
0 right?
plug it into 2 - (x/2)
h^-1 (2)= -2(2)+4
yeah that part is 0, but we're not done yet
oh, so it is just 2?
you take that result and plug it into h(x)
ultimately, the final result is 2 you'll find that \[\Large (h \circ h^{-1})(x) = x\] for any x value
2- 0/2=2-0
because the inverse essentially "undoes" what the original function does
Oh, I see.

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