Math Inverse Relations and Functions

6.4 Inverse Relations and Functions Relation: A mapping, or pairing of input values (set of all x values) with output values (set of all y values). Any set of ordered pairs. Inverse Relation: The new relation that results after reversing the input and output values of a given relation. Function: A mapping, or pairing of input values with output values such that every input has exactly one output. We test that it is a function by performing the Vertical Line Test. If a vertical line passes through exactly one point, then it is a function. Inverse Functions: Given two functions f and g and their equations, we can check to see if they are inverses of each other by performing the following compositions: f(g(x)) and g(f(x)). If both compositions equal x, then f and g are inverses of each other. We check whether an equation is a function using the Vertical Line Test. We can also check whether its inverse equation is a function by doing the Horizontal Line Test. Its inverse will be a function if a horizontal line that you draw passes through the original equation in exactly one point. STEPS TO FINDING THE EQUATION FOR THE INVERSE OF A RELATION: 1. Replace g(x) with y in the original equation. Find the inverse of 2. Interchange x and y. x = 3y + 1 3. Solve the equation for y. (Subtract 1 on both sides and divide both sides by 3) x = 3y + 1 g ( x) = 3x + 1 y = 3x + 1 −1 −1 x −1 = 3y x −1 = 3y 3 4. Replace y with g −1 ( x ) 3 x −1 y= 3 1 1 x− 3 3 Just because we have an equation for an inverse, does not make it an inverse function. This we have to test using either the compositions or the Horizontal Line Test. or g −1 ( x) = Graph the original. Determine if the original is a function and if its inverse is a function. Find the equation for the inverse relation and graph it as well. a. 1 y = − x+5 2 b. f ( x) = x 2 + 1 (#’s 29, 31, 34) Is the original relation a function? YES NO Is the original relation a function? YES NO Is the inverse a function? NO Is the inverse a function? NO YES Find equation for the inverse: (#’s 4-12 even) Find equation for the inverse: Find the equation for the inverse: (#’s 38, 39) a. f ( x) = 1 3 x +3 4 b. 2 f ( x) = − x 6 + 8 , x ≤ 0 5 YES Verifying Two Functions are Inverses Two functions f and g are inverse functions if and only if both of their compositions are the identity function (= x). You must show both of the following are true: f ( g (x )) = x and g ( f ( x )) = x f ( x) = 7 x − 4 a. 1 4 g ( x) = x + 7 7 h( x ) = b. 1 5 x 6 g ( x) = (6 x ) 5 1 (#’s 15-19 odd) 6.6 SOLVE RADICAL EQUATIONS An equation that has a radical and variables in the radicand. To solve this type of equation, raise each RADICAL EQUATIONS side of the equation to a power equal to the index of the radical to eliminate the radical. Or, if the power is a fraction, raise each side to its inverse in order to cancel that fraction. After solving for the variable, this is a solution that does not make a true statement when plugged into EXTRANEOUS XTRANEOUS SOLUTIONS the original problem. Because of this, it is not considered an answer to the problem. To understand how to solve these equations, consider the following questions: What happens when you square a square root? Cube a cube root? Rely on what you know about rational exponents and the index to help you. ( x) 2 ( x+6 ) 2 ( y) 3 3 ( 4x) 3 3  25  x      5 2 Solve x +1 + 2 = 4 x +1 + 2 = 4 Goal: Since our variable is stuck underneath the radical, our goal is to get the radical alone, remove the radical that is in the way, and then solve for the variable. Steps: 1. Get ONE radical isolated (alone) on one side of the equal sign. -2 -2 ( x +1 = 2 ) 2 x + 1 = ( 2) 2 x +1 = 4 -1 -1 2. Check to make sure that the problem is possible. Remember that the answer to a square root or any even root cannot be a negative number, but the answer to a cube root and any odd root can be a positive or negative number. 3. Remove the radical by squaring, cubing, etc each side of the equal sign. **Be careful when you square each side. Sometimes you may need to FOIL** See #2 below. x =3 4. Get the variable alone. 5. Check your answer(s) by plugging it/them back into the original. Make sure you do not have an extraneous solution. − 23 2 x + 5 + 7 = 15 1. 2. 2x + 7 = x + 2 4. 1 1 (5 x + 1)3 + 5 = 4 2 2 − 23 2 x + 5 = 8 2 x + 5 = −4 3 4 − 2 x 3 − 21= − 53 4 3. − 2 x 3 = −32 4 x 3 = 16 5. 6. 3 3x − 3 x + 6 = 0 3x + 4 − 1= x + 5