curve sketching(3)
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Curve Sketching
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Relative Extrema
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Definition. A function f has a relative
maximum value at a number c if there
exists an open interval I containing c such that ( ) ( ) f c f x≥ ∀ x in I .
a bc
f (c) f (c)
a bc
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Definition. A function f has a relative
minimum value at a number c if there
exists an open interval I containing c such that ( ) ( ) f c f x≤ ∀ x in I .
a b
f (c) f (c
)a bc c
If a function has either a relative maximum or relative minimum
value at c, then the function has a relative extremum at c.
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Illustration: Consider a function hose graph
is shon belo, at numbers does the function
have relative extrema!
A
C
B"he function has relative minima at
x # $ and x # % and a relative
maximum at x # &.
A, C ' relative minimum pts.
B ' relative maximum pt.
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Definition. If c is a number in the domain of
the function f , and if either f ()c* # + or f ()c*
does not exist, then c is called a critical
number of f .
Illustration $ Determine the critical numbers ofthe function defined b- ( ) 2 8 2. f x x x= − +
Solution
( )' 2 8 f x x= −
'( ) 0 f x =
ote that f () x* is + hen x # & and there is no value of x that
ill make f () x* undefined.
"hus, the onl- critical number of f is &.
4 x
⇒ =2 8 0 x
⇒ − =
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Illustration / Determine the critical numbers of
the function defined b- ( )
6 15 5
12 . f x x x= −
Solution
( ) 1 4
5 56 12'5 5
f x x x−
= − ( ) ( )45
45
6 26 25 5
x x x x
− −= − =
ote that f () x* is + hen x # / and it is undefined hen x # +.
"hus, the critical numbers of f are + and /.
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0onotonicit-
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( ) ( )1 2 f x f x<
1 2 1 2, . x x I x x∀ ∈ <such that
Definition. A function f is increasing on an
interval I if and onl- if
x1 x2
f ( x2)
f ( x1)
x1 x2
f ( x2)
f ( x1)
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( ) ( )1 2 f x f x>
1 2 1 2, . x x I x x∀ ∈ <such that
Definition. A function f is decreasing on an
interval I if and onl- if
x1 x2
f ( x1)
f ( x2)
x1 x2
f ( x1)
f ( x2)
Definition. A function f is said to be monotonic
on an interval I if it is either increasing or
decreasing on I .
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Illustration: Determine hich of the folloing graphs
represent an increasing or a decreasing function.
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[ ]b,a( )b,a
( )' 0 f x >
( )b,a[ ], .a b
( )' 0 f x < ( )b,a[ ], .a b
Theorem 1 Suppose a function f is
continuous on the closed interval anddifferentiable on the open interval .
a. If for all x in then f
isincreasing on
b. If for all x in then f isdecreasing on
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Theorem 2 The First-Derivative Test for
Relative Extrema
1et f be a function hich is continuous
at each number in some open interval
containing the number c and suppose thatexists at each number in except possibl-
at c.
( )b,a'f ( )b,a
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a. If for all values of x in some openinterval having c as its right endpoint,
and if for all values of x in someopen interval having c as its leftendpoint, then f has a relative
maximum value at c.
( )' 0 f x >
( )' 0 f x <
a bc
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b. If for all values of x in some openinterval having c as its right endpoint,
and if for all values of x in someopen interval having c as its leftendpoint, then f has a relative minimum
value at c.
( )' 0 f x <
( )' 0 f x >
a bc
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2rocedure in determining the relative
extrema of a function f $. Compute .
/. Determine the critical numbers of f )i.e.,
the value)s* of x for hich or
does not exist*.
3. Appl- the 4irst5Derivative "est.
( ) x 'f
( )' 0 f x = ( ) x 'f
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Example 1. 1et .
Determine the intervals on hich f is
increasing and the intervals here f isdecreasing. Also, determine the relative
extrema of f , if an-.
( ) 162 3
+−= x x x f
( ) 66' 2
−= x x f
( )' 0 f x = 26 6 0 1 x x⇒ − = ⇒ = ±
Solution
:Step 1. Find f ’( x).
Step2. Determine the critical numbers of f .
critical numbers: 1 and 1
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( )1!−∞− ( )1!1− ( )+∞!1
ep 3. "onsider the inter#als determined b$ 1 and
%hen determine the si&n of f () x* in each
interval and appl- "heorem &.3.$ and the
4irst5Derivative "est.
( ) x f ( ) x f '
( )1!−∞−
1−= x
( )1!1−
1= x
( )+∞!1
Conclusion
6 f is increasing
7 + f has a relativemaximum value
5 f is decreasing
53 + f has a relativeminimum value
6 f is increasing
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Interval here f is increasing ( ) ( ), 1 1,−∞ − ∪ +∞
( )1, 3−
( )1!1−Interval here f is decreasing
Relative minimum point :
( )1,5−
Relative maximum point :
( ) x f ( ) x f '
( )1!−∞−
1−= x
( )1!1−
1= x
( )+∞!1
Conclusion
6 f is increasing
7 + f has a relativemaximum value
5 f is decreasing
53 + f has a relativeminimum value
6 f is increasing
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Concavit-
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Definition. "he graph of a function f is said to be
concave upward at a point( )( )cf ,c if
( )c'f a. exists and
b. there is an open interval I containing
c such that for all values of x in I such
that , the point is above the tangent line to the graph at .
x ≠ ( )( ) x f , x ( )( )cf ,c
c c
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Definition. "he graph of a function f is said to be
concave downward at a point( )( )cf ,c if
( )c'f a. exists and
b. there is an open interval I containing
c such that for all values of x in I such
that , the point is belo the tangent line to the graph at .
x ≠ ( )( ) x f , x ( )( )cf ,c
c c
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Illustration: Determine hich of the folloing graphs
are concave upard or concave donard.
