Math- Lesson 8-5 Unit 4 review: a) Compositions o unctions b) Linear combinations o unctions c) Inverse Functions d) Quadratic Inequalities e) Rational Inequalities
1. Is the ollowing relation a unction (-, 5), (5, 6), (-, 6), (7, 6) No. Input value - has two output values.. Is the ollowing relation a unction Does the graph o the relation pass the vertical line test Yes. Each input value has eactly one output value.
Compositions o Functions () = () = Means: wherever you see an in the unction, replace it with a. 1. Replace the with a set o parentheses. () = ( ). Put the input value into the parentheses. () = (). Find the output value. () = 6
Your turn: 1 ( ) 1 ( 9) ( 4) 0 ( )
Function Notation () = + 1 ( 1) = 1 () = 6 + 1 (rule) (Input) (output) + 1 () () + 1 5 () = 5 () + 1 7 () = 7 1 ( 1) + 1 1 () + 1 6 + 1 I your input is an epression instead o a number: replace with parentheses and plug in the epression.
Function composition 1 g ( ) What does this mean Substitute in or in the unction (). ( g( )) What does this mean Substitute g() in or in the unction (). ( g( )) ( g( )) 1 Which means the same as ( ) ( ) 1 4 1
Compositions o Functions () = and g Let s use () as the input to g() g( ) g(..) (..) g( ) ( ) g( ) 4 1. Replace the with a set o parentheses.. Put the input value into the parentheses.. Find the output value.
Composition o Functions () = + 1 g() = + h() = + 5 ( g( )) ( ) 1 ( ) 1 + h( g( )) = ( ) + 5 = ( + ) + 5 h( ) g( h( )) = ( ) + 5 = ( + 1) + 5 = ( ) + = ( + 5) + ( ) ( ) 1 ( 1) 1
Compositions o Functions g ( g( )) Another way to write a composition o o g o OR g into.
Function compositions One more layer. ( g)() ( g()) g() ( 4) () (4) ( g()) 1 4 g The input to g() is. The output o g() is 4. The input to () is g(), so the input to () is 4.
Combining Functions Algebraically Multiplication by a number 1 ( 1) What is the domain o () 1. Division by zero No.. Square root o a negative number Domain o (): (-, ). Real world problem
Your turn: 4 Perorm the indicated operation: g 5 () = -g() = 5g(-) = (4 ) 1 6 ( 5 ) 10 6 g() 5( ) g( ) 18 5g( ) 5(18) 90
Perorm the indicated operation: 4 g 5 (1) Replace with parentheses, () plug in, () Simpliy. g() () = ( ) ( ) 5 4 = ( ) ( ) 5 8 1 1 4
Perorm the indicated operation: 5 14 g g() () ( ( ) 7)( ) 1 ( 7) Domain 1. Division by zero yes. Square root o a negative number. Real world problem Domain: X -, 7
Perorm the indicated operation: 4 1 g ( g)(-1) ( 1) ( 1) 4( 1) 1 4 g( 1) ( 1) 4 (-1) g(-1) 4 ( 4) 4 8 16
Vocabulary Inverse Relation: A relation that interchanges the input and output values o the original relation. Relation: (-, 5), (5, 6), (-, 6), (7, 6) Inverse Relation: (5, -), (6, 5), (6, -), (6, 7)
() = 1 Echange and y There s no y!!! Remember: y = () y = y This IS the inverse unction (written as: as a unction o y ) Rewrite it so that it is written as: y as a unction o ) Add (let and right) y Rearrange into slope intercept orm y This is the inverse o: y = - Graph both equations on your calculator. Also graph the line y=. Push zoom option 5 (zoom square) Are the two equations inverses o each other
I you have a graph; how can you tell i the inverse o the graphed unction is also a unction Horizontal Line Test: i the line intersects the graph more than once, then the Inverse o the unction is NOT a unction.
