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Second Derivative Test
Joseph Lee
Metropolitan Community College
Joseph Lee Second Derivative Test
Concave Upward and Concave Downward
A function f is concave upward on an open interval (a, b) iff ′′(x) > 0 for every x in (a, b).
A function f is concave downward on (a, b) if f ′′(x) < 0 for everyx in (a, b).
Joseph Lee Second Derivative Test
Point of Inflection
A point of inflection is any point where the function switchesconcavity.
Joseph Lee Second Derivative Test
Example 1.
Find any points of inflection and discuss the concavity of thefunction.
f (x) = x3 + x2 − 9x − 9
Solution.f ′(x) = 3x2 + 2x − 9
f ′′(x) = 6x + 20 = 6x + 2x = − 1
3
(−∞,−13) (−1
3 ,∞)
f ′′(x) < 0 f ′′(x) > 0
Concave downward on (−∞,−13)
Concave upward on (−13 ,∞)
Point of inflection at (−13 ,−
16027 )
Joseph Lee Second Derivative Test
Example 1.
Find any points of inflection and discuss the concavity of thefunction.
f (x) = x3 + x2 − 9x − 9
Solution.f ′(x) =
3x2 + 2x − 9
f ′′(x) = 6x + 20 = 6x + 2x = − 1
3
(−∞,−13) (−1
3 ,∞)
f ′′(x) < 0 f ′′(x) > 0
Concave downward on (−∞,−13)
Concave upward on (−13 ,∞)
Point of inflection at (−13 ,−
16027 )
Joseph Lee Second Derivative Test
Example 1.
Find any points of inflection and discuss the concavity of thefunction.
f (x) = x3 + x2 − 9x − 9
Solution.f ′(x) = 3x2 + 2x − 9
f ′′(x) =
6x + 20 = 6x + 2x = − 1
3
(−∞,−13) (−1
3 ,∞)
f ′′(x) < 0 f ′′(x) > 0
Concave downward on (−∞,−13)
Concave upward on (−13 ,∞)
Point of inflection at (−13 ,−
16027 )
Joseph Lee Second Derivative Test
Example 1.
Find any points of inflection and discuss the concavity of thefunction.
f (x) = x3 + x2 − 9x − 9
Solution.f ′(x) = 3x2 + 2x − 9
f ′′(x) = 6x + 2
0 = 6x + 2x = − 1
3
(−∞,−13) (−1
3 ,∞)
f ′′(x) < 0 f ′′(x) > 0
Concave downward on (−∞,−13)
Concave upward on (−13 ,∞)
Point of inflection at (−13 ,−
16027 )
Joseph Lee Second Derivative Test
Example 1.
Find any points of inflection and discuss the concavity of thefunction.
f (x) = x3 + x2 − 9x − 9
Solution.f ′(x) = 3x2 + 2x − 9
f ′′(x) = 6x + 20 = 6x + 2x =
− 13
(−∞,−13) (−1
3 ,∞)
f ′′(x) < 0 f ′′(x) > 0
Concave downward on (−∞,−13)
Concave upward on (−13 ,∞)
Point of inflection at (−13 ,−
16027 )
Joseph Lee Second Derivative Test
Example 1.
Find any points of inflection and discuss the concavity of thefunction.
f (x) = x3 + x2 − 9x − 9
Solution.f ′(x) = 3x2 + 2x − 9
f ′′(x) = 6x + 20 = 6x + 2x = − 1
3
(−∞,−13) (−1
3 ,∞)
f ′′(x) < 0 f ′′(x) > 0
Concave downward on (−∞,−13)
Concave upward on (−13 ,∞)
Point of inflection at (−13 ,−
16027 )
Joseph Lee Second Derivative Test
Example 1.
Find any points of inflection and discuss the concavity of thefunction.
f (x) = x3 + x2 − 9x − 9
Solution.f ′(x) = 3x2 + 2x − 9
f ′′(x) = 6x + 20 = 6x + 2x = − 1
3
(−∞,−13) (−1
3 ,∞)
f ′′(x) < 0 f ′′(x) > 0
Concave downward on (−∞,−13)
Concave upward on (−13 ,∞)
Point of inflection at (−13 ,−
16027 )
Joseph Lee Second Derivative Test
Example 1.
Find any points of inflection and discuss the concavity of thefunction.
f (x) = x3 + x2 − 9x − 9
Solution.f ′(x) = 3x2 + 2x − 9
f ′′(x) = 6x + 20 = 6x + 2x = − 1
3
(−∞,−13) (−1
3 ,∞)
f ′′(x) < 0 f ′′(x) > 0
Concave downward on (−∞,−13)
Concave upward on (−13 ,∞)
Point of inflection at (−13 ,−
16027 )
Joseph Lee Second Derivative Test
Example 1.
