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JenniferSmart1
Group Title
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?
 one year ago
 one year ago
JenniferSmart1 Group Title
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?
 one year ago
 one year ago

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JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
@satellite73
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
@Callisto
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Domain of... f(x) or f^(1) (x)?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
of \[f'(x)\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
i Mean \[f^{1}\]
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Try \(f^{1} (100)\), what would you get?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
it's an imaginary number according wolfram
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
Let's see if I understand this... \[f^{1}(f(x))=sin^{1}(sinx)=x \;\;\;\;where \;\;\;\;\frac{\pi}{2}\le x\le\frac{\pi}{2}\] \[f(f^{1}(x))=sin(sin^{1}x)=x \;\;\;\;where\;\;\;\;1\le x\le1\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
dw:1360683713662:dw
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
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...
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
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)
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I'll look at it again later...too tired to comprehend.
 one year ago
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