anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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jim_thompson5910
  • jim_thompson5910
is the function this? \[\Large h(x) = 2 - \frac{x}{2}\] OR is the function this? \[\Large h(x) = \frac{2-x}{2}\]
anonymous
  • anonymous
No, it is..
anonymous
  • anonymous
|dw:1442458455816:dw|

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jim_thompson5910
  • jim_thompson5910
ok so the first one I wrote
anonymous
  • anonymous
So yes, the first one you stated. Thank you.
jim_thompson5910
  • jim_thompson5910
think of it as \[\Large y = 2 - \frac{x}{2}\]
jim_thompson5910
  • jim_thompson5910
then swap x and y \[\Large x = 2 - \frac{y}{2}\] then solve for y
anonymous
  • anonymous
How
anonymous
  • anonymous
what about the h o h ^-1(2)
jim_thompson5910
  • jim_thompson5910
I would first multiply both sides by 2 that clears out the fraction
jim_thompson5910
  • jim_thompson5910
so you'll have 2x = 4 - y isolate y
anonymous
  • anonymous
-y=2x-4
anonymous
  • anonymous
then what happens with the h o h ^1 (2). That is my primary concern.
jim_thompson5910
  • jim_thompson5910
-y=2x-4 will turn into y = -2x+4
jim_thompson5910
  • jim_thompson5910
so the inverse is \[\Large h^{-1}(x) = -2x+4\]
jim_thompson5910
  • jim_thompson5910
\[\Large (h \circ h)^{-1}(2)\] is the same as \[\Large h(h^{-1}(2))\] the first task is to compute \(\Large h^{-1}(2)\)
anonymous
  • anonymous
Okay
jim_thompson5910
  • jim_thompson5910
sorry I meant to say \(\Large (h \circ h^{-1})(2)\)
jim_thompson5910
  • jim_thompson5910
anyways, what output do you get when you plug x = 2 into the inverse
anonymous
  • anonymous
0?
jim_thompson5910
  • jim_thompson5910
yep
anonymous
  • anonymous
Oh I see the -1
jim_thompson5910
  • jim_thompson5910
then that 0 is plugged into the h(x) function
jim_thompson5910
  • jim_thompson5910
\[\Large \Large h(\color{red}{h^{-1}(2)}) = h(\color{red}{0})\]
anonymous
  • anonymous
Thank you so much for the walkthrough, it really helped me. :)
jim_thompson5910
  • jim_thompson5910
so you found what h(0) is?
anonymous
  • anonymous
0 right?
jim_thompson5910
  • jim_thompson5910
plug it into 2 - (x/2)
anonymous
  • anonymous
h^-1 (2)= -2(2)+4
jim_thompson5910
  • jim_thompson5910
yeah that part is 0, but we're not done yet
anonymous
  • anonymous
oh, so it is just 2?
jim_thompson5910
  • jim_thompson5910
you take that result and plug it into h(x)
jim_thompson5910
  • jim_thompson5910
ultimately, the final result is 2 you'll find that \[\Large (h \circ h^{-1})(x) = x\] for any x value
anonymous
  • anonymous
2- 0/2=2-0
jim_thompson5910
  • jim_thompson5910
because the inverse essentially "undoes" what the original function does
anonymous
  • anonymous
Oh, I see.

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