Find any points of inflection and discuss the concavity of thefunction.
f (x) = x3 + x2 − 9x − 9
Solution.f ′(x) = 3x2 + 2x − 9
f ′′(x) = 6x + 20 = 6x + 2x = − 1
3
(−∞,−13) (−1
3 ,∞)
f ′′(x) < 0 f ′′(x) > 0
Concave downward on (−∞,−13)
Concave upward on (−13 ,∞)
Point of inflection at (−13 ,−
16027 )
Joseph Lee Second Derivative Test
Example 2.
Find any points of inflection and discuss the concavity of thefunction.
g(x) =x
x2 − 1
Solution.
g ′(x) =(x2 − 1)− x(2x)
(x2 − 1)2
=−x2 − 1
(x2 − 1)2
g ′′(x) =−2x(x2 − 1)2 + (x2 + 1)[2(x2 − 1)(2x)]
(x2 − 1)4
=−2x(x2 − 1) + 4x(x2 + 1)
(x2 − 1)3=
2x3 + 6x
(x2 − 1)3=
2x(x2 + 3)
(x2 − 1)3
Values where g ′′(x) = 0 or g ′′(x) undefined: −1, 0, 1
Joseph Lee Second Derivative Test
Example 2.
Find any points of inflection and discuss the concavity of thefunction.
g(x) =x
x2 − 1Solution.
g ′(x) =
(x2 − 1)− x(2x)
(x2 − 1)2
=−x2 − 1
(x2 − 1)2
g ′′(x) =−2x(x2 − 1)2 + (x2 + 1)[2(x2 − 1)(2x)]
(x2 − 1)4
=−2x(x2 − 1) + 4x(x2 + 1)
(x2 − 1)3=
2x3 + 6x
(x2 − 1)3=
2x(x2 + 3)
(x2 − 1)3
Values where g ′′(x) = 0 or g ′′(x) undefined: −1, 0, 1
Joseph Lee Second Derivative Test
Example 2.
Find any points of inflection and discuss the concavity of thefunction.
g(x) =x
x2 − 1Solution.
g ′(x) =(x2 − 1)− x(2x)
(x2 − 1)2
=
−x2 − 1
(x2 − 1)2
g ′′(x) =−2x(x2 − 1)2 + (x2 + 1)[2(x2 − 1)(2x)]
(x2 − 1)4
=−2x(x2 − 1) + 4x(x2 + 1)
(x2 − 1)3=
2x3 + 6x
(x2 − 1)3=
2x(x2 + 3)
(x2 − 1)3
Values where g ′′(x) = 0 or g ′′(x) undefined: −1, 0, 1
Joseph Lee Second Derivative Test
Example 2.
Find any points of inflection and discuss the concavity of thefunction.
g(x) =x
x2 − 1Solution.
g ′(x) =(x2 − 1)− x(2x)
(x2 − 1)2
=−x2 − 1
(x2 − 1)2
g ′′(x) =
−2x(x2 − 1)2 + (x2 + 1)[2(x2 − 1)(2x)]
(x2 − 1)4
=−2x(x2 − 1) + 4x(x2 + 1)
(x2 − 1)3=
2x3 + 6x
(x2 − 1)3=
2x(x2 + 3)
(x2 − 1)3
Values where g ′′(x) = 0 or g ′′(x) undefined: −1, 0, 1
Joseph Lee Second Derivative Test
Example 2.
Find any points of inflection and discuss the concavity of thefunction.
g(x) =x
x2 − 1Solution.
g ′(x) =(x2 − 1)− x(2x)
(x2 − 1)2
=−x2 − 1
(x2 − 1)2
g ′′(x) =−2x(x2 − 1)2 + (x2 + 1)[2(x2 − 1)(2x)]
(x2 − 1)4
=
−2x(x2 − 1) + 4x(x2 + 1)
(x2 − 1)3=
2x3 + 6x
(x2 − 1)3=
2x(x2 + 3)
(x2 − 1)3
Values where g ′′(x) = 0 or g ′′(x) undefined: −1, 0, 1
Joseph Lee Second Derivative Test
Example 2.
Find any points of inflection and discuss the concavity of thefunction.
g(x) =x
x2 − 1Solution.
g ′(x) =(x2 − 1)− x(2x)
(x2 − 1)2
=−x2 − 1
(x2 − 1)2
g ′′(x) =−2x(x2 − 1)2 + (x2 + 1)[2(x2 − 1)(2x)]
(x2 − 1)4
=−2x(x2 − 1) + 4x(x2 + 1)
(x2 − 1)3=
2x3 + 6x
(x2 − 1)3=
2x(x2 + 3)
(x2 − 1)3
Values where g ′′(x) = 0 or g ′′(x) undefined: −1, 0, 1
Joseph Lee Second Derivative Test
Example 2.
