8 the Slope Function Revised

MARY WARD CATHOLIC SECONDARY SCHOOL

MCV 4U

CALCULUS & VECTORS

UNIT 8

THE SLOPE FUNCTION

MCV 4U u 8 The Slope Function

Mary Ward C S S 2012 Author: H. Kopach

2

Unit 8

The Slope Function

EXPECTATIONS: A 1.1, A 1.2, A 1.3

Sketch, by hand, the slope function

Demonstrate that the slope of a secant to a curve represents an

average rate of change

Demonstrate that the slope of a tangent to a curve represents the

instantaneous rate change of the function at that point

Use graphing technology to determine instantaneous rate of

change demonstrate an understanding of secants & tangents

Use the Tangent Operation of the graphing calculator to draw a

tangent to a point on a curve

TEXT REFERENCE:

Calculus & Vectors (Nelson) 1.2, 1.6

Calculus & Advanced functions, MGH 3.1, 3.2, 3.3

ASEESSMENT / EVALUATION

ACTIVITY KTCA TIME

A The Slope Function T, C 1 h

B The Slope Function (Again) T, C 1 h

C Interactive Graphing Technology K, A 1 h

D More Slope Function K, A 1 h

E Using Calculator to Find Slope Function K, A 1 h

F From Secants to Tangents T, C 1 h

G Squeeze Theorem K, A 1 h

H From Secant To Tangent (Again) K, A 1 h

I Interactive Graphing Technology K, A 1 h

J Funny Tangents T, C 1 h

Seminar T, C 1 h

Total estimated time (in school and at home) 11 h +



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WHAT IS CALCULUS?

What is calculus?

1. The study of the functions and the slopes of functions (differential

calculus)

2. The study of the areas under these curves (integral calculus)

Calculus is a tool in the study of demographics (populations), motion,

economics, and business (in particular optimization”).

In this course we will concentrate on differential calculus.

Two mathematicians can be thought of as the

“fathers of calculus”. The Englishman Sir Isaac

Newton (1642 – 1727) is a familiar name from prior

studies in science. We will be using the Newton

Quotient in the study of limits.

The German mathematician was Gottfried Wilhelm

Leibniz (1647 – 1716). We will be using Leibniz notation

in the study of derivatives.

These two men, who lived in the eighteenth century,

were bitter rivals but their legacy lives on.



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THE SLOPE FUNCTION

EXAMPLE

Sketch the graph of y = x2 in your graphing

calculator using an appropriate window.

Using the and Tangent operations,

find the slope of the tangent line to the curve at

x = -4

From the equations of the tangent line

y = -8x -16 we know that the slope is -8.

We will define an ordered pair with the independent

variable being x, and the dependent variable being

m.

Thus (x,m) = (-4, -8)

Repeat finding the slope of the tangent for different

values of x :

{ 4, 2,0,1,2}x

Record your results in a table of values. Enter these

into your graphing calculator as lists



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Deselect the equation y = x2. Graph the data using

the STAT PLOT

feature and a suitable window. The new graph, the

slope function, appears to be linear.

Turn DIAGNOSTICS on. Using Regression Analysis, find

the equation of the line of best fit.

The new function, the slope function, is clearly a

linear function (since r = 1 there is a “100 % fit”) with

the equation y = 2x

The original function has even symmetry, whereas

the slope function has odd symmetry.

Assignment A. THE SLOPE FUNCTION

Using your graphing calculator, repeat the above steps for the functions:

y = x, y = x3, y x

Summarize your results in the following categories :

a) graph of original function; symmetry

b) table of values (x,m)

c) graph of slope function; symmetry

d) equation of slope function (obtained by regression analysis)

(answers at the back of the unit guide).



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THE SLOPE FUNCTION (AGAIN)

We will start with the graph of a function, and by examining

the hills,

valleys,

intervals of increase,

intervals of decrease, and

concavity,

draw a second graph directly below the first one.

The second function we will call the slope function.

When we talk about the slope of a graph y = f(x) at a point x, what we

will mean is the slope of the tangent to the graph at the point. It is a

convenient use of language.

