sketching graphs

8/3/2019 Sketching Graphs

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©2011 Michael Aryee

Worksheet

We can usually recognize when a function is increasing or decreasing by looking at the graph of its

derivative. We can obtain information about the behavior of a given function f by studying the behavior

of the different derivatives of that function. If f is a given function, then its derivative f (also called

the first derivative of the function) is also a function.

The first derivative f provides information about the steepness, or the slope of a tangent line to the

curve at any instant. If f is positive, then the function f must be increasing. This means that the

function f is changing by having some quantity added to it so that f is actually growing. Similarly, if f

is negative, then the function f must be decreasing. This means that f is changing by having some

quantity subtracted from it so that f is actually shrinking.

Increasing function

If we wish to find the range of values of x for which a function is increasing, just find the range of

values of x where the derivative is positive.

That is, find the range of values of x for which

)( x f > 0.

Decreasing function

If we wish to find the range of values of x for which a function is decreasing, just find the range of

values of x where the derivative is negative.


)( x f < 0.

Maximum and minimum points of a graph

A positive derivative indicates that the function is increasing.

A negative derivative indicates that the function is decreasing.

In this section, we will learn about how to use of the derivative of a function to detect whether a

function is increasing or decreasing. In addition, we learn how to find the absolute maximum,absolute minimum, local maximum and local minimum of a function.




o A function f ( x) is said to be an increasing function on the interval (a, b) if for x1 < x2,f( x1) < f( x2) for any two numbers x1 and x2. At a point where a curve is rising, the slope of the

tangent line is positive.

o A function f (x) is said to be a decreasing function on the interval (a, b) if for x1 < x2, f(x1) >

f(x2) for any two numbers x1 and x2. At a point where a curve is falling, the slope of the tangentline is negative.

HORIZONTAL TANGENT LINE

If )( x f = 0, then the function is said to be stationary or constant at that point. At this point, the

slope of the curve is zero.

A tangent line to the curve at this point is parallel to the x-axis and it is called a horizontal tangent line. The curve is neither increasing nor decreasing at this point.

The slope of a curve at a point is the same as the slope of the tangent line to the curve at thatpoint. Similarly, the slope of a curve at a point is the same as the instantaneous rate at which the

height of the curve is changing at that point.

In general, if f is a function that is continuous on the closed interval [a, b], and differentiable on the

open interval (a, b), then we can conclude that:

1. If )( x f > 0 for all x in (a, b), then f is increasing on the interval [a, b].

2. If )( x f < 0 for all x in (a, b), then f is decreasing on the interval [a, b].

3. If )( x f = 0 for all x in (a, b), then f is constant on the interval [a, b]




THE SLOPE OF A CURVE

In general, the slope of the tangent line is the same as the slope of the curve at the point of

tangency. Therefore to find the slope of a curve at a point, just find the instantaneous rate of

change of the function that the curve represents and substitute the value of the independent

variable into the derivative.

Assume y = f(x) is the equation of the curve, then the slope m at the point (x 1 , y1), where (x1 , y1)

is a point on the curve, is given by

m = )(1

x f or m =1 x x

dx

dy

EXAMPLE 1

Find the slope of the curve )( x f = x4

+ 2 x3 – x + 3 at the point x = -1.

Determine where the curve is rising or falling as it passes through this point.

Solution

We must first obtain the derivative of the function using the following format:

Therefore,

o The slope of the curve at the point (-1, 3) is 9.

o The curve is rising as it passes through that point.

x )( x f )( x f

x x4

+ 2 x3 – x + 3 4 x

3+ 6 x

2 – 1

-1 f(-1) = (-1)4 + 2(-1)3 – (-1)+ 3

= 3)1( f = 4(-1)

3+ 6(-1)

2 – 1

= 9




Many of our applications of derivatives can be reduced to finding minimum and maximum

values of a function.

We will begin by differentiating between two types of minimum or maximum values, namely:o absolute (or global) and

o relative (or local).We call the minimum and maximum points of a function the extrema of the function.