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Theorem 1et f be a function hich is
differentiable on some open intervalcontaining c.
( )'' 0 f c >
( )( )cf ,c
( )'' 0 f c <
( )( )cf ,c
a. If , the graph of f is concave
upard at .
b. If , the graph of f is concavedonard at .
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2oint of Inflection
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Definition. A point is a point of inflection of
the graph of the function f
ifa. the graph has a tangent line at and
b. there exists an open interval I containing
c such that if x is in I , then either
i. henever , and
henever 8 or
ii. henever
, and
henever .
( )( )cf ,c
( )'' 0 f x < c x <
( )'' 0 f x > x >
c x <
x >
)c, f )c**
)c, f )c**( )'' 0 f x >
( )'' 0 f x <
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Illustration: In 4igures &.3.9)a* and &.3.9)b*, e see portions of
the graphs of to different functions, each having a point of
inflection at .( )( )cf ,c
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Theorem ! The Secon"-Derivative Testfor Relative Extrema1et c be a critical number of a function f athich .Suppose further that f '’
exists for each value of x in some openinterval containing c.
a. If ,then f has a relative
maximum value at c. b. If , then f has a relativeminimum value at c.
( )' 0 f c =
( )'' 0 f c <
( )'' 0 f c >
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Example 2. 1et .
Determine the intervals here the graph
of f is concave upard and concave
donard. Also, determine the inflection
points of the graph of f , if an-.
( ) 162 3
+−= x x x f
( )'' 12 0 f x x= = 0 x⇒ =
Solution
:Step 1. Find f ’’( x) and e:uate to ;ero to find the
possible <point)s* of inflection=.
( ) 66'
2−=
x x f
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( ), 0−∞ ( )0, +∞
tep 2. "onsider the inter#als determined b$ .
%hen determine the si&n of f (() x* in each
interval and appl- "heorem &.3.3 and the
Second5Derivative "est.
( ) x f ( ) x f '
( ), 0−∞
0 x =
( )0, +∞
Conclusion
5 graph of f is concavedonard
$ 59 + graph of f has point ofinflection
6 graph of f is concaveupard
( )'' 12 0 f x x= =( ) 66'
2
−= x x f
( )'' f x
( ) 162 3
+−= x x x f
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( ) x f ( ) x f '
( ), 0−∞
0 x =
( )0, +∞
Conclusion
5 graph of f is concavedonard
$ 59 + graph of f has point ofinflection
6 graph of f is concaveupard
( )'' f x
Interval here graph of
f is concave upard ( )0, +∞
( )0,1
( ), 0−∞
Inflection point :
Interval here graph of f is concave donard
( ) 162 3 +−= x x x f
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In order to make the graphing of a given
function f easier, e summari;e the things e
have done as follos.
Step 2. "onsider all the critical numbers of f
and thesolution to the euation f ’’( x) * .
+rranðese numbers in their natural order.Step 3. ,btain the inter#als determined b$
these numbers.
Step 1. Sol#e for f ’) x* and f
’’( x).
Step -. +ppl$ theorems 2 and 3 and the rstand
second deri#ati#e tests for relati#e
e/trema.
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Example . Sketch the graph of the
function f defined b- ( ) 1623
+−= x x x f
( )'' 12 f x x=
0 x⇒ =
Solution:
Step 1. ( ) 66' 2
−= x x f
Step 2. 12 0 x =2
6 6 0 x − =
1 x⇒ = ±
Arrange these numbers in the real number line.
-1 # 1
( )1!−∞− ( )(!1− ( )1!( ( )+∞!1 Step 3.
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Step -.
( ) x f ( ) x f ' ( ) x f 0
( )1!−∞− +
1−= x (
( )(!1−
(= x
( )1!(1= x 3− (
( )+∞!1 +
Conclusion
( ) 162 3
+−= x x x f ( )'' 12 f x x=( ) 66' 2
−= x x f
1 −
−
(
+−
−
+
+
−
−
f has relative maximum value
f is decreasing8 graph is concave donard
f is decreasing8 graph is concave upard
f is increasing8 graph is concave upard
f is increasing8 graph is concave donard
f has relative minimum value
graph has a point of inflection
Step
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Step .
f ↓, CD f ↑, CD
f ↑, C> f ↓, C>
+ $5$
( ) x f
( )1!−∞−
1−= x
( )(!1−
(= x
( )1!(
1= x 3−( )+∞!1
Conclusion
1
f has relative maximum value
f is decreasing8 graph is concave donard
f is decreasing8 graph is concave upard
f is increasing8 graph is concave upard
f is increasing8 graph is concave donard
f has relative minimum value
graph has a point of inflection
)5$,7*
)+,$*
)$,53*
( )xf ( )xf ' ( )xf0
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( ) x f ( ) x f ' ( ) x f
( )1!−∞− +
1−= x (
( )(!1−
(= x
( )1!(
1= x 3− (
( )+∞!1 +
Conclusion
1 −
−
(
+−
−
+
+
−
−
f has relative maximum value
f is decreasing8 graph is concave donard
f is decreasing8 graph is concave upard
f is increasing8 graph is concave upard
f is increasing8 graph is concave donard
f has relative minimum value
graph has a point of inflection
f ↓
f ↑
CD
C>
(-1,1)
( ) ( ), 1 1,−∞ − ∪ +∞
Rel min pt
( ), 0−∞
( )0, +∞
Rel max pt
(1,-3)
(-1,5)
(0 1)
( ) 32 6 1 f x x x= − +