How do you find the domain of the function f and of its inverse function f-1? f(x) = cos(x - 2) + 7
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Domain = "all valid inputs for x". Or to say it another way "what values of x will NOT give you 1/0 or square root of a negative or a log of a number less than or equal to 0?"
the domain is all real numbers. this function is just about as not one to one as you can get, so it does not have an inverse.
if you choose you can restrict the domain of cosine to say \[(0,\pi)\] and then it will be one to one and have an inverse. perhaps that is what they are asking you to do, to restrict the domain to give an inverse
thank you both. I didnt know it was possible for a function not to have an inverse.
in which case you would say "restrict the domain to \[(2,\pi+2)\] and then this thing would be one to one
Have you tried graphing the function?
i dont really know how to graph it
sure. if the function is not one to one then it does not have a function for an inverse. in some cases you can restrict the domain of your function to make it one to one. for example \[f(x)=cos(x)\] is certainly not one to one but if you restrict it for \[(0,\pi)\] then it is.
TO GRAPH ANYTHING 1) go to http://www.wolframalpha.com 2) type " plot y = cos(x-2) + 7 There it is. :)
if you do that in your case here, your function says :subtract 2, take the cosine, then add 7. your inverse would say to do the opposite things in the opposite order : subtract 7, take the inverse cosine, add 2.
Thank you mathteacher1729 and satellite73!!!!! a lot! I'll try doing the other problem like this and if i need help I'll let u know
i will write out a possible inverse for you if you like. it would just do exactly the opposite of what this one does, as i wrote in english above
\[f^{-1}(x)=cos^{-1}(x-7)+2\]
the domain of your original function would be \[(2, \pi+2)\] and the range would be [6,8] making the domain of your inverse function [6,8] and your range [2, pi+2)
okay and then the inverse would be (-2, pi -2)?
I'm sorry..I'm slow in math
hello?
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