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
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 4thFiona D'ArcyGillian away- Free time to work on graphing calculator assignment (Due Wednesday January 12th)If you need a graphing calculator, you can download graphmatica online @http://www.graphmatica.com/
=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
Monday January 10th
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
Average rate of change: Rate of change over an interval.
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]
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.
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.
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
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
Tuesday January 11th
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.
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.
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 PiggottInverse 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 4thFiona D'ArcyGillian away- Free time to work on graphing calculator assignment (Due Wednesday January 12th)If you need a graphing calculator, you can download graphmatica online @http://www.graphmatica.com/
=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
Monday January 10th
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
Average rate of change: Rate of change over an interval.
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]
- 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.
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.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.
http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/derivative/slope.html
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
Tuesday January 11th
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.