When the graph is curved, then the slope

m of the curve will change from point to

point. That is m, will depend on x:

m is a function of x.

This “slope function” has a special name.

It is called the derivative of the function f

and is written f ’(x).

Note that f ’(x) = m(x)

Thus, the derivative of the function f is the function f ‘ (say f

prime)which associates to each x the slope m of the f -graph at x.

If f (t) represents a quantity (volume, revenue, distance) which is changing

with time t, then f’ (t) gives the instantaneous rate of change of this

quantity.

For example, if s =s (t) is distance traveled at time t, then v= s ’(t) is

velocity at time t.



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EXAMPLE 1

At the right is the graph y = f (x). Directly beneath, draw the

graph of the derivative f’ (x). Use vertical lines to show the

relationship between the graphs

at certain special points.

Solution

Just keep in mind that the

height of the f ‘-graph is

the slope of the

f -graph.

x value Original function, f Slope function, f’

x = 0 is increasing slope is positive

(0, )x a increases, concaves

down

slope is positive, but

decreasing

x = a Hill point the slope here becomes zero

( , )x a b decreases slope is negative

x = b Inflection point (change

in concavity from

concave-down, to

concave-up)

minimum Valley point

( , )x b c decreases, concaves up slope is negative, but

increasing

x = c Valley Point the slope here becomes zero

( , )x c d Increases, concaves up slope is positive

x = d Inflection point (change

in concavity from

concave-up, to

concave-down)

maximum Hill point

( , )x d approaches the slant

asymptote

the slope approaches a

horizontal asymptote

(constant positive value)



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EXAMPLE 2

In each case blow, you are given the graph of a function f (x).

The (b) –graph is obtained from the (a) –graph by reflecting the right half

in the y-axis.

Directly beneath each graph, draw the graph of the derivative f ‘(x), using vertical lines to show the relationship between the graphs at certain

special points.

Discuss any symmetry properties of the pairs of graphs.

Solution

Below each graph, we have drawn the graph of the derivative. There

are a couple of features worthy of note.

A Corner.

The (b)-graph has a corner at x = 0 and this causes a jump in the

derivative graph at the point. A jump in slope of f means a jump in

height of f ‘. Note the little open circles (“donut holes”) which indicate

that f ‘(x) has no value at x = 0. The derivative is not defined at

corners.



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Symmetry.

Each of the two top graphs exhibit some

nice symmetry, but of a different kind. The

(b)-graph is bilaterally symmetric about

the

y-axis,

and the (a)-graph exhibits what is called

antisymmetry .

The interesting thing is that the two derivative graphs exhibit the same

two types of symmetry but in the opposite order: the slope graph on

the left is bilaterally symmetric, and the slope graph on the right is

antisymmetric.

Is there a general result here: if f has one kind of symmetry, f ‘ has the

other?

Bilateral symmetry

Mirror symmetry

Symmetry about the y-axis

f(x) = f(-x)

Antisymmetry

Rotational symmetry

Symmetry about the origin

f(x) = - f(-x)



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Assignment B. THE SLOPE FUNCTION (AGAIN)

Sketch each of the twelve graphs (a-l) and directly below each graph draw the slope

function using the concepts from Examples 1 – 2. Then match with the correct graph

(A –L) of its derivative.



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(answers at the back of the unit guide).

Assignment C : USING GRAPHING TECHNOLOGY

Check the internet for the slope function: click and drag interactive technology.

Unfortunately websites change with time. When you find a suitable application, please

share with your classmates.



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Now that you understand the basics, you can use a graphing calculator

to find the slope function from a given function.

Assignment D. MORE SLOPE FUNCTION

Calculus & Vectors ( Green Book) Calculus & Advanced Functions (Blue Book)

p 75 # 15 p 194 # 1 - 4

Assignment E. USING CALCULATOR TO FIND SLOPE FUNCTION

1. Enter the function y = x3 – 2x

2 – 5x + 5 into

[ 5,5], [ 10,10]x y

Graph.

2. Sketch the cubic function into your notebook, labeling hill & valley points. Use

green to sketch the intervals where the function is increasing, and red to sketch

where the function is decreasing.

