Question

i need to:
Find f^-1(x) if f(x) =3sin(2x)cos(x)

Answer 1

There's no inverse

Answer 2

x = 3 (2 sin(x) cos(y)) cos(y), and express in terms of cos(y) only. You will then have a cubic equation in sin(y):
\[\sin ^{3}x - \sin x + y/6 = 0,\]
written in standard form. Solve for sin y, then take the arcsine of the result.
The exact solution will be complicated!

Answer 3

It is
\[ f ^{-1}(x) = \sin ^{-1}(\sqrt[3]{-y/12 + (1/2) \sqrt{-4 +y ^{2}/36}}\]
\[+\sqrt[3]{-y/12- (1/2)\sqrt{-4+y ^{2}/36}}).\]

Answer 4

We need to determine its domain.

Answer 5

BTW, those y's are supposed to be x's.

Answer 6

I though the original function failed the horizontal line test

Answer 7

How can the inverse exist if that's the case, lol

Answer 8

You need to restrict the domain of the original function for it to have a valid inverse.

Answer 9

It will be a very small domain

Answer 10

Graph the original function and look for intervals where the function is strictly increasing and where it's strictly decreasing.

Answer 11

like not enough for it to be even considered the graph of the original function, I'm sorry, I'm going with no inverse

Answer 12

You'd have to say that one domain is from -.615 to .615 and that's just one domain, you have to list all of them

Answer 13

good luck with that

Answer 14

I agree that the problem is quite complicated. It is what it is, and not all problems will be easy.

Answer 15

Here's another wrinkle:
Since the equation is cubic in sin x, there are two other roots. It would be nice if they were complex so we could disregard them. Are they?

Answer 16

There is a lot to this problem.

Answer 17

I think the question was meant to be graphed first and checked to see if it passes the horizontal line test

Answer 18

If so, then I believe we are to consider all real numbers to be the domain of the original function.

Answer 19

You would then be correct.

Answer 20

I don't see why we wouldn't....It didn't give us a restricted domain to begin with.

Answer 21

True.