The Absolute Extrema

An absolute maximum or absolute minimum point is a point that is higher or lower

than all the possible values of the function.

Absolute Maximum

Suppose y is a continuous function of x given by y = f( x), defined over some given

interval [a, b], then there will be some point c in the interval such that

f(c ) > f( x) for all x on the interval [a, b].

The value of f(c) is called the absolute maximum on [a, b]. We say the function has an

absolute maximum at x = c.

Absolute Minimum

Suppose y is a continuous function of x given by y = f( x), defined over some given

interval [a, b], then there will be some point c in the interval such that

f(c ) < f( x) for all x on the interval [a, b].

The value of f(c) is called the absolute minimum on [a, b]. We say the function has anabsolute minimum at x = c.

The maximum and minimum points of a graph




The Relative Extrema

On the other hand, when there are other points in the domain of the function we are working on,

other than the absolute maximum or minimum, which are higher or lower than the points

surrounding them, we call the higher or lower value relative (or local) maximum or minimum.

Relative Maximum

The relative (or local) maximum point of a function is the point of the function that ishigher than the neighboring points just to the right or to the left of the graph.

We say that f(x) has a relative (or local) maximum at x=c

if f(c ) > f( x) for every x in some open interval around x=c.

Relative Minimum

The relative (or local) minimum point of a function is the point of the function that islower than the neighboring points just to the right or to the left of the graph. We say that f(x) has a relative (or local) maximum at x=c

if f(c ) < f( x) for every x in some open interval around x=c.




ANOTHER GRAPHICAL LOOK AT EXTREMES

SITUATION WHERE THERE IS A RELATIVE MAXIMUM OR MINIMUM




ANOTHER SITUATION WHERE THERE IS A RELATIVE MAXIMUM OR MIMMUM

It is also possible that )( x f does

not exist but there may be a

maximum or a minimum.

This is when there is no jumps inthe graph but the graph changes

suddenly from steep downhill to

a steep uphill or vice versawithout being flat for a moment.

An example of such a graph is

given here.

SITUATION WHERE THERE IS NO RELATIVE MAXIMUM OR MINIMUM

It is also possible that a graph may have no maximum or minimum points. If the slope of thecurve jumps from positive to negative or vice versa without the curve being flat for a moment,

then the slope of the function at that point would be undefined . From the diagram below, as x

passes through the value of zero, )( x f changes from negative to positive. However, both

)( x f and )( x f can never take on x = 0 so that it is no minimum point. Also the curve is never

flat.




To find the maximum or minimum points of a curve, first find all the critical points of the

function

To find the critical points of a function:

First, find the first derivative of the function. Then,

a) Find all values c such that )(c f = 0. If the derivative is a rational function, set numerator

to zero and solve for x.

b) Find all values c for which the derivative does not exist but f(c) is defined. To do this, set

the denominator of the derivative to zero and solve for x.

Secondly, find the maximum or minimum.

Let c-

(any number smaller than c) be a test point to the left of c and let c+

(any number largerthan c) be a test point to the right of c. Then,

a) f(c) is a maximum if )(

c f > 0 (increasing), )(

c f < 0 (decreasing), and )(c f is defined at

x = c.

In general, if the sign of )( x f changes from negative to positive at c, and

)(c f is defined at c, then f(x) has a relative (or local ) minimum at x = c.

b) f(c) is a minimum if )(

c f < 0 (decreasing), )(

c f > 0 (increasing), and )(c f is defined at

x = c.If the sign of )( x f changes from positive to negative at c, and )(c f is

defined at c, then f(x) has a relative (or local ) maximum at x = c.

If the critical value cannot be found in the domain of the original function, then that particular

critical value cannot be classified as a maximum or a minimum. Maximum and minimum valuesmust always exist in the domain of the original function.

The critical points of a function f are the points in the domain of the f where either

a) the tangent line to the curve is horizontal, that is, )( x f = 0, or

b) the points at which )( x f does not exist.

