Mathematics
10 Online
OpenStudy (anonymous):
refresh my memory?
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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
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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?...
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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
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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 :)
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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?
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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|