Question

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!!

Answer 1

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

OR

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

Answer 2

No, it is..

Answer 3

ok so the first one I wrote

Answer 4

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

Answer 5

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

Answer 6

then swap x and y

\[\Large x = 2 - \frac{y}{2}\]

then solve for y

Answer 7

How

Answer 8

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

Answer 9

I would first multiply both sides by 2

that clears out the fraction

Answer 10

so you'll have 2x = 4 - y

isolate y

Answer 11

-y=2x-4

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

\[\Large (h \circ h)^{-1}(2)\]

is the same as

\[\Large h(h^{-1}(2))\]

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

Answer 16

Okay

Answer 17

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

Answer 18

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

Answer 19

0?

Answer 20

yep

Answer 21

Oh I see the -1

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

so you found what h(0) is?

Answer 26

0 right?

Answer 27

plug it into 2 - (x/2)

Answer 28

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

Answer 29

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

Answer 30

oh, so it is just 2?