Math 1304 Calculus I
4.05 - 4.06 Curve Sketching
Major Concerns for Curve Sketching
A.
B.
C.
D.
E.
F.
G.
H.
Domain
Intercepts
Symmetry
Asymptotes
Intervals of Increase or Decrease
Local Maximum and Minimum
Concavity and Points of Inflection
Sketch the Curve
Recall: Graphing Method
• Start with a function f(x).
• Compute formulas for f’(x) and f”(x).
• Find all points of discontinuity, points where the first derivative is
zero or undefined (critical points), and all points where the second
derivative is zero (possible points of inflection)
• Make a table of f(x), f’(x), and f”(x) for all these interesting points,
making room for test points in between interesting points.
• Use the first derivative to find increasing/decreasing behavior
• Use the second derivative to determine concavity and points of
inflection.
• Use the first and second derivative tests to find all local extrema.
• Plot this information and sketch the graph.
Example
• Sketch: f[x] = -3 + 4 x2 - x4
Example
• Sketch: f[x] = -3 + 4 x2 - x4
Solution
Proceed with each of the following
A.
B.
C.
D.
E.
F.
G.
H.
Domain
Intercepts
Symmetry
Asymptotes
Intervals of Increase or Decrease
Local Maximum and Minimum
Concavity and Points of Inflection
Sketch the Curve
Domain
f[x] = -3 + 4 x2 - x4 is a polynomial
A. So domain is: (-Infinity, Infinity)
Intercepts
y intercept:
f[x] = -3 + 4 x2 - x4
x=0  y = -3
So (0, -3) is the y intercept.
x intercept:
f[x] = -3 + 4 x2 - x4
y=0  -3 + 4 x2 - x4 = 0
Set x4 - 4 x2 + 3 = 0
(x2 – 1)(x2 – 3) = 0
x = ± 1 or x = ± √3
So x intercepts are: (-1,0) or (1,0) or (-√3,0) or (√3,0)
Symmetry
• The polynomial is even, so we have
symmetry across the y axis.
Asymptotes
• There are no vertical asymptotes
• Horizontal asymptotes
Limit as x -Infinity is –Infinity
Limit as x Infinity is -Infinity
Therefore there are no horizontal asymptotes.
Now use the derivatives
f[x]= -3 + 4 x^2 - x^4
f'[x]= 8 x - 4 x^3
f''[x]= 8 -12 x^2
Critical Points
f’[x]==0
 8 x - 4 x^3==0
Solutions:
{{x->0.},{x->-1.41421},{x->1.41421}}
Inflection Points
f’’[x]==0
8 -12 x^2==0
Possible solutions:
{{x->-0.816497},{x->0.816497}}
End Points
End Pts: x==-2.||x==2.
Points of Interest
x==-2.||x==-1.41421||x==0.816497||x==0.||x==0.816497||x==1.4142
1||x==2.
Number of Real Pts of Interest: 7
Make a Table of Values
Fill in the Table
Transfer to a graph
Sketch the Graph
This uses the curve drawing in PowerPoint.
Sketch
Compare