anonymous
  • anonymous
What is the inverse of f if….
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[f(3) \sqrt[3]{x-2}\]
bibby
  • bibby
\[f(x)=\sqrt[3]{x-2}\] ?
anonymous
  • anonymous
\[f(3)=\sqrt[3]{x-2}\]

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anonymous
  • anonymous
i forgot to put an = sign the first time
bibby
  • bibby
are we finding f'(3) either way to get the inverse, let's replace f(x) with y \[y = \sqrt[3]{x-2}\] switch x and y \[x = \sqrt[3]{y-2}\] cube both sides \[x^3 = {y-2}\] \[x^3+2=y\]
anonymous
  • anonymous
\[x^{3}+2=y^{3}\] is that right?.. i have no idea I'm so stressed
bibby
  • bibby
How'd you get y^3? Remember that square roots and squares are inverse operations, right? \[(\sqrt{x})^2 = x\] \[(\sqrt[3]{x})^3 = x\]
anonymous
  • anonymous
oh i got confused i thought you said for ME to cube both sides lol. I'm sorry I'm extremely tired and have to get a bunch of work done by like tomorrow. ughhh. my bad
bibby
  • bibby
lol no worries, just always go over your work a bazillion times so that you can prove to yourself that your answer is right. ANYHOW, the inverse is now f(x) =x^2+2 f(3) = 3^2+2 = ??
anonymous
  • anonymous
wait so the answer is f(x) =x^2+2 ?
bibby
  • bibby
whoops./ you generally say inverse with a little dash f'(x) so the equation for the inverse is f'(x) = x^2 + 2 then you plug in 3 into the inverse equation (assuming that's the question)
anonymous
  • anonymous
heres the actual question with the answer choices so you can see
bibby
  • bibby
god damn. I kept saying x^2 +2 but I meant x^3 + 2
anonymous
  • anonymous
hahah so is it the 3rd option then?
bibby
  • bibby
it is, it is...
anonymous
  • anonymous
thank youuu! i have another question! do you mind helping me??
bibby
  • bibby
are you gonna post it here or make a new question? I don't mind
anonymous
  • anonymous
ill post it here! and thanks! (:
anonymous
  • anonymous
bibby
  • bibby
uh you'd multiply by the conjugate. pardon my handwriting
1 Attachment
anonymous
  • anonymous
i dont understand...
bibby
  • bibby
to simplify radicals you multiply by the conjugate pair which means flipping the sign
bibby
  • bibby
why do you have that answer selected?
anonymous
  • anonymous
what do you mean that answer selected?
bibby
  • bibby
what are the answer choices?
anonymous
  • anonymous
the ones with the bubble next to them on the link i posted!
bibby
  • bibby
rofl I didn't see it scrolled down a bit.
anonymous
  • anonymous
hahahah its all good! (:
anonymous
  • anonymous
so what is the answer?
myininaya
  • myininaya
Perform the operation bibby suggested and then simplify.
bibby
  • bibby
you can use wolfram if you don't like learning http://www.wolframalpha.com/input/?i=%28sqrt%283%29-sqrt%286%29%29%2F%28sqrt%283%29%2Bsqrt%286%29%29
anonymous
  • anonymous
its not that i don't like learning its that i am in a hurry and i have no idea whats going on.
myininaya
  • myininaya
\[\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}}\] Do this multiplication here.
myininaya
  • myininaya
Why are you in a hurry? Is this a test?
anonymous
  • anonymous
idk how to..
anonymous
  • anonymous
i had to go out of town so I'm behind on a bunch of work due like really soon
myininaya
  • myininaya
So you don't know how to multiply something like this (a+b)(c+d)?
myininaya
  • myininaya
A lot of Americans like to use this thing called the foil method.
anonymous
  • anonymous
i have heard of it but idk how to really use it
myininaya
  • myininaya
(a+b)(c+d)=ac+ad+bc+bd
anonymous
  • anonymous
all you did was distribute those right?
myininaya
  • myininaya
|dw:1388451041982:dw|
myininaya
  • myininaya
yep|dw:1388451059969:dw|
myininaya
  • myininaya
(a+b)(c+d)=a(c+d)+b(c+d) =ac+ad+bc+bd
myininaya
  • myininaya
So if you had \[(\sqrt{3}-\sqrt{6})(\sqrt{3}-\sqrt{6})\] what do you have after multiplying?
anonymous
  • anonymous
idk how to group square roots together or multiply them together
myininaya
  • myininaya
Do that foil method then we will work on simplifying
anonymous
  • anonymous
hold on its gonna take awhile to plug in
myininaya
  • myininaya
You don't have to use what I gave you as a formula. Just distribute. Take each of the terms in the first grouping and multiply to each term in the second grouping
anonymous
  • anonymous
oh wow my computer plugged in the wrong numbers so thats not even the right question, this is the right question...
myininaya
  • myininaya
Well I will show you how to do the first one and you try this one.
anonymous
  • anonymous
i tried solving it and i think its the second answer choice
myininaya
  • myininaya
\[\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}}\] This was the operation you were asked to perform. I was having you look at the numerator first. So looking at the numerator if we multiply the top out we would get: \[\frac{\sqrt{3} \sqrt{3}-\sqrt{3} \sqrt{6}-\sqrt{6} \sqrt{3}+\sqrt{6} \sqrt{6}}{(\sqrt{3}+\sqrt{6})(\sqrt{3}-\sqrt{6})}\]
myininaya
  • myininaya
Do you see how I got that top part?
anonymous
  • anonymous
yes
myininaya
  • myininaya
If a, b>=0 then \[\sqrt{a} \cdot \sqrt{b} =\sqrt{a \cdot b} \]
myininaya
  • myininaya
so basically \[\sqrt{3} \cdot \sqrt{3} =3 \text{ and } \sqrt{3} \cdot \sqrt{6}=\sqrt{18}=\sqrt{9 \cdot 2} =\sqrt{9} \cdot \sqrt{2}=3\sqrt{2}\]
myininaya
  • myininaya
combine like terms on top and multiply the bottom out i have to go for now
anonymous
  • anonymous
ok.
anonymous
  • anonymous
This has to be the longest problem eve
anonymous
  • anonymous
ever* lol my name is eve:P
anonymous
  • anonymous
hahaha hey eve! and yah i know…i have a lot of stuff i need help withhh ughh
anonymous
  • anonymous
Hai!
anonymous
  • anonymous
hahah whats up!
anonymous
  • anonymous
not mch jst listening to music i was about to leave lol
anonymous
  • anonymous
hahaha ok then! lol
anonymous
  • anonymous
bye

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