Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

What is the inverse of f if….

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

i forgot to put an = sign the first time
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 \[x^3 = {y-2}\] \[x^3+2=y\]
\[x^{3}+2=y^{3}\] is that right?.. i have no idea I'm so stressed
How'd you get y^3? Remember that square roots and squares are inverse operations, right? \[(\sqrt{x})^2 = x\] \[(\sqrt[3]{x})^3 = x\]
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
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 = ??
wait so the answer is f(x) =x^2+2 ?
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)
heres the actual question with the answer choices so you can see
god damn. I kept saying x^2 +2 but I meant x^3 + 2
hahah so is it the 3rd option then?
it is, it is...
thank youuu! i have another question! do you mind helping me??
are you gonna post it here or make a new question? I don't mind
ill post it here! and thanks! (:
uh you'd multiply by the conjugate. pardon my handwriting
1 Attachment
i dont understand...
to simplify radicals you multiply by the conjugate pair which means flipping the sign
why do you have that answer selected?
what do you mean that answer selected?
what are the answer choices?
the ones with the bubble next to them on the link i posted!
rofl I didn't see it scrolled down a bit.
hahahah its all good! (:
so what is the answer?
Perform the operation bibby suggested and then simplify.
you can use wolfram if you don't like learning
its not that i don't like learning its that i am in a hurry and i have no idea whats going on.
\[\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}}\] Do this multiplication here.
Why are you in a hurry? Is this a test?
idk how to..
i had to go out of town so I'm behind on a bunch of work due like really soon
So you don't know how to multiply something like this (a+b)(c+d)?
A lot of Americans like to use this thing called the foil method.
i have heard of it but idk how to really use it
all you did was distribute those right?
(a+b)(c+d)=a(c+d)+b(c+d) =ac+ad+bc+bd
So if you had \[(\sqrt{3}-\sqrt{6})(\sqrt{3}-\sqrt{6})\] what do you have after multiplying?
idk how to group square roots together or multiply them together
Do that foil method then we will work on simplifying
hold on its gonna take awhile to plug in
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
oh wow my computer plugged in the wrong numbers so thats not even the right question, this is the right question...
Well I will show you how to do the first one and you try this one.
i tried solving it and i think its the second answer choice
\[\frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}+\sqrt{6}} \cdot \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{6}}\] 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: \[\frac{\sqrt{3} \sqrt{3}-\sqrt{3} \sqrt{6}-\sqrt{6} \sqrt{3}+\sqrt{6} \sqrt{6}}{(\sqrt{3}+\sqrt{6})(\sqrt{3}-\sqrt{6})}\]
Do you see how I got that top part?
If a, b>=0 then \[\sqrt{a} \cdot \sqrt{b} =\sqrt{a \cdot b} \]
so basically \[\sqrt{3} \cdot \sqrt{3} =3 \text{ and } \sqrt{3} \cdot \sqrt{6}=\sqrt{18}=\sqrt{9 \cdot 2} =\sqrt{9} \cdot \sqrt{2}=3\sqrt{2}\]
combine like terms on top and multiply the bottom out i have to go for now
This has to be the longest problem eve
ever* lol my name is eve:P
hahaha hey eve! and yah i know…i have a lot of stuff i need help withhh ughh
hahah whats up!
not mch jst listening to music i was about to leave lol
hahaha ok then! lol

Not the answer you are looking for?

Search for more explanations.

Ask your own question