Transcript CHAPTER 3 SECTION 3.6 CURVE SKETCHING

CHAPTER 3
SECTION 3.6
CURVE SKETCHING
Graph the following functions on your
calculator and make a sketch.
What do you notice about the graph of a
function where there is a factor that
appears twice?
1) y=(x+1)(x-2)2
2) y= x(x+3)2
3) y=(x+1)(x-3)2
* Where there is a factor that appears twice (double
root) the graph “bounces” at that zero.
Graph the following functions on your calculator
and make a sketch.
What do you notice about the graph of a
function where there is a factor that appears
three times?
1) y=x3(x-3)
2) y= (x+2)(x-1)3
3) y=(x-2)(x+1)3
* Where there is a factor that appears three times
(triple root) the graph “wiggles” at that zero.
Graph the following functions on your
calculator and make a sketch.
What do you notice about the graph of a
function where there is a factor that appears
four times?
1) y=x4(x-2)
2) y= (x-1)4(x+1)
3) y=(x+1)4(x-1)
* Where there is a factor that appears four times the
graph “bounces” at that zero.
What do you think the graph of a function would look like at
the zero that corresponds to a factor that appears n times?
• If a linear factor appears once – the graph “goes
through” the x-axis at that zero
• If a linear factor appears an even number of times- the
graph “bounces” at that zero.
• If a linear factor appears an odd number of times
(greater than 1) – the graph “wiggles” at that zero.
What will the graph look like?
f(x)= x(x-5)3(x+4)
What will the graph look like?
y= (x-1)2 (x+3)(x+1)5
Write an equation for the graph.
-5
5
f ( x)  x( x  5)( x  4)
2
Write an equation for the graph.
-5
5
f ( x)  ( x  3)( x  2) ( x  6)
3
Write an equation for the graph.
-5
f ( x)  ( x  4) ( x 1)( x  3)
2
2
DISAIMIS
O
M
A
I
N
N
T
E
R
C
E
P
T
S
Y
M
M
E
T
R
Y
S
Y
M
P
T
O
T
E
S
N
T
E
R
V
A
L
S
A
X
M
I
N
N
F
L
E
C
T
I
O
N
K
E
T
C
H
Strategy
•
•
•
•
•
•
USE ALGEBRA FIRST WITH A T-CHART
Determine domain of function
Find y-intercepts, x-intercepts (zeros)
Check for vertical, horizontal asymptotes
Determine values for f '(x) = 0, critical points
Determine f ''(x)
– Gives inflection points
– Test for intervals of concave up, down
• Plot intercepts, critical points, inflection points
• Connect points with smooth curve
• Check sketch with graphing calculator
GUIDELINES FOR
SKETCHING A CURVE
• The following checklist is intended as a
guide to sketching a curve y = f(x) by
hand.
– Not every item is relevant to every function.
– For instance, a given curve might not have
an asymptote or possess symmetry.
– However, the guidelines provide all the information
you need to make a sketch that displays the most
important aspects of the function.
A. DOMAIN
• It’s often useful to start by determining
the domain D of f.
– This is the set of values of x for which f(x)
is defined.
B. INTERCEPTS
• The y-intercept is f(0) and this tells us
where the curve intersects the y-axis.
• To find the x-intercepts, we set y = 0
and solve for x.
– You can omit this step if the equation is difficult
to solve.
C. SYMMETRY—EVEN
FUNCTION
•If f(-x) = f(x) for all x in D, that is, the
equation of the curve is unchanged when x
is replaced by -x, then f is an even function
and the curve is symmetric about the yaxis.
– This means that our work is cut in half.
C. SYMMETRY—EVEN
FUNCTION
• If we know what the
curve looks like for x
≥ 0, then we need
only reflect about the
y-axis to obtain the
complete curve.
C. SYMMETRY—EVEN
FUNCTION
• Here are some
examples:
–
–
–
–
y = x2
y = x4
y = |x|
y = cos x
C. SYMMETRY—ODD
FUNCTION
• If f(-x) = -f(x) for all x in D, then f
is an odd function and the curve is
symmetric about the origin.
C. SYMMETRY—ODD
FUNCTION
• Again, we can
obtain the complete
curve
if we know what it
looks like for x ≥ 0.
– Rotate 180°
about the origin.
C. SYMMETRY—ODD
FUNCTION
• Some simple
examples of odd
functions are:
–
–
–
–
y=x
y = x3
y = x5
y = sin x
C. SYMMETRY—
PERIODIC FUNCTION
• If f(x + p) = f(x) for all x in D, where p
is a positive constant, then f is called
a periodic function.
• The smallest such number p is called
the period.
– For instance, y = sin x has period 2π and y = tan x
has period π.
C. SYMMETRY—
PERIODIC FUNCTION
• If we know what the graph looks like in an
interval
of length p, then we can use translation to
sketch
the entire graph.
D. ASYMPTOTES—
HORIZONTAL
• Recall from Section 2.6 that, if either
lim f ( x)  ,L
lim f ( x)  L or
x 
x 
then the line y = L is a horizontal asymptote
of the curve y = f (x).
f ( x)   (or -∞), then
– If it turns out that lim
x 
we do not have an asymptote to the right.
– Nevertheless, that is still useful information
for sketching the curve.
D. ASYMPTOTES—VERTICAL
Equation 1
• Recall that the line x = a is a vertical
asymptote if at least one of
the following statements is true:
lim f ( x)  
lim f ( x)  
x a
x a
lim f ( x)  
lim f ( x)  
x a
x a
D. ASYMPTOTES—VERTICAL
• For rational functions, you can locate
the vertical asymptotes by equating
the denominator to 0 after canceling any
common factors.
– However, for other functions, this method
does not apply.
D. ASYMPTOTES—VERTICAL
• Furthermore, in sketching the curve, it is
very useful to know exactly which of the
statements in Equation 1 is true.
– If f(a) is not defined but a is an endpoint of lim f ( x)
x a
the domain of f, then you should compute
f ( x) , whether or not this limit is infinite.
or xlim