3. Draw a set of x & y axes directly below your sketch. Using the techniques learned

in Assignment B, draw the slope function.

4. Use your calculator. Find the Derivative Function, and enter into your calculator

as Y2 to draw the slope function on your calculator.

5. Compare your sketch (obtained in # 3) to that obtained from the graphing

calculator (in # 4). They should be the same.

6. Repeat steps # 1 – 5 for the three functions from assignment A

y = x, y = x3, y x



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FROM SLOPES OF SECANTS TO SLOPES OF TANGENTS

The word secant comes from the Latin word to cut. A secant is a straight

line, which cuts a curve at two points. The most important feature of a

secant is its slope.

Consider the function y = x2. We know that

P (1, 1) is a point on the curve since it satisfies

the equation. We will consider P (1, 1) a

fixed point of reference.

We know that Q (2, 4) is also on the curve.

We will draw the secant line PQ.

We will call Q the traveling point. Q will

move along the curve closer and closer to P.

With each new Q, we get a different secant.

The slope of each secant will be evaluated

using the slope formula.

As Q comes nearer to P (the fixed point of reference), the slope of the

secant line becomes closer and closer to 2

Q (2, 4)

P (1, 1)



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Assignment F. FROM SECANTS TO TANGENT

(Graphing calculator activity)

1. Enter y = x2 into the equations editor Y=

2. Go to STAT EDIT 1: Edit

3. Into L1 enter different x values for the traveling

point Q :

2, 1.5, 1.25, 1.1, 1.01, 1.001, and 1.0001

4. Into L2 enter y values for Q. Go to the extreme

top of the column. You want y = x2 so enter L12.

You should see corresponding y values

generated.

The y values should get closer to the y value of the fixed point

(y = 1).

5. Into L3 we will calculate ∆y or y2 – y1 . Go to the extreme top, and

enter L2 – 1

6. Into L4 we will calculate ∆x or x2 – x1 . Go to the

extreme top, and enter L1 – 1. What happens to

the numbers in this column?

7. Into L5 we will calculate the slope y

mx

. Go to

the extreme top, and enter 3 4L L . What happens to the numbers

in this column? Explain.

8. State your observations and conclusion. Consider the traveling point

Q moving closer and closer to the fixed point P (1,1).

In # 3 above, the traveling point Q started at (2, 4), to the right of the

fixed point P (1, 1). We generated a FORWARD SECANT.

Now we will consider a BACKWARD SECANT: Q will be to the left of the

fixed point P.



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9. Into List 1 L1 enter different x values for Q :

0, 0.5, 0.75, 0.9, 0.99, 0.999, 0.9999 and 0.99999

10. Repeat steps 4 – 8.

11. Now GRAPH the function, using a suitable window. Find the slope

of the tangent line by using the TANGENT OPERATION on your

calculator: 2nd PRGM DRAW 5: Tangent. Enter 1 (x=1) to draw the

tangent line at x=1. From the equation of the tangent line, what is

the slope of the tangent?

Thus, we can use the slope of a secant line to determine the slope of a

tangent by sliding the secant closer and closer to a tangent line. The main

idea behind this procedure is that of a limit. As the travelling point gets

closer and closer to the fixed point we can say that the slope of the

secant is approaching a limiting value.

The word tangent comes from the Latin word tangens, which means

touching. When the backward secant rotates counter clockwise or the

forward secant rotates clockwise, about the fixed reference point P, a

tangent is obtained.

