Presentation on theme: "Composition Table of Contents Word Problems with Compositions….……………………Slides 2 - 5 Compositions with a given value…………………..………Slides 6 – 8 Compositions."— Presentation transcript:
Composition Table of Contents Word Problems with Compositions….……………………Slides 2 - 5 Compositions with a given value…………………..………Slides 6 – 8 Compositions without a value………….………………….Slides 9 – 10 Domains of Compositions………………………….……….Slides 11 – 14 Review of FOIL………………………………………………………..…Slide 15 Practice…………………………………………………..…………..Slides 16 - 18
Composition of Functions Composite Functions are basically two step functions…You need the solution to the first part to plug into the second part. You see composite functions mostly when you’re shopping. Everyone has seen the specials “50% off the ticketed item and then take an extra 20% off!” or something similar. You might think 50% then another 20% is 70%, but that’s not true. First y0u need to find out what 50% of the first is THEN take 20% off of that number. Here’s an example of why it matters….
Let’s say the item was originally $130 and now is on sale for 50% off plus another 20% off. 1)What is 50% of 130? Well, that’s easy! That’s 65. 2) Now take 20% off 65. 20% of 65 is 13. So you need to subtract the 13 from 65. The total price is $52 If you just found 70% of 130, you would have a total price of $39… that’s a big difference! When you apply a function rule on the result of another function rule, you compose the function. Composition is defined as, read as “f of g of x”.
Example: The cost of gasoline for an automobile trip depends on the number of gallons that you buy. Let’s assume gas everywhere costs $2.99 per gallon. The number of gallons that you buy depend on the number of miles that you drive. Let’s assume your vehicle can drive 20 miles on 1 gallon of gas. A chain of relationships like this is an example of a composition of functions. Let x = the number of miles driven; f(x)= number of gallons needed; g(x) = cost per gallon. x= 300 milesf(x) = x/20g(x)= $2.99x How would you calculate the cost of gasoline on a 300 mile trip? Miles I will drive How many gallons I need since I can go 20mi / gallon Cost of gas per gallon
x= 300 milesf(x) = x/20g(x)= $2.99x x = 300 miles f(300) = 300/20 = 15 gallons g(15) = 2.99(15) = $44.97 This represents g(f(x)): The number of miles goes inside the number gallon function. The number of gallons needed goes inside of the cost function. I need 15 gal to go 300 miles Cost of 15 gals if each gal cost 2.99
What if it’s not a word problem…. Evaluate g(f(2)) when f(x) = 3x – 5 and g(x) = x 2 + 2.
Below is a visual to help you understand how composition works. So, to evaluate g(f(2)) when f(x) = 3x – 5 and g(x) = x 2 + 2 First find f(2) -> f(2) = 3(2) – 5 = 6 – 5 = 1 Then plug that answer in for f(2) in g(f(2))…so g(f(2)) = g(1) = 1 2 + 2 = 1 + 2 = 3
Use the model below to help you find So, to evaluate g(f(5)) when f(x) = 3x – 5 and g(x) = x 2 + 2 First find f(5) -> f(5) = 3(5) – 5 = 15 – 5 = 10 Then plug that answer in for f(5) in g(f(5))…so g(f(5)) = g(10) = 10 2 + 2 = 100 + 2 = 102
Using the same functions lets find a composite function when there is not a numerical value for x. What is g(f(x))? Remember, you read it from the inside out…So we’re going to take the f(x) and plug the whole expression into the g(x) wherever there is a variable. Plug in (3x – 5) for x
Example: Find g(f(x)) and f(g(x)). Let f(x) = 3x 2 + 2. and g(x) = 2x Remember: Begin in the inside first! Find g(f(x)) Find f(g(x)) Plug in 3x 2 + 2 for x Plug in 2x for x
The domain of f(g(x)) is the set of all values of x in the domain of g whose range values g(x) are in the domain of f. The domain of g(f(x)) is the set of all values of x in the domain of f whose range values f(x) are in the domain of g. Note: If your initial functions are just polynomials, then their domains are "all x", and so will be the domain of the composition. When dealing with denominators (where you can't divide by zero) or square roots (where you can't have a negative) the domains have restrictions.
Example: Given f(x)= 1/x and g(x) = x + 1, find the domains of (f o g)(x) and (g o f )(x). f(x) is a written in fraction form which means no zeros in the denominator. (f o g)(x) = therefore the domain is all real numbers except -1. (g o f )(x) = therefore the domain is all real numbers except 0.
Find the domain of g(f(x)) if f(x) = 2x + 2and g(x) = 1/x. 1)Find g(f(x)) by putting f(x) into g(x) g(f(x)) = 2) To find the domain, what numbers can you plug in for x? Since the x is in the denominator, you can plug in anything BUT numbers that make the denominator = 0. So… 2x + 2 ≠ 0 2x ≠ -2 x ≠ -1 The domain is all Real numbers EXCEPT -1.
Find the domain of f(g(x)) if f(x) = 2x + 2 and g(x) = 3x. 1)Find f(g(x)) by putting g(x) into f(x) f(g(x)) = 2(3x) + 2 = 6x + 2 2) To find the domain, what numbers can you plug in for x? You can plug in any number for x and find a y value, so…. The domain is All Real numbers. If there are no restrictions (like radicals or denominators) to worry about, the domain is often (not always!) all real numbers.
FOIL (2x – 8)(-3x + 4) = 2x(-3x) + 2x(4) – 8(-3x) – 8(4) = -6x^2 + 32x - 32 -6x^2 + 8x + 24x – 32 = -6x^2 + 32x - 32 more than 2 terms The same thing applies to more than 2 terms in one or both (2x – 8)(-4x^2 -3x + 4) = 2x(-4x^2) + 2x(-3x) + 2x(4) – 8(-4x^2) - 8(-3x) – 8(4) = -8x^3 -6x^2 + 8x + 32x^2 + 24x – 32 = -8x^3 + 26x^2 + 32x - 32
Determine whether the following functions are inverses of one another by finding No Yes
Given: f(x) = 4x^2 -3x + 8 and g(x) = 2x – 6 Find (f o g)(x): (f o g)(x) means: Put the expression for g(x) into all x’s in f(x) Since, f(x) is 4(x)^2 – 3(x) + 8 (f o g)(x) = 4( 2x-6)^2 – 3( 2x – 6) + 8 (f o g)(x) = 4( 2x-6 )( 2x-6 ) – 3( 2x – 6) + 8 (use FOIL for (2x – 6)^2 ) (f o g)(x) = 4( 4x^2 -24x + 36) – 3( 2x – 6) + 8 (f o g)(x) = 16x^2 - 96x + 144 – 6x + 18 + 8 16x^2 - 102x + 170 (f o g)(x) = 16x^2 - 102x + 170
Given: f(x) = 4x^2 -3x + 8 and g(x) = 2x – 6 Find (g o g)(x): (ignore f(x) for this problem) (g o g)(x) means: Put the expression for g(x) into all x’s in g(x) Since, g(x) is 2( x ) – 6 (g o g)(x) = 2( 2x – 6) – 6 (g o g)(x) = 4x – 12 - 6 4x - 18 (g o g)(x) = 4x - 18