a
D. ASYMPTOTES—SLANT
• Slant asymptotes are discussed
at the end of this section.
E. INTERVALS OF INCREASE
OR DECREASE
• Use the I /D Test.
• Compute f’(x) and find the intervals
on which:
– f’(x) is positive (f is increasing).
– f’(x) is negative (f is decreasing).
F. LOCAL MAXIMUM AND
MINIMUM VALUES
• Find the critical numbers of f (the numbers c
where f’(c) = 0 or f’(c) does not exist).
• Then, use the First Derivative Test.
– If f’ changes from positive to negative at
a critical number c, then f(c) is a local maximum.
– If f’ changes from negative to positive at c,
then f(c) is a local minimum.
F. LOCAL MAXIMUM AND
MINIMUM VALUES
• Although it is usually preferable to use the
First Derivative Test, you can use the
Second Derivative Test if f’(c) = 0 and
f’’(c) ≠ 0.
• Then,
– f”(c) > 0 implies that f(c) is a local minimum.
– f’’(c) < 0 implies that f(c) is a local maximum.
G. CONCAVITY AND POINTS
OF INFLECTION
• Compute f’’(x) and use the Concavity
Test.
• The curve is:
– Concave upward where f’’(x) > 0
– Concave downward where f’’(x) < 0
G. CONCAVITY AND POINTS
OF INFLECTION
• Inflection points occur
where the direction of concavity
changes.
H. SKETCH AND CURVE
• Using the information in items A–G,
draw the graph.
– Sketch the asymptotes as dashed lines.
– Plot the intercepts, maximum and minimum points,
and inflection points.
– Then, make the curve pass through these points,
rising and falling according to E, with concavity
according to G, and approaching the asymptotes
H. SKETCH AND CURVE
• If additional accuracy is desired near
any point, you can compute the value of
the derivative there.
– The tangent indicates the direction in which
the curve proceeds.
Example 1
• Use the guidelines to sketch
2
the curve
2x
y 2
x 1
Example 1
• A. The domain is:
{x | x2 – 1 ≠ 0} = {x | x ≠ ±1}
= (-∞, -1) U (-1, -1) U (1, ∞)
• B. The x- and y-intercepts are both 0.
Example 1
• C. Since f(-x) = f(x), the function
is even.
– The curve is symmetric about the y-axis.
Example 1
• D.
2
2x
2
lim 2
 lim 