Find the inverse o: Echange and y y y This IS the inverse unction, but it is written in the orm as a unction o y Rewrite it so that it is written as: y as a unction o ) Add (let and right) y cubed root both sides y Simpliy 1 Is the inverse o:
What unction would undo a: 4 1 ) ( ) [0,, ) ( 4 ) ( g 1 ) ( g 5 4 ) ( h ) ( 1 h 5 ) ( k ) ( 1 k 1... 4. 4 5 5 5 1
4 y 4 y 1 ( 4) 4 y y ( 4) ( y )( 4)
6 1 1 y 6 y 1 y 6 y 1 y 6y 6 y y 6y y y 9y 6 y( 9) 6 6 ( y 1)( 6) y multiply this out! y 6 ( 9)
Using an inverse unction to solve an equation. Ticket prices in the NFL can be modeled by: P 5t 0.19 where t is the number o years since 1995. (price as a unction o time since 1995) During what year was the price o a ticket $50.85 P 5t 50.85 5 t 0.19 0.19 1 0. 19 50.85 5 50.85 5t t 0.19 t 6 1995 6 001
1. Find the boundary numbers: (Solve the equation) 0 0 1 1-4. The solution is usually either: 1) Between the boundary numbers or 0 = ( 4)( + ) = 4, - - 4 ) Outside o the boundary numbers - 4
The solution is usually either: 1) Between the boundary numbers or 0 1-4 ) Outside o the boundary numbers 0-4. Test a value to see i it is a solution. Zero is oten the best number to test. 0 1 0 (0) (0) 1 Is 0 a solution (does it make the inequality true The shaded part o the graph is the solution we must pick the option that shades the number 0. 4
9 10 0 4 16 6 0 ) 1)( 1)( ( 0 8 4 0 numbers) ( 0 positive numbers) ( 0 negative ) ( 0 ) ( 0
y 4 1 0 4 1 Graph the general shape o the equation. Positive lead coeicient, even degree Up on let and right No even multiplicities + - + What do you notice about the graph and the signs or each interval o - 6 Same: you could solve the inequality by looking at the sign o y rom the graph!!!
0 4 10 9 0 ( 9)( 1) 1. Find the real zeroes o the polynomial equation. 0 ( )( )( 1)( 1). Build Sign (+/-) Chart ( ) ( ) (+) (+) (+) (, ) 1 1 ( )( )( 1)( 1), -1,1, ( 4) ( )( )( )( ) ( 4) ( ). odd multiplicities sign change 4. Solve: 0 ( )( )( 1)( 1) (, ) ( 1, 1) (, )
Solving Single Variable Rational Inequalities We will solve inequalities similar to the ollowing: 0 ( ( 6)( 0 ( positive )( 0 ( ) ) 1) numbers) Solution: -values that make the rational epression positive 0 7 6 0 ( negative 0 ( ) 10 16 numbers) Solution: -values that make the rational epression negative
6 0 16 y ( ( ) 4)( Passes thru means opposite sign on each side o the zero. ) Zero o the denominator: (Either vertical asymptote OR hole). VA at = -4, 4 (, 4) ( 4,) (,4) (4, ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) 4) ( ) ( )( ) ( ) ( 4,) (4, ) 0 ( positive 0 ( ) numbers) Solution: -values that ( ) (+) ( ) (+) make the rational epression positive 4 4 1) Zero o the numerator: -intercept: (Passes thru at = ) ( ) ( )( ) ( )
Combinations Row 1 1 1 1 1 1 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 0 15 6 1
Your turn: 1. Find the coeicients o the binomial 7 epansion o ( a b) Eponent 0. 1 1 1 1. 1 1 1 1 4.. 1 4 6 4 1 1 5 10 10 5 1 5. 1 6 15 0 15 6 1 6 1 7 1 5 5 1 7 1 7.. 1 8 8 56 70 56 8 8 1
Look or a pattern: ( a b) 0 1 ( a b) ( a b) ( a b) ( a b) 1 4 1 1a 1b 1 1 1a a b 1b 1 1 1a a b a b 1b 4 4 1a 4a b 6a b 4ab 1b 1 ( a b) 5 5 4 4 5 1a 5a b 10a b 10a b 5ab 1b The sum o the eponents o each term is the same as the eponent o the binomial.
( a b) 0 1 ( a b) ( a b) ( a b) ( a b) 1 4 1a 1b 1 1 1a a b 1b 1 1 1a a b a b 1b 4 1 1 4 1a 4a b 6a b 4a b 1b In standard orm, the eponents o the binomial s let side term ( a ) get 1 number smaller as you move rom let terms to the right.
We can use this pattern to ind the binomial epansion o powers o binomials. ( ) 1 1 a b 1a a b 1b ( ) 1 () () 4 4 ( a b) ( ) ( a b) ( ) 4 4 1a a b ab 1b () () () 6 1 8 4 4 1a 4a b 6a b 4ab 1b 4 4 4 () 6 () 4() 8 4 () 16 4