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2.3 Curve Sketching (Introduction)
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We have four main steps for sketching curves:
1. Starting with f(x), compute f’(x) and f’’(x).2. Locate all relative maximum and minimum
points and make a partial sketch.3. Examine concavity of f(x) and locate
inflection points.4. Consider other properties of the graph such
as intercepts and complete the sketch.
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Locating Relative Extreme Points
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The tangent line has a slope of zero at relative maximum and relative minimum points. So, to find relative extreme points, we find values of x so that f’(x) = 0.
Look for possible relative extreme points of f(x) by setting f’(x) = 0 and solving for x.
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Is the point a relative maximum point or a relative minimum point?
How can we tell?
• Check concavity at relative extreme point using second derivative.
• Examine slope of nearby points on either side using the first derivative.
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Locating Inflection Points
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An inflection point can only occur at a value of x for which f’’(x) = 0 because the curve is concave up when f’’(x) is positive and concave down when f’’(x) is negative.
Look for possible points of inflection by setting f’’(x) = 0 and solving for x.
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563)(Graph 2 xxxf1. Find relative extreme points…find x where f’(x) = 0
563)(' 2 xxdx
dxf
rule sum via563 2
dx
dx
dx
dx
dx
d
rule multipleconstant via563 2
dx
dx
dx
dx
dx
d
rulepower via0)1(6)2(3 x66)(' xxf
066 x1
66
x
x Relative extreme point at (-1, f(-1)=-8)
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2. Check concavity at relative extreme point, x = -1.
66)(' xxf
66)(')('' xdx
dxf
dx
dxf
rule sum via66dx
dx
dx
d
rule multipleconstant via66dx
dx
dx
d
rulepower via0)1(6
6)('' xf
So, the second derivative is 6 (concave up) for all x, including x = -1.
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563)(Graph 2 xxxf
(-1, -8)Concave up
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2436152)(Graph 23 xxxxf1. Find relative extreme points…find x where f’(x) = 0
2436152)(' 23 xxxdx
dxf
rule sum via24 36152 23
dx
dx
dx
dx
dx
dx
dx
d
rulemult const via24 36152 23
dx
dx
dx
dx
dx
dx
dx
d
rulepower via0 )1(36)2(15)3(2 2 xx
36306)(' 2 xxxf
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36306)(' 2 xxxf
0 36306 2 xx0 )65(6 2 xx
0 652 xx0 2)3)(x-x(
3x
0 3-x
2x
0 2x
Relative extreme point at (3, f(3)= 3)
Relative extreme point at (2, f(2)= 4)
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2. Check concavity at relative extreme points, x = 2, 3.
36306)(')('' 2 xxdx
dxf
dx
dxf
rule sum via36306 2
dx
dx
dx
dx
dx
d
rulemult const via36 306 2
dx
dx
dx
dx
dx
d
3012)('' xxf
36306)(' 2 xxxf
rulepower via0 )1(30)2(6 x
down concave , 630)2(12)2('' f
up concave , 630)3(12)3('' f
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3. Find inflection points, f’’(x) = 0
3012)('' xxf
03012 x
5.22
5
12
30
3012
x
x
Inflection point at (2.5, f(2.5)=3.5)
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(2, 4)Concave down
(3, 3)
Concave up
(2.5, 3.5)Inflection point
2436152)(Graph 23 xxxxf