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Mathematics 10 Online
OpenStudy (anonymous):

refresh my memory?

OpenStudy (anonymous):

OpenStudy (kc_kennylau):

(x,y) represents that g(x)=y

OpenStudy (kc_kennylau):

-1 means inverse

OpenStudy (kc_kennylau):

For example, \(g(4)=-5\), so \(g^{-1}(-5)=4\)

OpenStudy (anonymous):

so g^-1 (6) = 9

OpenStudy (kc_kennylau):

well done :)

OpenStudy (anonymous):

would h^-1 be : -2x-9 ?

OpenStudy (kc_kennylau):

nope :)

OpenStudy (kc_kennylau):

Try to make x the subject of \(h(x)=2x+9\) :)

OpenStudy (anonymous):

x = 9/2?...

OpenStudy (kc_kennylau):

ok in other words make \(x\) the subject of \(y=2x+9\) :)

OpenStudy (anonymous):

subject?

OpenStudy (kc_kennylau):

that means \(x=...\)

OpenStudy (anonymous):

isn't it x = 9/2?

OpenStudy (anonymous):

-9/2

OpenStudy (kc_kennylau):

almost...

OpenStudy (kc_kennylau):

hint: the inverse of \(j(x)=18x+67\) would be \(x=\dfrac{j(x)-67}{18}\) or \(j^{-1}(x)=\dfrac{x-67}{18}\) (The middle one is the step)

OpenStudy (kc_kennylau):

(I know your book doesn't teach you this way)

OpenStudy (anonymous):

sorry xD .. its 2:31 am... okay so its (x-9)/2 ?

OpenStudy (kc_kennylau):

yep :)

OpenStudy (kc_kennylau):

Do you need to know the presentation?

OpenStudy (kc_kennylau):

Like are you preparing this for an exam that's not in the form of MC

OpenStudy (anonymous):

no so h^-1 would be equal to = (x-9)/2

OpenStudy (kc_kennylau):

yes

OpenStudy (anonymous):

what would be the other form?

OpenStudy (kc_kennylau):

No the other form is just an intermediate form so it's useless and will not be tested and is actually extra

OpenStudy (anonymous):

ok

OpenStudy (kc_kennylau):

Here's what I do in my head:|dw:1386498967203:dw|

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