emma97
What is the inverse of f if….
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emma97
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\[f(3) \sqrt[3]{x-2}\]
bibby
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\[f(x)=\sqrt[3]{x-2}\]
?
emma97
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\[f(3)=\sqrt[3]{x-2}\]
emma97
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i forgot to put an = sign the first time
bibby
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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
\[x^3 = {y-2}\]
\[x^3+2=y\]
emma97
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\[x^{3}+2=y^{3}\] is that right?.. i have no idea I'm so stressed
bibby
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How'd you get y^3?
Remember that square roots and squares are inverse operations, right?
\[(\sqrt{x})^2 = x\]
\[(\sqrt[3]{x})^3 = x\]
emma97
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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
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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 = ??
emma97
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wait so the answer is f(x) =x^2+2 ?
bibby
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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)
emma97
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heres the actual question with the answer choices so you can see
bibby
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god damn. I kept saying x^2 +2 but I meant x^3 + 2
emma97
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hahah so is it the 3rd option then?
bibby
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it is, it is...
emma97
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thank youuu! i have another question! do you mind helping me??
bibby
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are you gonna post it here or make a new question? I don't mind
emma97
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ill post it here! and thanks! (:
emma97
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bibby
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uh you'd multiply by the conjugate. pardon my handwriting
emma97
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i dont understand...
bibby
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to simplify radicals you multiply by the conjugate pair which means flipping the sign
bibby
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why do you have that answer selected?
emma97
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what do you mean that answer selected?
bibby
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what are the answer choices?
emma97
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the ones with the bubble next to them on the link i posted!
bibby
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rofl I didn't see it scrolled down a bit.
emma97
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hahahah its all good! (:
emma97
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so what is the answer?
myininaya
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Perform the operation bibby suggested and then simplify.
emma97
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its not that i don't like learning its that i am in a hurry and i have no idea whats going on.
myininaya
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\[\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}}\]
Do this multiplication here.
myininaya
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Why are you in a hurry? Is this a test?
emma97
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idk how to..
emma97
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i had to go out of town so I'm behind on a bunch of work due like really soon
myininaya
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So you don't know how to multiply something like this (a+b)(c+d)?
myininaya
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A lot of Americans like to use this thing called the foil method.
emma97
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i have heard of it but idk how to really use it
myininaya
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(a+b)(c+d)=ac+ad+bc+bd
emma97
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all you did was distribute those right?
myininaya
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|dw:1388451041982:dw|
myininaya
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yep|dw:1388451059969:dw|
myininaya
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(a+b)(c+d)=a(c+d)+b(c+d)
=ac+ad+bc+bd
myininaya
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So if you had \[(\sqrt{3}-\sqrt{6})(\sqrt{3}-\sqrt{6})\]
what do you have after multiplying?
emma97
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idk how to group square roots together or multiply them together
myininaya
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Do that foil method then we will work on simplifying
emma97
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hold on its gonna take awhile to plug in
myininaya
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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
emma97
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oh wow my computer plugged in the wrong numbers so thats not even the right question, this is the right question...
myininaya
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Well I will show you how to do the first one and you try this one.
emma97
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i tried solving it and i think its the second answer choice
myininaya
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\[\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}}\]
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
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Do you see how I got that top part?
emma97
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yes
myininaya
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If a, b>=0 then \[\sqrt{a} \cdot \sqrt{b} =\sqrt{a \cdot b} \]
myininaya
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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
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combine like terms on top
and multiply the bottom out
i have to go for now
emma97
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ok.
Chooch146
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This has to be the longest problem eve
Chooch146
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ever*
lol my name is eve:P
emma97
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hahaha hey eve! and yah i know…i have a lot of stuff i need help withhh ughh
Chooch146
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Hai!
emma97
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hahah whats up!
Chooch146
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not mch jst listening to music i was about to leave lol
emma97
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hahaha ok then! lol
Chooch146
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bye