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Loujoelou
Group Title
Let f(x) = x^2 – 81. Find f–1(x).
Can someone doublecheck what I have?
 one year ago
 one year ago
Loujoelou Group Title
Let f(x) = x^2 – 81. Find f–1(x). Can someone doublecheck what I have?
 one year ago
 one year ago

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hba Group TitleBest ResponseYou've already chosen the best response.0
Show me what ya got :)
 one year ago

Loujoelou Group TitleBest ResponseYou've already chosen the best response.0
Okay so I know first to substitute f(x) with y and then to reverse the x & y variables which would give me x=y^281. Then I use a root on both sides and get \[\pm x=y9\] and finally I add 9 on both sides to get \[\pm 9 \sqrt{x} = y\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
how u got y 9 ??
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Interesting :)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
but incorrect :P
 one year ago

Loujoelou Group TitleBest ResponseYou've already chosen the best response.0
well we have y^281, idk factoring it would be (y+9)(y9) but am I supposed to do that?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
no, what you did was \(\sqrt{y^281}=y9\) that is incorrect. whats to be done : let y= x^281 add 81 to both sides. then take square root of both sides...
 one year ago

Loujoelou Group TitleBest ResponseYou've already chosen the best response.0
oh okay :) so it'd be x+81= y^2 and once we square root both sides would the answer come out to \[\pm 9 \sqrt{x}=y\]
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
@hartnn Bache mai halka haat rakho :)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
taking square root on both sides of \(x+81=y^2\) will give you \(\sqrt{x+81}=\sqrt{y^2}=y\) to get the inverse function as \(\sqrt{x+81}\) got this ? hba, i didn't get you.
 one year ago

Loujoelou Group TitleBest ResponseYou've already chosen the best response.0
okay I get it :) thx a ton! :)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
welcome ^_^
 one year ago
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