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bibbyBest ResponseYou've already chosen the best response.3
\[f(x)=\sqrt[3]{x2}\] ?
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
i forgot to put an = sign the first time
 3 months ago

bibbyBest ResponseYou've already chosen the best response.3
are we finding f'(3) either way to get the inverse, let's replace f(x) with y \[y = \sqrt[3]{x2}\] switch x and y \[x = \sqrt[3]{y2}\] cube both sides \[x^3 = {y2}\] \[x^3+2=y\]
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
\[x^{3}+2=y^{3}\] is that right?.. i have no idea I'm so stressed
 3 months ago

bibbyBest ResponseYou've already chosen the best response.3
How'd you get y^3? Remember that square roots and squares are inverse operations, right? \[(\sqrt{x})^2 = x\] \[(\sqrt[3]{x})^3 = x\]
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
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
 3 months ago

bibbyBest ResponseYou've already chosen the best response.3
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 = ??
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
wait so the answer is f(x) =x^2+2 ?
 3 months ago

bibbyBest ResponseYou've already chosen the best response.3
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)
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
heres the actual question with the answer choices so you can see
 3 months ago

bibbyBest ResponseYou've already chosen the best response.3
god damn. I kept saying x^2 +2 but I meant x^3 + 2
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
hahah so is it the 3rd option then?
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
thank youuu! i have another question! do you mind helping me??
 3 months ago

bibbyBest ResponseYou've already chosen the best response.3
are you gonna post it here or make a new question? I don't mind
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
ill post it here! and thanks! (:
 3 months ago

bibbyBest ResponseYou've already chosen the best response.3
uh you'd multiply by the conjugate. pardon my handwriting
 3 months ago

bibbyBest ResponseYou've already chosen the best response.3
to simplify radicals you multiply by the conjugate pair which means flipping the sign
 3 months ago

bibbyBest ResponseYou've already chosen the best response.3
why do you have that answer selected?
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
what do you mean that answer selected?
 3 months ago

bibbyBest ResponseYou've already chosen the best response.3
what are the answer choices?
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
the ones with the bubble next to them on the link i posted!
 3 months ago

bibbyBest ResponseYou've already chosen the best response.3
rofl I didn't see it scrolled down a bit.
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
hahahah its all good! (:
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
Perform the operation bibby suggested and then simplify.
 3 months ago

bibbyBest ResponseYou've already chosen the best response.3
you can use wolfram if you don't like learning http://www.wolframalpha.com/input/?i=%28sqrt%283%29sqrt%286%29%29%2F%28sqrt%283%29%2Bsqrt%286%29%29
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
its not that i don't like learning its that i am in a hurry and i have no idea whats going on.
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
\[\frac{\sqrt{3}\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}\sqrt{6}}{\sqrt{3}\sqrt{6}}\] Do this multiplication here.
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
Why are you in a hurry? Is this a test?
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
i had to go out of town so I'm behind on a bunch of work due like really soon
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
So you don't know how to multiply something like this (a+b)(c+d)?
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
A lot of Americans like to use this thing called the foil method.
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
i have heard of it but idk how to really use it
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
(a+b)(c+d)=ac+ad+bc+bd
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
all you did was distribute those right?
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
dw:1388451041982:dw
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
yepdw:1388451059969:dw
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
(a+b)(c+d)=a(c+d)+b(c+d) =ac+ad+bc+bd
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
So if you had \[(\sqrt{3}\sqrt{6})(\sqrt{3}\sqrt{6})\] what do you have after multiplying?
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
idk how to group square roots together or multiply them together
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
Do that foil method then we will work on simplifying
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
hold on its gonna take awhile to plug in
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
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
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
oh wow my computer plugged in the wrong numbers so thats not even the right question, this is the right question...
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
Well I will show you how to do the first one and you try this one.
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
i tried solving it and i think its the second answer choice
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
\[\frac{\sqrt{3}\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}\sqrt{6}}{\sqrt{3}\sqrt{6}}\] 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: \[\frac{\sqrt{3} \sqrt{3}\sqrt{3} \sqrt{6}\sqrt{6} \sqrt{3}+\sqrt{6} \sqrt{6}}{(\sqrt{3}+\sqrt{6})(\sqrt{3}\sqrt{6})}\]
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
Do you see how I got that top part?
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
If a, b>=0 then \[\sqrt{a} \cdot \sqrt{b} =\sqrt{a \cdot b} \]
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
so basically \[\sqrt{3} \cdot \sqrt{3} =3 \text{ and } \sqrt{3} \cdot \sqrt{6}=\sqrt{18}=\sqrt{9 \cdot 2} =\sqrt{9} \cdot \sqrt{2}=3\sqrt{2}\]
 3 months ago

myininayaBest ResponseYou've already chosen the best response.1
combine like terms on top and multiply the bottom out i have to go for now
 3 months ago

Chooch146Best ResponseYou've already chosen the best response.1
This has to be the longest problem eve
 3 months ago

Chooch146Best ResponseYou've already chosen the best response.1
ever* lol my name is eve:P
 3 months ago

emma97Best ResponseYou've already chosen the best response.1
hahaha hey eve! and yah i know…i have a lot of stuff i need help withhh ughh
 3 months ago

Chooch146Best ResponseYou've already chosen the best response.1
not mch jst listening to music i was about to leave lol
 3 months ago
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