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Inverse Functions

Determine one-on-one functions and thier inverse

Determine if the funtion is a one-on-one function. If so, then list the inverse function of the coordinates by switching the coordinates: f={(3,2),(8,6),(2,4),(16,18)} Since each x-value cooresponds to a unique y value, and each y-value corresponds to a unique x value, it is a one-on-one function. To find the inverse of the function, simply switch the coordinates of the ordered pairs: f^-1={(2,3),(6,8),(4,2),(16,17)}
Determine whether the following function is a one-on-one function. If so, find the inverse by switching inputs and outputs: Month of 2013: (input) Sept. Oct. Nov. Dec. % rate of smokers in MI (output): 45, 45, 40, 42 The function is not a one-on-one function. Each output of the table DOES NOT correspond to a unique input. The output 45 corresponds to both inputs Sept. and Oct. Due to the fact that this is not a one-on-one function, there is no inverse.
Given the one-on-one function F(x)=x^3+8, find the following: a. f(-4) b. f^-1(-56) For f(-4), substitute x=-4 into the original equation: a. f(x)=x^3+8=(-4)^3+8=-56 b. f^-1(-56)=-4 (you do not need to necessarily have to solve for f^-1)
Find the inverse of the one-on-one function: f(x)=x-6/4 First, replace f(x) with y so that the function reads Y=X-6/4. Next, interchange x and y so that the function now reads x=y-6/4. Now solve for y: y=4x+6 so f^-1(x)=4x+6
Find the inverse of the one-on-one function: f(x)=7/4x+3 f^-1(x)=7-3x/4x (Don't forget to flip the sign from positive to a negative)
Find the inverse of the one-on-one function: f(x)=x-7/13 Replace f(x) with y so that y=x-7/13. Then interchange x and y so that the function now reads x=y-7/13. Solve for y to get your final answer: f^-1(x)=13x+7
Applying the horizontal line test on the graph of a function serves what purpose? To determine if the graph is a function of a one-on-one function. If the horizontal line (horizontal line test) intersects the graph of a function only one time, then the function is in fact a one-on-one function.
Created by: Antczak123