2
2
x  x  1
x 
1  1/ x
Therefore, the line y = 2 is a horizontal
asymptote.
Example 1
• Since the denominator is 0 when x = ±1,
we compute the following limits:
2x2
2x2
lim 2

lim 2
 
x 1 x  1
x 1 x  1
2
2
2x
2x
lim 2
 
lim 2

x 1 x  1
x 1 x  1
– Thus, the lines x = 1 and x = -1 are vertical
asymptotes.
Example 1
• This information
about limits and
asymptotes enables
us to draw the
preliminary sketch,
showing the parts of
the curve near the
asymptotes.
Example 1
4 x( x  1)  2 x  2 x
4 x
f '( x) 
 2
2
2
2
( x  1)
( x  1)
2
......
2



1
......
...... ......
.... ...
..... f '(x)
f '(2) .86 1 f '(.5)  3.56 0 f '(.5) 3.56 1 f '(2) .89
...... Increasin g ....
...... Increasin g ..... ...... decreasin g ...... ...... decreasin g ....
1
0
1
f (x)
Example 1
• F. The only critical number is x = 0.
• Since f’ changes from positive to negative
at 0, f(0) = 0 is a local maximum by the
First Derivative Test. (1 and -1 are not in
the domain!!!!!!)
Example 1
4( x2  1)2  4 x  2( x 2  1)2 x 12 x 2  4
f ''( x) 
 2
2
4
( x  1)
( x  1)3
   
   


f (2)  1.93 1
'


f '(2)  .86 1
Increa sin g
   Concave UP 







1
'













f
( x)
'
'
'
f (.5)  16.6
0
f (.5)  16.6 1
f (2)  1.93
 
 
   f ' ( x)
f '(.5)  3.56
0
f '(.5)  3.56 1
f '(2)  .89
Increa sin g
Increa sin g
Increa sin g
 Concave DOWN   Concave DOWN   Concave UP    f ( x)
1
0
1
Example 1
• It has no point of inflection since 1 and -1
are not in the domain of f.!!!!!!!!!!!
Example 1
• H. Using the
information in E–G,
we finish the
sketch.
Example 2
• Sketch the graph of:
f ( x) 
2
x
x 1
Example 2
• A. Domain = {x | x + 1 > 0}
= {x | x > -1}
= (-1, ∞)
• B. The x- and y-intercepts are both 0.
• C. Symmetry: None
Example 2
x2
lim

• D. Since x x  1
, there is no
horizontal asymptote.
• Since
•
+ and f(x) is
as
x
→
-1
x 1  0
2
lim
always positive, we have x1
and so the line x = -1 is a vertical
asymptote
x

x 1 ,
Example 2
2 x x  1  x 1/(2 x  1) x(3x  4)
f '( x) 

x 1
2( x  1)3/ 2
• E.
2
• We see that f’(x) = 0 when x = 0 (notice
that
-4/3 is not in the domain of f).
– So, the only critical number is 0.
Example 2
• As f’(x) < 0 when -1 < x < 0 and f’(x) >
0 when x > 0, f is:
– Decreasing on (-1, 0)
– Increasing on (0, ∞)
Example 2
• F. Since f’(0) = 0 and f’ changes from
negative to positive at 0, f(0) = 0 is
a local (and absolute) minimum by
the First Derivative Test.
Example 2
2( x  1) (6 x  4)  (3 x  4)3( x  1)
f ''( x) 
3
4( x  1)
• G.
3x 2  8 x  8

