You have given me an equation, not a function
The concept of domain and range applies to functions
was your function
f(x) = 3cos(3x+π) - 4 ?
if so , the domain is any real number
the range is -1 to -7
were you looking for a solution to your equation?
cos(3x+π) = 4/3
BUT, the cosine of anything lies between -1 and 1
so your equation has no solution.
How would I go about finding the domain and range of 3cos(3x+π) = 4? Thanks
2 answers
The question(s) were as follows:
Given: f(x) = 3cos(3x+π) - 4 (as you thought)
Find:
a) the inverse of F(x)
b) the domain and range of F(x)
c) the domain and range of F -1(x)
I got cos-1[(x-4)/9] - π =y for the inverse, although I'm not sure I did the problem right (I set x to y and y to x and then solved for y).
For cos(3x+π) = 4/3, don't I have to finish solving cos(3x+π) before I decide that there is no solution? I have been doing these and other problems for so long today that I am gettin more confused - not less, lol! Thanks for your help.
Given: f(x) = 3cos(3x+π) - 4 (as you thought)
Find:
a) the inverse of F(x)
b) the domain and range of F(x)
c) the domain and range of F -1(x)
I got cos-1[(x-4)/9] - π =y for the inverse, although I'm not sure I did the problem right (I set x to y and y to x and then solved for y).
For cos(3x+π) = 4/3, don't I have to finish solving cos(3x+π) before I decide that there is no solution? I have been doing these and other problems for so long today that I am gettin more confused - not less, lol! Thanks for your help.