## anonymous one year ago 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!!

• This Question is Open
1. jim_thompson5910

is the function this? $\Large h(x) = 2 - \frac{x}{2}$ OR is the function this? $\Large h(x) = \frac{2-x}{2}$

2. anonymous

No, it is..

3. anonymous

|dw:1442458455816:dw|

4. jim_thompson5910

ok so the first one I wrote

5. anonymous

So yes, the first one you stated. Thank you.

6. jim_thompson5910

think of it as $\Large y = 2 - \frac{x}{2}$

7. jim_thompson5910

then swap x and y $\Large x = 2 - \frac{y}{2}$ then solve for y

8. anonymous

How

9. anonymous

what about the h o h ^-1(2)

10. jim_thompson5910

I would first multiply both sides by 2 that clears out the fraction

11. jim_thompson5910

so you'll have 2x = 4 - y isolate y

12. anonymous

-y=2x-4

13. anonymous

then what happens with the h o h ^1 (2). That is my primary concern.

14. jim_thompson5910

-y=2x-4 will turn into y = -2x+4

15. jim_thompson5910

so the inverse is $\Large h^{-1}(x) = -2x+4$

16. jim_thompson5910

$\Large (h \circ h)^{-1}(2)$ is the same as $\Large h(h^{-1}(2))$ the first task is to compute $$\Large h^{-1}(2)$$

17. anonymous

Okay

18. jim_thompson5910

sorry I meant to say $$\Large (h \circ h^{-1})(2)$$

19. jim_thompson5910

anyways, what output do you get when you plug x = 2 into the inverse

20. anonymous

0?

21. jim_thompson5910

yep

22. anonymous

Oh I see the -1

23. jim_thompson5910

then that 0 is plugged into the h(x) function

24. jim_thompson5910

$\Large \Large h(\color{red}{h^{-1}(2)}) = h(\color{red}{0})$

25. anonymous

Thank you so much for the walkthrough, it really helped me. :)

26. jim_thompson5910

so you found what h(0) is?

27. anonymous

0 right?

28. jim_thompson5910

plug it into 2 - (x/2)

29. anonymous

h^-1 (2)= -2(2)+4

30. jim_thompson5910

yeah that part is 0, but we're not done yet

31. anonymous

oh, so it is just 2?

32. jim_thompson5910

you take that result and plug it into h(x)

33. jim_thompson5910

ultimately, the final result is 2 you'll find that $\Large (h \circ h^{-1})(x) = x$ for any x value

34. anonymous

2- 0/2=2-0

35. jim_thompson5910

because the inverse essentially "undoes" what the original function does

36. anonymous

Oh, I see.