Question

find (f-1)'(x) for f(x)=cos2x on the interval of (0,pi/2)

Answer 1

I have already found f(-1)'(x)=\[\frac{ -1 }{ 2\sqrt{1-x^2} }\]

Answer 2

yeah the derivative of the inverse. This problem I have though is, what am I supposed to do with the interval?

Answer 3

Originally, we have domain = (0,pi/2) , range = ...?

Answer 4

doesn't say, the only thing they give me is the function and on 0,pi/2

Answer 5

oh wait sorry, we can find the range. The range is 1,-1?

Answer 6

(1, -1), the brackets are important.
For inverse function, the range will be the domain..

Answer 7

ok but im still not sure what I have to be answering here. Am I supposed to plug in the values for x and find the difference like integrals?

Answer 8

No, I don't think you need to..

Answer 9

so why would they give me the interval? do I just find the inverse of the derivative?

Answer 10

Not quite sure...
@hartnn Please give a hand here!!

Answer 11

so would this be all?

Answer 12

ignore that part of the question

Answer 13

the words "on the interval of \([0,\frac{\pi}{2}]\)" are necessary to make the function \(\cos(2x)\) a one to one function
if the interval were larger, it would not be one to one and therefore would not have an inverse

Answer 14

it is like saying "let \(f(x)=\frac{1}{x-1}, x\neq 1\)"
just restricting the domain to make the function well defined

Answer 15

ah ok, so my final answer is correct then?

Answer 16

no

Answer 17

oh, maybe
hold on before i say something stupid

Answer 18

yeah you are right, sorry about that

Answer 19

haha ok thanks so much! It always messes me up when they add something that im not going to use

Answer 20

yeah i know
often people think the extra restrictions are part of the question, when it is really part of the definition of the function
a result of bad notation really

Answer 21

haha ok thanks again! and thanks for your help too @Callisto !

Answer 22

Thanks @satellite73 and thanks for your question as well - good revision :)