4( x  1)5/ 2
3/ 2
2
– Note that the denominator is always positive.
– The numerator is the quadratic 3x2 + 8x + 8,
which is always positive because its discriminant
is b2 - 4ac = -32, which is negative, and the coefficient
of x2 is positive.
1/ 2
Example 2
• So, f”(x) > 0 for all x in the domain of f.
• This means that:
– f is concave upward on (-1, ∞).
– There is no point of inflection.
Example 2
• H. The curve is
sketched here.
• Sketch the graph of:
cos x
f ( x) 
2  sin x
Example 3
• A. The domain is R
• B. The y-intercept is f(0) = ½.
The x-intercepts occur when cos x =0,
that is, x = (2n + 1)π/2, where n is an
integer.
Example 3
• C. f is neither even nor odd.
– However, f(x + 2π) = f(x) for all x.
– Thus, f is periodic and has period 2π.
– So, in what follows, we need to consider only 0 ≤ x ≤ 2π
and then extend the curve by translation in part H.
• D. Asymptotes: None
Example 3
• E.
(2  sin x)( sin x)  cos x(cos x)
f '( x) 
(2  sin x) 2
2sin x  1

2
(2  sin x)
• Thus, f’(x) > 0 when 2 sin x + 1 < 0
sin x < -½  7π/6 < x < 11π/6

Example 3
• Thus, f is:
– Increasing on (7π/6, 11π/6)
– Decreasing on (0, 7π/6) and (11π/6, 2π)
Example 3
• F. From part E and the First Derivative
Test, we see that:
– The local minimum value is f(7π/6) = -1/
3
– The local maximum value is f(11π/6) = -1/ 3
Example 3
• G. If we use the Quotient Rule again
and simplify, we get:
2 cos x(1  sin x)
f ''( x)  
3
(2  sin x)
• (2 + sin x)3 > 0 and 1 – sin x ≥ 0 for all x.
• So, we know that f’’(x) > 0 when cos x < 0,
that is, π/2 < x < 3π/2.
Example 3
• Thus, f is concave upward on (π/2, 3π/2)
and concave downward on (0, π/2) and
(3π/2, 2π).
• The inflection points are (π/2, 0) and
(3π/2, 0).
Example 3
• H. The graph of the
function restricted
to 0 ≤ x ≤ 2π is
shown here.
Example 3
• Then, we extend it,
using periodicity,
to the complete
graph here.
SLANT ASYMPTOTES
• Some curves have asymptotes that
are oblique—that is, neither
horizontal nor vertical.
SLANT ASYMPTOTES
• For rational functions, slant asymptotes
occur when the degree of the numerator is
one more than the degree of the
denominator.
– In such a case, the equation of the slant
asymptote can be found by long division—
as in following example.
Example 6
• Sketch the graph of:
3
x
f ( x)  2
x 1
Example 6
SLANT ASYMPTOTES
• A. The domain is: R = (-∞, ∞)
• B. The x- and y-intercepts are both 0.
• C. As f(-x) = -f(x), f is odd and its graph is
symmetric about the origin.
Example 6
SLANT ASYMPTOTES
• Since x2 + 1 is never 0, there is no vertical
asymptote.
• Since f(x) → ∞ as x → ∞ and f(x) → -∞ as
x → - ∞, there is no horizontal asymptote.
Example 6
SLANT ASYMPTOTES
• However,3long division gives:
x
x
f ( x)  2
 x 2
x 1
x 1
1
x
x
f ( x)  x   2

 0 as x  
1
x 1
1 2
x
– So, the line y = x is a slant asymptote.
Example 6
• E.
SLANT ASYMPTOTES
3x ( x  1)  x  2 x x ( x  3)
f '( x) 
 2
2
2
( x  1)
( x  1)2
2
2
3
2
2
• Since f’(x) > 0 for all x (except 0), f is
increasing on (- ∞, ∞).
Example 6
SLANT ASYMPTOTES
• F. Although f’(0) = 0, f’ does not
change sign at 0.
– So, there is no local maximum or minimum.
Example 6
SLANT ASYMPTOTES
(4 x3  6 x)( x 2  1) 2  ( x 4  3 x 2 )  2( x 2  1)2 x
2
4
• G. f ''( x) 
( x  1)
2 x(3  x 2 )

