- anonymous

What is the inverse of f if….

- chestercat

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- anonymous

\[f(3) \sqrt[3]{x-2}\]

- bibby

\[f(x)=\sqrt[3]{x-2}\]
?

- anonymous

\[f(3)=\sqrt[3]{x-2}\]

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## More answers

- anonymous

i forgot to put an = sign the first time

- 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

\[x^{3}+2=y^{3}\] is that right?.. i have no idea I'm so stressed

- 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

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

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

wait so the answer is f(x) =x^2+2 ?

- 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

heres the actual question with the answer choices so you can see

##### 1 Attachment

- bibby

god damn. I kept saying x^2 +2 but I meant x^3 + 2

- anonymous

hahah so is it the 3rd option then?

- bibby

it is, it is...

- anonymous

thank youuu! i have another question! do you mind helping me??

- bibby

are you gonna post it here or make a new question? I don't mind

- anonymous

ill post it here! and thanks! (:

- anonymous

##### 1 Attachment

- bibby

uh you'd multiply by the conjugate. pardon my handwriting

##### 1 Attachment

- anonymous

i dont understand...

- bibby

to simplify radicals you multiply by the conjugate pair which means flipping the sign

- bibby

why do you have that answer selected?

- anonymous

what do you mean that answer selected?

- bibby

what are the answer choices?

- anonymous

the ones with the bubble next to them on the link i posted!

- bibby

rofl I didn't see it scrolled down a bit.

- anonymous

hahahah its all good! (:

- anonymous

so what is the answer?

- myininaya

Perform the operation bibby suggested and then simplify.

- 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

its not that i don't like learning its that i am in a hurry and i have no idea whats going on.

- 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

Why are you in a hurry? Is this a test?

- anonymous

idk how to..

- anonymous

i had to go out of town so I'm behind on a bunch of work due like really soon

- myininaya

So you don't know how to multiply something like this (a+b)(c+d)?

- myininaya

A lot of Americans like to use this thing called the foil method.

- anonymous

i have heard of it but idk how to really use it

- myininaya

(a+b)(c+d)=ac+ad+bc+bd

- anonymous

all you did was distribute those right?

- myininaya

|dw:1388451041982:dw|

- myininaya

yep|dw:1388451059969:dw|

- myininaya

(a+b)(c+d)=a(c+d)+b(c+d)
=ac+ad+bc+bd

- myininaya

So if you had \[(\sqrt{3}-\sqrt{6})(\sqrt{3}-\sqrt{6})\]
what do you have after multiplying?

- anonymous

idk how to group square roots together or multiply them together

- myininaya

Do that foil method then we will work on simplifying

- anonymous

hold on its gonna take awhile to plug in

- 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

oh wow my computer plugged in the wrong numbers so thats not even the right question, this is the right question...

##### 1 Attachment

- myininaya

Well I will show you how to do the first one and you try this one.

- anonymous

i tried solving it and i think its the second answer choice

- 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

Do you see how I got that top part?

- anonymous

yes

- myininaya

If a, b>=0 then \[\sqrt{a} \cdot \sqrt{b} =\sqrt{a \cdot b} \]

- 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

combine like terms on top
and multiply the bottom out
i have to go for now

- anonymous

ok.

- anonymous

This has to be the longest problem eve

- anonymous

ever*
lol my name is eve:P

- anonymous

hahaha hey eve! and yah i know…i have a lot of stuff i need help withhh ughh

- anonymous

Hai!

- anonymous

hahah whats up!

- anonymous

not mch jst listening to music i was about to leave lol

- anonymous

hahaha ok then! lol

- anonymous

bye

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