Combing+-+Lesson+Summaries

Dec 13 - Operations with Functions



December 14th Edited by : Samantha Guest from Waterloo In class problem sets !

December 15th Edited by : Samantha Notes on Domain and Range, Function Operations and Notation

Dec. 17 - Function Compostion OR Composite Functions Gwynne Finlay I am so sorry but I do not know how to make the fony superscript so my squares, etc. are the big numbers after the variables. f 0 g ("0" is actually a little 'o' that looks like the dot when you multiply but it's not filled in) ("0" means 'composed of' and is a new operation) f 0 g = f(g(x)) eg. for f(x)=2x+1 and g(x)=x2 a) f 0 g = f(g(x)) =2(x2)+1

b) g 0 f = g(f(x)) =(2x+1)2

f 0 g does not equal g 0 f (ORDER MATTERS!!!!!!!!!!!)

c) f 0 f =2(2x+1)+1 =4x+3

d) f 0 g 0 g =f(g(g(x))) =f((x2)2) =f(x4) =2(x4)+1 =2x+1

eg. f={(-2,3)(-1,1)(0,0)(1,-1)(2,-3)} g={(-3,1)(-1,-2)(0,2)(2,2)(1,1)}

a) f(1)=0 b)g(o)=2

c)(g o f)(-1) ﻿ =g(f(-1)) =g(1) *this is g(1) because f(-1) is 1 =1 eg. a) f(-2)=5 b) g(0)=2
 * = x ||= f (x) ||= g(x) ||
 * = -2 ||= 5 ||= -1 ||
 * = -1 ||= 4 ||= 3 ||
 * = 0 ||= 3 ||= 2 ||
 * = 1 ||= 2 ||= 0 ||
 * = 2 ||= 1 ||= 1 ||

c)(f 0 g)(2)=f(g(2)) d)(g 0 f)(1)=g(f(1)) =f(1) =g(2) =2 =1

e)(g 0 f 0 g)(2)=g(f(g(2))) =g(f(1)) =g(2) =1

eg. For: f(x)=x2+1 and g(x)=3x a) find (f 0 g)(X) this can also been seen as f 0 g or f(g(x)) =f(g(x)) =f(3x)2+1 =9x2+1

b) find (f 0 g)(1) =9(1)2+1 =f(g(1)) =10 =f(3(1)) =f(3) =(3)2+1 =9+1 =10

=Monday January 3rd= Robert Piggott

__Inverse Functions__

Given:

f(x)=2x-1 and g(x)=x^2+3

The inverse is:

y=2x-1 x=2y-1 x+1=2y f(x)^-1=(x+1)/2

y=x^2+3 x=y^2+3 x-3=y^2 g(x)^-1=(x-3)^0.5

Therefore:

f^-1 o g^-1 = [[(x-3)^0.5]+1]/2(f o g)^-1 = 2(x^2+3)-1x=2(y^2+3)-1x+1=2(y^2+3)[(x+1) -3]/2=y^2([(x+1 -3]/2)^0.5= (f o g)^-1Therefore f^-1 o g^-1 does not equal (f o g)^-1Given:f(x)=3x-1 g(x)=(x+1)/31. f(x) is the inverse of g(x)2. f o g = 3(__x+1__)/3 -1y=x 3. g o f = (3x-1)+1/3y=x4. No matter which way you composes these two functions they will always become y=xThis true for any two functions that are the inverses of each other - the round trip theorem==
 * Tuesday January 4th**
 * Fiona D'Arcy**Gillian away- Free time to work on //g////raphing calculator assignment// (Due Wednesday January 12th)If you need a graphing calculator, you can download graphmatica online @[]

==Wikispaces is spazzing, it keeps mashing up our entries. It is determined to bring us together.=

Friday January 7th= Robert Piggott

Mixing/Composing Functions test

**Fiona D'Arcy**
=__Rate of Change__= Slope = Rate of change of one thing/ Rate of change of another thing E.g kilometres/ hour E.g farts/ month E.g # of times Mimi checks her test/ second E.g muffins eaten/ minute on Mondays

Rate of Change = slope = rise/run = __/\__y/ __/\__x

For a linear function, the rate of change of the variables is constant, and the interval size over which it is measured is irrelevant. Interval: section across which slope is measured in terms of x (a.k.a run)

If function is non-linear (curved)?

Use a secant: line joining two points on a curve



We can calculate the r of c (slope) of the secant (rise/run). Average r of c over the interval P -> Q. **This is an approximation!!!** Algebraically: r of c= __/\__ y/ __/\__ x f(x)= y.... avg r of c = __/\__f(x) / __/\__x avg r of c = [f(x1)- f(x2)] / x2-x1 over the interval [x1, x2]
 * Average rate of change**: Rate of change over an interval.

To find the instantaneous r of c, use the same formulas as for average, but use values that are really really really close to the exact x value for which you are trying to find the r of c. This works for all kinds of functions.
 * By shrinking the interval P-> Q we are getting better and betterapproximations of the r of c of the fxn.
 * If Q gets reallly really really close to P, we can assume that the interval P -> gets really really really close to zero.
 * We can find thr slope of the curve at P. This is **instantaneous rate of change.**

Example: f(x)= -4.9x^2 + 14x +1 Find the average r of c for 0 __<__ x __<__ 0.5 seconds avg r of c = [f(x)-f(x)]/x2-x1 = {[-4.9(0.5)^2 + 14(0.5) +1] - [-4.9(0)^2 +14(0) + 1]} / 0.5 - 0 = 11.55m/s

Example: f(x)= -4.9x^2 + 14x +1 Find the instantaneous r of c @ x=3 *use 3, 3.001, or 2.999 inst r of c = [f(x)-f(x)]/x2-x1 = {[-4.9(3)^2 + 14(3) +1] - [-4.9(2.999)^2 +14(2.999) + 1]} / 3 - 2.999 = -15.4m/s


 * Do not round until final answer

For more information on rate of change, visit the link below which has detailed explanations, more images, and practice questions. []

Homework: Worksheet-> //Rate of Change Problems// -Essay due friday, presentations for bonus on Monday -Quiz Wednesday: 2 questions, 10 marks. Two different functions, calculating average or instantaneous rate of change

**Fiona D'Arcy**
Work period- Time to do Portfolio Project, bonus essay or presentation, graphing calculator assignment, studying for quiz on Wednesday, studying for exams.