CRITICAL POINTS OF A FUNCTION




EXAMPLE 2

Given the function f(x) = 27 x - x3, - 7 < x < 7.

Find (a) the critical values and

(b) determine the intervals for which the function is increasing and decreasing.Find (c) the relative maximum and minimum values.

Also, find (d) the absolute maximum and minimum values.

Solution We must first obtain the derivative of the function using the following format:

Determine the signs of f

At this point, we have several x-values: the endpoints (x = -7 and x = 7), the critical values(x = -3 and x = 3). Arrange these values in order from smallest to largest .

(If there are no endpoints, then arrange starting from the lowest endpoint (-infinity) to the highest

endpoint (+infinity)).

x )( x f )( x f

x f(x) = 27 x - x3 )( x f = 27 - 3x2

)( x f = 3(9 - x2)

)( x f = 3(3 - x )(3 + x)

Critical values:

a) Set )( x f = 0 and solve for x.

3(3 - x )(3 + x) = 0

x = - 3 or x = 3

b) There are no points at which

)( x f does not exist.

Endpoint

- 7

f(-7) = 27(-7) – (-7)3

= 154

Endpoint

7

f(7) = 27(7) – (7)3

= -154

Critical Value

3

f(3) = 27(3) – (3)3

= 54

Critical Value

-3

f(-3) = 27(-3) – (-3)3

= -54




Arrange values in order from smallest to largest:

x = -7 (endpt) -3(crit) 3(crit) 7(endpt)

Use the arranged values above and test points to determine the signs of f and by creating a

table such as the example below.

Thus, the function is:

o increasing on (-3, 3) and decreasing on [-7, -3) (3, 7].

Also, the function f has:

o local minimum at the point (-3, -54), local maximum at (3, 54),

o absolute maximum at (-7, 154) and absolute minimum at (7, -154).




In this section, we will learn about how to use of the second derivative of a function to detectintervals for which the function is concave upward or concave downward. In addition, we

learn how to find the absolute maximum, absolute minimum, local maximum and local

minimum of a function using the second derivative.

So far we know that the first derivative f provides information about the steepness, or the slope

of a tangent line to the curve at any instant. What information does the second derivative

f provides?

Suppose f is a differentiable function, then, when we take the derivative of the first derivative, we

obtain a new function, f , called the second derivative of f . The second derivative function f ,

can help us to gain information about the behavior of the first derivative function, f .

The second derivative can tell us more about whether the first derivative f is increasing (i.e.,

f > 0), whether the first derivative f is decreasing (i.e., f < 0), or whether the first derivative

f is constant (i.e., f = 0). Thus, f is the rate of growth of f .

If f is positive, then the function f must be increasing. This means that the function f is

changing by having some quantity added to it so that f is actually growing. A differentiable

function is concave upward on an interval if its derivative f is increasing on that interval (that

is, if f > 0).

Similarly, if f is negative, then the function f must be decreasing. This means that f is

changing by having some quantity subtracted from it so that f is actually shrinking. A

differentiable function is concave downward on an interval if its derivative f is decreasing on

that interval (that is, if f < 0).

Concave upward

If we wish to find the range of values of x for which a function is concave upward, just find the

range of values of x where the second derivative f is positive.


)('' x f > 0.

Using the second derivative to determine concavity

A positive second derivative indicates that the first derivative function is increasing.




Concave downward

If we wish to find the range of values of x for which a function is concave downward, just find the

range of values of x where the second derivative f is negative.


)('' x f < 0.

Graphical illustration

Lets now explore the meaning of the phrase concave upward and concave downward. To do this

lets consider the following diagrams.

In both Figure 1 and Figure 2, f is decreasing for every x in the given interval (a, b) since the

first derivative, which is the same as the slope of the tangent line, decreases as we move from left

to right. In figure 1, for instance, the slope of the tangent line to the curve decreases from higher

positive values towards zero, and in figure 2, the slope decreases from zero towards highernegative values.

A negative second derivative indicates that the first derivative function is decreasing.




