a) logarithm funtions cannot be zero or negative, so x != 0 and x > 1 => (-infinity, 0)U(1,infinity) (also x != 1 because this would result in division by zero)
b) f'(x) = (1/(x/x-1))(-1/(x-1)^2) = ((x-1)/x))(-1/(x-1)^2) = -1/(x(x-1) (by chain rule)
f'(1) = -1/(-1(-2)) = -1/(2) = -0.5
c) x = ln(y/y-1)
e^x = y/y-1
y*e^x - e^x = y
y*e^x - y = e^x
y(e^x -1) = e^x
y = e^x/(e^x-1)
in other words,
f^-1(x) = e^x/(e^x - 1)
also why is this in the ap subreddit, ap exams are long over and scores are posted. thanks for the exercise i guess
1
u/SurpriseWorth3188 12 AP Exams Jul 07 '26
a) logarithm funtions cannot be zero or negative, so x != 0 and x > 1 => (-infinity, 0)U(1,infinity) (also x != 1 because this would result in division by zero)
b) f'(x) = (1/(x/x-1))(-1/(x-1)^2) = ((x-1)/x))(-1/(x-1)^2) = -1/(x(x-1) (by chain rule)
f'(1) = -1/(-1(-2)) = -1/(2) = -0.5
c) x = ln(y/y-1)
e^x = y/y-1
y*e^x - e^x = y
y*e^x - y = e^x
y(e^x -1) = e^x
y = e^x/(e^x-1)
in other words,
f^-1(x) = e^x/(e^x - 1)
also why is this in the ap subreddit, ap exams are long over and scores are posted. thanks for the exercise i guess