AP Calculus Section 3.4
Concavity and
the Second Derivative Test
Student Notes
Definitions, Theorems, and Examples
Definition
Let f be differentiable on an open interval I. The
graph of f is concave upward on I if f’ is increasing
on the interval and concave downward on I if f’ is
decreasing on the interval.
Concave upward. f’ is increasing. The
graph of f lies above its tangent lines.
Concave downward. f’ is decreasing.
The graph of f lies below its tangent
lines.
The graph of a function f is concave upward at the point (c, f(c))
if f(c) exists and if for all x in some open interval containing c,
the point (x, f(x)) on the graph of f lies above the corresponding
point on the graph of the tangent line to f at c.
Imagine holding a ruler along the tangent line through the point
(c, f(c)): if the ruler supports the graph of f near (c, f(c)), then
the graph of the function is concave upward.
The graph of a function f is concave downward at the point
(c,f(c)) if f(c) exists and if for all x in some open interval
containing c, the point (x, f(x)) on the graph of f lies below the
corresponding point on the graph of the tangent line to f at c.
In this situation, the graph of f supports the ruler.
You try:
Theorem 3.7
Test for Concavity
Let f be a function whose second derivative exists on an
open interval I.
1.) If f’’(x) > 0 for all x in I, then the graph of f is concave
upward in I.
2.) If f’’(x) < 0 for all x in I, then the graph of f is concave
downward in I.
To apply Theorem 3.7, locate the x-values at
which f’’(x) = 0 or f’’ does not exist. Then, use
these x-values to determine test intervals. Finally,
test the sign of f’’(x) in each of the test intervals.
A straight line is neither concave
upward nor concave downward.
Can you explain why???
Example 1 (#3)
Determine the open intervals on which 𝑓 𝑥 =
downward.
Interval
Test Value
Sign of f’’(x)
Conclusion
24
𝑥 2 +12
is concave upward or concave
Example 2 (#10)
Determine the open intervals on which y = 𝑥 +
concave downward.
Interval
Test Value
Sign of f’’(x)
Conclusion
2
𝑠𝑖𝑛𝑥
on (−𝜋, 𝜋) is concave upward or
Definition
Let f be a function that is
continuous on an open interval
and let c be a point in the
interval. If the graph of f has a
tangent line at this point (c,
f(c)), then this point is a point of
inflection of the graph of f if the
concavity of f changes from
upward to downward (or
downward to upward) at the
point.
The graph of a function crosses its
tangent line at a point of inflection.
Example 3 A (#16)
Determine the points of inflection and discuss the concavity of the graph of 𝑓 𝑥 =
𝑥 3 (𝑥 − 4).
Interval
Test Value
Sign of f’’(x)
Conclusion
Example 3 B (#24)
Determine the points of inflection and discuss the concavity of the graph of 𝑓 𝑥 =
𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 on [0, 2𝜋].
Interval
Test Value
Sign of f’’(x)
Conclusion
Theorem 3.9
Second Derivative Test
Let f be a function such that f’(c)=0 and the second
derivative of f exists on an open interval containing c.
1.) If f’’(c) > 0, then f has a relative minimum at (c, f(c)).
2.) If f’’(c) < 0, then f has a relative maximum at (c,
f(c)).
If f’’(c) = 0, the test fails. f may have a relative
maximum, a relative minimum, or neither. In these
cases, you can use the First Derivative Test.
Example 4 A (#34)
Use the Second Derivative Test to find all relative extrema of
1
𝑔 𝑥 = − 8 (𝑥 + 2)2 (𝑥 − 4)2 .
Example 4 b (#40)
Use the Second Derivative Test to find all relative extrema of
𝑓 𝑥 = 2𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠2𝑥 on [0, 2𝜋].
Assignment – Section 3.4
p. 195-197 #5, 9, 15, 19, 26, 35, 37,
67 or 68 (you choose)