2.3 curve sketching (introduction). we have four main steps for sketching curves: 1.starting with...
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2.3 Curve Sketching (Introduction)
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.
Locating Relative Extreme Points
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.
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.
Locating Inflection Points
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.
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)
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.
563)(Graph 2 xxxf
(-1, -8)Concave up
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
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)
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
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)
(2, 4)Concave down
(3, 3)
Concave up
(2.5, 3.5)Inflection point
2436152)(Graph 23 xxxxf