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Big Ideas Examples
Solving Equations: Compose Both Sides with Inverse Functions in Reverse Order
Linear
5x – 3 = 2x + 9
First, put the equation in a form where we can use this
3x – 3 = 9
approach. To evaluate the left side we would use the
3x
functions 3x and x – 3 in that order. We use the inverse
 3  3  9  3
3x = 12
3x 
3

function x + 3 first, then the inverse function
12
3
x
.
3
Note that there is “cancellation” at each step.
x=4
Quadratic
 2x  3  5  7
2
Evaluate: 1) x  3 , 2) x 2 , 3)  2x , 4) x  5
Solve: 1) x  5 , 2)
x
, 3)
2
x , 4) x  3
Radical
5 2x  3  1  9
Evaluate: 1) 2x, 2) x – 3, 3)
Solve: 1) x  1 , 2)
x , 4) 5x , 5) x  1
x
x
, 3) x 2 , 4) x  3 , 5)
5
2
Exponential
4e x / 2  3  5  7
Evaluate: 1)
x
, 2) x  3 , 3) e x , 4) 4x , 5) x  5
2
Solve: 1) x  5 , 2)
x
, 3) ln( x) , 4) x  3 , 5) 2x
4
1
Functions as Mappings
I.
Definitions
A. A relation is a mapping f : X  Y that maps elements of the set X to elements
of the set Y.
B. A function is a relation where no element x in X is mapped to two elements in Y.
C. The domain of a relation is the set of elements of X mapped to an element of Y.
D. The range of a relation is set of elements of Y that are mapped to by an element of
the domain of the relation.
E. The implied domain of a function f : X  Y with mapping rule x  f (x) is
x X |
II.
f ( x)  Y .
Consequences of Using This Definition
A. Definition of a discrete function:
B. Addition of real numbers is a function. Why?
C. What happens if X (Y) is not a set of numbers?
D. Is differentiation of functions a function?
E. How would you define the graph of a function?
F. What are other consequences?
2
Modeling
Converting Graphs to Tables to Equations: Translation of Axes
Linear
y'
6
y = f(x)
4
2
-10
-5
5
10
x'
-2
-4
-6
x
2
3
4
x2
0
1
2
y
1
3
5
Equation: y  1   2x  2
y 1
0
2
4
(Point-Slope Form)
3
Quadratic
8
y'
6
x'
4
2
y = f(x)
-10
-5
5
10
-2
-4
x
y
0
1
2
3
5
3
Equation: y  5   2 x  1
2
x 1
1
0
1
y5
2
0
2
(Vertex Form)
4
Square Root
6
4
y'
2
y = f(x)
-10
-5
5
10
-2
-4
x'
-6
x
3
2
y
4
2
x3
0
1
y4
0
2
Equation: y  4  2 x  3
5
Absolute Value
4
y'
2
-10
-5
5
10
y = f(x)
-2
x'
-4
-6
x
y
2
3
4
3
 2.5
2
x2
0
1
2
y3
0
0.5
1
Equation: y  3  0.5x  2
6
Reciprocal
8
y'
6
y = f(x)
4
2
x'
-10
-5
5
10
-2
-4
-6
-8
-10
y
x
5
2
5
3
4
2
Equation: y  2 
(center)
x5
0
1
3
y2
0
3
1
3
x5
7
Other Transformations
Reflection Across An Axis
8
6
(x, y)
(-x, y)
4
2
-10
-5
5
10
-2
-4
(x, -y)
-6
-8
-10
Reflection of y  f (x ) Across x-Axis:  y  f (x)
Reflection of y  f (x ) Across y-Axis: y  f ( x)
Example: y  2 x 2  3x  4
8
Stretch Away From x-Axis; Compression Toward x-Axis
8
Stretch
(x, 2y)
6
(x, y)
4
2
-10
-5
Compression
5
-2
Reflection &
Compression
(x, 0.5y)
10
(x, -0.5y)
-4
-6
-8
Reflection &
Stretch
(x, -2y)
Stretch of y  f (x ) away from x-axis: y  a f (x) where a  1
Compression of y  f (x ) toward x-axis: y  a f (x) where a  1


Example: y  3   2 sin  x  
6

9
Problem Solving
Polya’s four-step heuristic works in any context
1.
2.
3.
4.
Understand the problem
Devise a plan
Carry out the plan
Look back
Absolute Value as Distance From the Origin
1. 1-space: x 
x2
2. 2-space: x, y   x 2  y 2
3. 3-space: x, y, z   x 2  y 2  z 2
4. Distance between two points: Translate one point to origin. Now it is an absolute value
problem. Space does not matter; process is the same.
5. “Circle”: Locus of points equidistant from a fixed point called the center.
6. 1-space circle with center at origin: x  x 2  r or x 2  r 2
7. 1-space circle with center at h: Translate center to origin; x  h 
x  h2
 r or
x  h 2  r 2
8. 2-space circle with center at h, k :
9. 3-space circle with center at h, k , l  :
10. How would you define disks in 1-space, 2-space, 3-space?
11. What is the connection to vectors in different spaces?
10