The more you “squeeze” the domain, the closer the slope of the secant

gets to the slope of the tangent. This gives rise to the

“Squeeze” or “Sandwich” Theorem

For a function f (x) which is increasing

Slope of backward secant < slope of tangent < slope of forward secant

1.9999 < 2 < 2.0001



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H. FROM SECANT TO TANGENT (Again)

Calculus & Vectors (green book) Calculus & Advanced Function (blue book)

Read p 12 - 15 Read pp 120 - 127

p 127

# 1

#2 e

#3

p 19 # 7

# 20 (use graphing calculator)

Assignment G. USING SQUEEZE THEOREM TO FIND THE TANGENT

The fixed reference point P (1, 3) lies on the curve y= 2x+ x2

1. If Q is the traveling point (x, 2x+x2), find the slope of

the secant line PQ for the following values of x:

2, 1.5, 1.1, 1.01, 1.001 and 1.0001

2. Using the results of part (a), use the Squeeze Theorem

to determine the value of the slope of the forward

tangent line to the curve at P (1,3)

3. Repeat by placing Q to the left of P. Find the

Backward Tangent

4. Using the slope from part (b)and (c) find the equation

of the tangent line to the curve at P(1,3)

5. Verify your answer in (d) by using Tangent operation

on your graphing calculator

6. Sketch the curve, labeling intercepts and vertex.

Draw one backward secant (in red) one forward

secant (in green) and tangent line (in blue)



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Assignment I : USING GRAPHING TECHNOLOGY

Check the internet for the secants to tangents: click and drag interactive technology.

Unfortunately websites change with time. When you find a suitable application, please

share with your classmates.

FUNNY TANGENTS

Crucial insight can be obtained form a study of pathology: examples in

which things go wrong greatly aid our understanding of normal behavior

EXAMPLE 1: A Corner.

A ball, falling vertically, hits a cement floor with speed 5m/s and

bounces vertically upwards, leaving the floor with speed 2m/s. Draw

the distance – time graph of the ball and discuss the slope of the

graph of the bounce point. So we don’t have to worry about the

deformation of the ball, and so on, assume that the ball has zero

radius (i.e., is a single point), and that the change in velocity at the

bounce is instantaneous.

Solution

The s-t graph is made up of

two parabolas: the ball

changes from one to the

other at the bounce point.

The slope of the graph is the

velocity of the graph. If we

take the positive direction to

be up, then the velocity of

the ball is -5 as it arrives at the

floor, and +2 as it leaves the

floor. These are the slopes of

the two component

parabolas at the bounce

point.



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How are we to describe the slope of the s-t graph at the bounce

point? It seems reasonable to say that, going forward in time, the

slope is 2, and going backward in time the slope is -5. Thus, at the

bounce points, we have two “one-sided” slopes, and they are

different.

In General

The forward slope of the graph at a point is the slope going

“forward” from the point. If we calculate the slope by taking

the limiting slopes of secants, we look only at secants going

forward from the point.

The backward slope of the graph at a point is the slope going

“backward” from the point. If we calculate the slope by

taking the limiting slopes of secants, we look only at secants

going backward from the point.

If the forward and the backward slopes are different at a

point, we say that we have a corner.

a corner is formed when two straight lines join at an angle, whereas a

cusp is formed when two curves join at an angle. (The angle is the one

between their tangents at the point of contact.)



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EXAMPLE 2 : A Corner

Consider the absolute value

function

f(x) = |x3 - 8|.

This can be rewritten as a

piecewise function

3

3

( 8), 2( )

( 8), 2

x xf x

x x

Enter the absolute value function into your

graphing calculator. Use the nDeriv feature to

sketch the derivative function

Before you graph the slope function, set your

calculator to dot mode.

You will get a jump discontinuity. If your

calculator is in connected mode, the graph

will appear (mistakenly) to be continuous.

Hence

2

2

3 , 2'( )

3 , 2

x xf x

x x

show that f'(2) Direct calculations

does not exist. In fact, we have left

and right derivatives with

The phenomenon this function shows at x=2 is called a corner.



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EXAMPLE 3 : A Cusp

Consider with the function 2

3y x

at x = 0. Note that this function is

also defined for all x, positive and

negative.

Solution

Find the Derivative Function. Remember to

use Dot Mode.

There is a discontinuity in the slope function

graph. However, there is a vertical asymptote

at x =0.

This function shows asymptotic behaviour.

You can find the values of the derivative

function to the right and left of x =0 by

evaluating the function for x values

immediately to the right and left of 0.

So we have



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We can also evaluate the slope of the secant algebraically.