( x 2  1)3
– Since f’’(x) = 0 when x = 0 or x = ±
we set up the following chart.
3,
Example 6
SLANT ASYMPTOTES
• The points of inflection are:

–
3,  34 3
– (0, 0)
–

3,
3
4
3


Example 6
SLANT ASYMPTOTES
• H. The graph of f
is sketched.
First derivative:
y is positive
Curve is rising.
y is negative
Curve is falling.
y is zero
Possible local maximum or
minimum.
Second derivative:
y is positive
Curve is concave up.
y is negative
Curve is concave down.
y is zero
Possible inflection point
(where concavity changes).

Example:
Graph
y  x  3x  4   x  1 x  2 
3
2
2
There are roots at x  1 and x  2 .
y  3x  6x
2
Set
y  0
0  3x 2  6 x
Possible extreme at x  0, 2 .
We can use a chart to organize our thoughts.
First derivative test:
y

0
0  x2  2 x
0  x  x  2
x  0, 2

0
0
2
y 1  3 12  6 1  3
y  1  3  1  6  1  9
2
y 3  3  32  6  3  9

negative
positive
positive

Example:
Graph
y  x  3x  4   x  1 x  2 
3
2
2
There are roots at x  1 and x  2 .
y  3x  6x
2
Set
y  0
0  3x 2  6 x
Possible extreme at x  0, 2 .
First derivative test:
y
0  x2  2 x
0  x  x  2

0

0

0
2
maximum at
x0
minimum at
x2
x  0, 2


Example:
Graph
y  x  3x  4   x  1 x  2 
3
2
2
NOTE: On the AP Exam, it is not sufficient to simply draw
the chart and write the answer. You must give a written
explanation!
y  3x2  6x
First derivative test:
y

0
0

0

2
There is a local maximum at the point (0,4) because y  0
for all x in (, 0)and y  0 for all x in the interval (0,2) .
There is a local minimum at (2,0) because y  0 for all x in
the interval (0,2) and y  0 for all x in (2, ) .

Example:
Graph
y  x  3x  4   x  1 x  2 
3
2
2
There are roots at x  1 and x  2 .
y  3x2  6x
Possible extreme at x  0, 2 .
Or you could use the second derivative test:
y  6 x  6
y  0  6  0  6  6
y  2  6  2  6  6
Because the second derivative at
x = 0 is negative, the graph is
concave down and therefore (0,4) is a
local maximum.
Because the second derivative at
x = 2 is positive, the graph is
concave up and therefore (2,0) is a
local minimum.

Example:
Graph
y  x  3x  4   x  1 x  2 
3
2
2
We then look for inflection points by setting the second
derivative equal to zero.
y  6 x  6
0  6x  6
6  6x
1 x
Possible inflection point at x  1 .
y

0

1
y  0  6  0  6  6
negative
y  2  6  2  6  6
positive
There is an inflection point at x = 1 because the second
derivative changes from 
negative
to positive.
inflection
point at x  1

Make a summary table:
x
y
y
y
1
0
9
12
0
4
0
6
1
2
3
0
falling, inflection point
2
0
0
6
local min
3
4
9
12
rising, concave down
local max
rising, concave up
5
4
3
2
1
0
-2
-1
0
-1
1
2
3
4
p
x 2  2x  4
Sketch f  x  
x 2
Include intercepts, asymptotes, extrema, inflection points, and intervals of concavity.
x 2  2x  4
Sketch f  x  
x 2
VA at x = 2
Include intercepts, asymptotes, extrema, inflection points, and intervals of concavity.
 x  2 2x  2   x 2  2x  4  1
f 'x 
2
x