In both Figure 3 and Figure 4, f is increasing for every x in the given interval (a, b) since the

first derivative, which is the same as the slope of the tangent line, increases as we move from leftto right. In figure 3, for instance, the slope of the tangent line to the curve increases from highernegative values towards zero, and in figure 4, the slope increases from zero towards higher

positive values.

An inflection point

A point P on a curve is called an inflection point if f is continuous there and the curve changes

from concave upward to concave downward or from concave downward to concave upward.




TEST FOR CONCAVITY

Clearly, we can lean about the concavity of a given function by studying the behavior of its first

derivative function, f . We now know that graph is concave down if f is decreasing (i.e.,

f < 0), on the interval and a graph is concave up if f is increasing (i.e., f > 0), on the

interval. Thus, in order to identify the intervals for which a graph is concave down or concave up,it is easier to use the second derivative as the test for concavity.

SECOND DERIVATIVE TEST FOR MAXIMUM AND MINIMUM VALUES

For a given function, at the point where the tangent line is horizontal, a relative maximum occurs

where the curve is concave down, and a relative minimum occurs where the curve is concave up.We can therefore use the sign of the second derivative to determine relative maximum orminimum at the point where the tangent line is horizontal.

o If )(c f = 0 and )(c f > 0, then f has a local or relative minimum at c.

o If )(c f = 0 and )(c f < 0, then f has a local or relative maximum at c.

Suppose f is a function whose second derivative f exists on an open interval I .

a) If f ( x) < 0 for every x in the open interval I , then the graph of f is concave

downward on that interval.

b) If f ( x) > 0 for every x in the open interval I , then the graph of f is concave

upward on that interval.c) If f ( x) = 0 for every x in the open interval I , then the graph of f is linear and

there is no concavity defined on that interval.




EXAMPLE 3

Given the function f(x) = 2 x3

+ 3 x2 – 36 x + 4. Find the (a) the critical values, and (b) the relative

maximum and minimum values, if any (c) the intervals for which the function is increasing and

for which the function is decreasing; (d) the point of inflection; and (e) the intervals for which the

function is concave up or concave down.Solution

x )( x f )(' x f )('' x f

x )( x f =

2 x3 + 3 x2 – 36 x + 4

Domain:(- , )

)(' x f = 6 x2

+ 6 x – 36

= 6( x2 + x - 6)

= 6( x - 2)( x + 3)

Critical values:

a) Set )( x f = 0 and solve

for x.

6( x - 2)( x + 3) = 0

x = - 3 or x = 2

b) There are no points at

which )( x f does not exist.

)('' x f = 12 x + 6

Points of inflection

a) Set f ( x) = 0 and solve

for x.

6(2 x + 1) = 0 x = - ½

b) There are no points at

which f ( x) does not exist .

Critical

Value -3 )3( f = 85 )3( f = 0 f (-3) = - 30 < 0

Maximum at (-3, 85)

Critical

Value 2 )2( f = - 40 )2( f = 0 f (2) = 30 > 0

Minimum at (2, - 40)

Inflection

point - ½ )(21 f = - 13

a) The critical points occurredat x = -3 and x = 2.

b) There is a maximum at (-3,85) and a minimum at (2, -40)

c) The curve is increasing on

(- , -3) (2, )and decreasing on (-3, 2).

d) The point of inflection

occurred at (- ½, -13).

e) The curve is concave down

on the interval ( - , - ½ )and concave up on (- ½, )




Question #1 Find the (a) the critical values, and (b) the relative maximum and minimum values, if

any (c) the intervals for which the function is increasing and for which the function is

decreasing; (d) the point of inflection; and (e) the intervals for which the function is

concave up or concave down, and (f) sketch the graph of f.

)( x f = x3

–

9 x2

–

24 x –

19

Questions








)( x f = 2 x3

–

9 x2

–

12 x








)( x f = –

2 + 12 x –

x3

.








)( x f = 4 x3

–

3 x4

.








)( x f = 1 + 6 x2

–

2 x3

.




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Multiple choice Questions




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