SOLUTION: Determine the domain of {{{ f^-1 }}}(x) for each given function f(x).
1. f(x)= 1 + 2cos(pi x + pi), -1 is less than or equal to x is less than or equal to 0
2. f(x) = {{{ 3co
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-> SOLUTION: Determine the domain of {{{ f^-1 }}}(x) for each given function f(x).
1. f(x)= 1 + 2cos(pi x + pi), -1 is less than or equal to x is less than or equal to 0
2. f(x) = {{{ 3co
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Question 417568: Determine the domain of (x) for each given function f(x).
1. f(x)= 1 + 2cos(pi x + pi), -1 is less than or equal to x is less than or equal to 0
2. f(x) = (5x/6)-pi, -infinity is less than x is less than infinity Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Determine the domain of (x) for each given function f(x).
1. f(x)= 1 + 2cos(pi x + pi), -1 <= x <= 0
Interchange x and y to get:
x = 1 + 2cos(pi*y+pi)
Solve for "y":
cos(piy + pi) = (x-1)/2
cos^-1[(x-1)/2)] = pi*y + pi
y = [cos^-1[(x-1)/2)] - pi]/pi
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Domain ?
-1 <= (x-1)/2 <= 1
-2 <= (x-1) <= 2
-1 <= x <= 3
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2. f(x) = (5x/6)-pi, -infinity is less than x is less than infinity
Same procedure.
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Cheers,
Stan H.
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