Here's the question you clicked on:
JenniferSmart1
I found the inverse of \[f(x)=-2cos(3x)\] \[f^{-1}(x)=\frac 13 cos^{-1}(-\frac x 2)\] how do I find the domain? Is it all real numbers?
Domain of... f(x) or f^(-1) (x)?
i Mean \[f^{-1}\]
Try \(f^{-1} (100)\), what would you get?
it's an imaginary number according wolfram
I suppose the function should give real output if you input a real number. Okay, when you put x= 100 into f^(-1) (x), it doesn't give you real output. So, it's not in the domain of y. Consider cosine function, range of cosine function is -1 ≤ cos x ≤ 1. Consider the inverse of cosine function, domain of the inverse is the range of the cosine function, so the domain of cos^(-1)x is [-1, 1]. Now, you have cos^(-1) (x/2) in your inverse function, hmm.. how would you get the domain of the inverse function?
Let's see if I understand this... \[f^{-1}(f(x))=sin^{-1}(sinx)=x \;\;\;\;where \;\;\;\;-\frac{\pi}{2}\le x\le\frac{\pi}{2}\] \[f(f^{-1}(x))=sin(sin^{-1}x)=x \;\;\;\;where\;\;\;\;-1\le x\le1\]
|dw:1360683713662:dw|
I"m trying to picture this. What's bothering me is that fact the domain and range of one function change to the range and domain of the next function....I'm trying to keep them in order somehow
the cosine gives numbers between -1 and 1 there is no real number x, where cos(x)= 2 (for example) so x = acos(2) has no real solution. You would want to restrict the domain to -1 to +1 for the case of acos(x/2) -1 ≤ x/2 ≤ 1 or -2 ≤ x ≤ 2
but the restriction of the inverse of cosine is \[0\le x \le \pi\] but since it's x/2 do I multiply or divided that by two? or am i totally on the wrong path...
you mean the restriction on the cosine is 0 to pi the restriction on the domain of the inverse cos(x) is -1 ≤ x ≤ 1
the restriction on the domain of the inverse cos(x) is -1 ≤ x ≤ 1 if you are given acos(x/2) then rename the variable to y= x/2 you know the domain is restricted to -1 ≤ y ≤ 1 but with y= x/2 -1 ≤ x/2 ≤ 1 solve for x, (2 separate relations), to get -2 ≤ x ≤ 2 that is the domain in terms of x for acos(x/2)
I find gauss's law easier to understand than this for whatever reason. I'm very visual and I can't visualize this....I would much rather draw unit circles and sine graphs, and modify that to represent the current problem that we're working on.
I'll look at it again later...too tired to comprehend.