2
 


2x 2  6 x  4  x 2  2x  4
x 2  4x
 x  2
2
 x  2

2
x  x  4
 x  2
2
crit #'s: x  0,4,2
 x  2  2x  4   x 2  4x  2  x  21
f ''  x  
4
 x  2
2
2  x  2  x  2   x 2  4x 



4
 x  2
2  x 2  4 x  4  x 2  4 x 
2  4
8


crit #'s: x  2

3
3
3
 x  2   x  2
 x  2
2
Intervals:
Test values:
f ’’(test pt)
f(x)
f ’(test pt)
f(x)
Asymptotes:
Intercepts:
x 2  2x  4
f x 
x 2
Intervals:
f 'x 
x  x  4
 x  2
 ,0 0,2 2,4  4,
1

3
5
f ’’(test pt)
1



f(x)
down
down
up
up
f ’(test pt)




f(x)
inc
dec
dec
inc
Test values:
rel max
f 0  2
x 0
0, 2
No inf
pts
2
f ''  x  
8
 x  2
3
Intercepts:
x 2  2x  4  0
No x-int
4
y
2
y  2
rel min
x4
 4,6 f  4  6
 x  2  4x 
 x 2  2x  4 
 lim 

lim 


2
x

x 
 1 x 
2 1 2 4
 x 2 
Asymptotes:
Oblique Asymptote at y  x
VA at x = 2
1
2
0
0
4
remainder
10
5
-10
10
-5
-10
10
5
-10
10
-5
-10
Example
Sketch: f ( x)  x3  6x2  9x  1
1. Domain: (−∞, ∞).
2. Intercept: (0, 1)
f ( x)   and lim f ( x)  
3. lim
x 
x 
4. No Asymptotes
5. f ( x)  3x2 12 x  9; f inc. on (−∞, 1) U (3, ∞), dec. on (1, 3).
6. Relative max.: (1, 5); relative min.: (3, 1)
7. f ( x)  6 x  12; f concave down (−∞, 2); up on (2, ∞).
8. Inflection point: (2, 3)
3
2
f
(
x
)

x

6
x
 9x  1
Sketch:
1,5
2,3
3,1
 0,1
Example
2x  3
Sketch: f ( x) 
x3
1. Domain: x ≠ −3
2. Intercepts: (0, −1) and (3/2, 0)
2x  3
2x  3
 2 and lim
2
3. lim
x  x  3
x  x  3
4. Horizontal: y = 2; Vertical: x = −3
9
f is increasing on (−∞,−3) U (−3, ∞).
5. f ' ( x) 
2
( x  3)
6. No relative extrema.
 18
7. f ' ' ( x) 
( x  3)3
8. No inflection points
f is concave up on (−∞,−3) and
f is concave down on (−3, ∞).
2x  3
Sketch: f ( x) 
x3
y=2
 3 
  ,0
 2 
 0, 1
x = −3
2
f
x

x
 6x  5
Sketch  
Include intercepts, asymptotes, extrema, inflection points, and intervals of concavity.
10
5
-10
10
-5
-10
2
f
x

x
 6x  5
Sketch  
Include intercepts, asymptotes, extrema, inflection points, and intervals of concavity.
Compare to
g  x   x 2  6x  5
g '  x   2x  6
crit #'s: x  3
Rel min at (3, -4)
What’s the difference between f(x) and g(x)?
The abs val makes all the
negative parts of g(x) reflect
above the x-axis.
Parabola opening upward (no asymptotes)
g ''  x   2
crit #'s: none
10
Always concave up
x-int
x 2  6x  5  0
 x  5 x  1  0
x  1, 5
5
y-int
y 5
-10
10
g(x)
-5
-10
f  x   x 2  6x  5
g  x   x 2  6x  5
x-int: x = 1 and 5
10
y-int: y = 5
f(x)
5
Rel max at (3, 4)
Rel min at (1, 0)
Rel min at (5, 0)
Inf pts none
conc up  ,1 and 5, 
conc down 1,5
-10
10
-5
-10