The slope of the secant between 0 and h is

2 2

3 3

2

3

1

3

0

1

h

h

h

h

h

When h is close to 0, the denominator is

close to zero, and the expression is very

large.

Positive values of h give large positive

values.

Let h = 0.000001

1 1

3 3

1 1

(0.000001)

100

h

Negative values of h give large negative

values. Let h = 0.000001

1 1

3 3

1 1

( 0.000001)

100

h

for h positive, the limit is plus infinity 0

lim '( )h

f h

for h negative, the limit is minus infinity 0

lim '( )h

f h

Thus the forward and the backward slopes are both infinite, but with

opposite sign.

This is a special kind of corner known as a cusp.



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EXAMPLE 4- A JUMP

The Heaviside step function is a discontinuous function whose value is zero

for negative argument and one for positive argument. The function is used

in the mathematics of control theory and signal processing to represent a

signal that switches on at a specified time and stays switched on

indefinitely. It was named after the English electrical engineer Oliver

Heaviside.

0, 0,[ ]

1, 0,

xH x

x

Your job is to examine the slope situation at the jump-point x = 0

Write expressions for the slopes of the general forward and backward

secants from x = 0. By taking the limit as h approaches zero, calculate the

forward and backward slopes of the graph at x = 0.

Solution

Let h be a value of x near x = 0.

The expression for the general secant from x=0 is

( ) (0) ( )H h H H h

h h

Since H(0) = 0 we get forward secants by taking h positive, and

backward secants by taking h negative.

Let h be a value of x on the positivie x axis.

Forward (h > 0) ( ) 1H h

h h

http://en.wikipedia.org/wiki/Continuous_function

http://en.wikipedia.org/wiki/Function_(mathematics)

http://en.wikipedia.org/wiki/0_(number)

http://en.wikipedia.org/wiki/1_(number)

http://en.wikipedia.org/wiki/Control_theory

http://en.wikipedia.org/wiki/Signal_processing

http://en.wikipedia.org/wiki/England

http://en.wikipedia.org/wiki/Polymath

http://en.wikipedia.org/wiki/Oliver_Heaviside

http://en.wikipedia.org/wiki/Oliver_Heaviside



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As h approaches zero, this expression approaches plus infinity. Short

forward secants are nearly vertical.

Let h be a value of x on the negative x axis.

Backward (h < 0) ( ) 0

0H h

h h

Backward secants always have slope zero.

We conclude that, at x = 0, the forward slope is + ∞ and the

backward slope is 0.

EXAMPLE 5 : A Vertical Tangent

The function1

3( )f x x is defined for all x

(every number, positive or negative, has a

unique cube root).

Note the vertical tangent at x = 0

Write down the general expression for the slope of the secant

between x = 0 and x = h. Simplify it and take the limit as h

approaches zero from above and from below. Thus find the forward

slope and the backward slope at the origin. Use this to help you

draw a picture of the graph of the function.

Solution

The formula ( ) (0)f h f

h

will give the slope of the secant between 0

and h

for h both positive and negative.



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In this case we get

1 1

3 3

1

3

2

3

0

1

h

h

h

h

h

where we have simplified the expression so we can see what

happens as h approaches zero.

If h is close to 0, the denominator is close to zero, and the expression

is very large.

Furthermore, the expression is always

positive (since

2

3h is a square) and so the

end behaviour, as h approaches zero,

through both positive and negative

values, is plus infinity. The slope at the

origin must be + ∞.

Note the vertical asymptote for the

slope function at x = 0

Assignment J. FUNNY TANGENTS

Calculus & Vectors (Green book)

Calculus & Advanced Functions (blue

book)

Read p 60 (bottom)

p 196

# 13 (Must use Case 1, Case 2)

#16

#17

p 21 # 18

Draw each function and the corresponding

slope function directly below

p 52 # 7, 8, 11



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Answers to Assignments :

A

Linear regression : y =0

quadratic regression: y = x2

power regression: y = 0.5 x-0.5

B Solutions : a-E b-K c-D e-F f-J g-C h-L i-A j-L k-B l-H

F 6. 0x 7. Slope of secant 2



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8 the Slope Function Revised

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