Find any points of inflection and discuss the concavity of thefunction.
g(x) =x
x2 − 1Solution.
g ′(x) =(x2 − 1)− x(2x)
(x2 − 1)2
=−x2 − 1
(x2 − 1)2
g ′′(x) =−2x(x2 − 1)2 + (x2 + 1)[2(x2 − 1)(2x)]
(x2 − 1)4
=−2x(x2 − 1) + 4x(x2 + 1)
(x2 − 1)3=
2x3 + 6x
(x2 − 1)3=
2x(x2 + 3)
(x2 − 1)3
Values where g ′′(x) = 0 or g ′′(x) undefined: −1, 0, 1
Joseph Lee Second Derivative Test
Example 2.
Find any points of inflection and discuss the concavity of thefunction.
g(x) =x
x2 − 1Solution.
g ′(x) =(x2 − 1)− x(2x)
(x2 − 1)2
=−x2 − 1
(x2 − 1)2
g ′′(x) =−2x(x2 − 1)2 + (x2 + 1)[2(x2 − 1)(2x)]
(x2 − 1)4
=−2x(x2 − 1) + 4x(x2 + 1)
(x2 − 1)3=
2x3 + 6x
(x2 − 1)3=
2x(x2 + 3)
(x2 − 1)3
Values where g ′′(x) = 0 or g ′′(x) undefined: −1, 0, 1
Joseph Lee Second Derivative Test
Example 2.
Find any points of inflection and discuss the concavity of thefunction.
g(x) =x
x2 − 1Solution.
g ′(x) =(x2 − 1)− x(2x)
(x2 − 1)2
=−x2 − 1
(x2 − 1)2
g ′′(x) =−2x(x2 − 1)2 + (x2 + 1)[2(x2 − 1)(2x)]
(x2 − 1)4
=−2x(x2 − 1) + 4x(x2 + 1)
(x2 − 1)3=
2x3 + 6x
(x2 − 1)3=
2x(x2 + 3)
(x2 − 1)3
Values where g ′′(x) = 0 or g ′′(x) undefined: −1, 0, 1Joseph Lee Second Derivative Test
Example 2. (continued)
Determine the open intervals on which the graph is concaveupward or concave downward.
g(x) =x
x2 − 1
Solution.
g ′′(x) =2x(x2 + 3)
(x2 − 1)3
(−∞,−1) (−1, 0) (0, 1) (1,∞)
g ′′(x) < 0 g ′′(x) > 0 g ′′(x) < 0 g ′′(x) > 0
Concave upward on (−1, 0) ∪ (1,∞)Concave downward on (−∞, 1) ∪ (0, 1)Point of inflection at (0, 0)
Joseph Lee Second Derivative Test
Example 2. (continued)
Determine the open intervals on which the graph is concaveupward or concave downward.
g(x) =x
x2 − 1
Solution.
g ′′(x) =2x(x2 + 3)
(x2 − 1)3
(−∞,−1) (−1, 0) (0, 1) (1,∞)
g ′′(x) < 0 g ′′(x) > 0 g ′′(x) < 0 g ′′(x) > 0
Concave upward on (−1, 0) ∪ (1,∞)Concave downward on (−∞, 1) ∪ (0, 1)Point of inflection at (0, 0)
Joseph Lee Second Derivative Test
Example 2. (continued)
Determine the open intervals on which the graph is concaveupward or concave downward.
g(x) =x
x2 − 1
Solution.
g ′′(x) =2x(x2 + 3)
(x2 − 1)3
(−∞,−1) (−1, 0) (0, 1) (1,∞)
g ′′(x) < 0 g ′′(x) > 0 g ′′(x) < 0 g ′′(x) > 0
Concave upward on (−1, 0) ∪ (1,∞)Concave downward on (−∞, 1) ∪ (0, 1)
Point of inflection at (0, 0)
Joseph Lee Second Derivative Test
Example 2. (continued)
Determine the open intervals on which the graph is concaveupward or concave downward.
g(x) =x
x2 − 1
Solution.
g ′′(x) =2x(x2 + 3)
(x2 − 1)3
(−∞,−1) (−1, 0) (0, 1) (1,∞)
g ′′(x) < 0 g ′′(x) > 0 g ′′(x) < 0 g ′′(x) > 0
Concave upward on (−1, 0) ∪ (1,∞)Concave downward on (−∞, 1) ∪ (0, 1)Point of inflection at (0, 0)
Joseph Lee Second Derivative Test
Second Derivative Test
Let c be a critical number of f .
If f ′′(c) > 0, then f has a relative minumum at c .
If f ′′(c) < 0, then f has a relative maximum at c .
If f ′′(c) = 0, then the Second Derivative Test is inconclusive.
Joseph Lee Second Derivative Test
Example 3.
Use the second derivative test to identify any extrema.
f (x) = −x2 − 2x + 3
Solution.f ′(x) = − 2x − 2
0 = 2(x + 1)
x = − 1
f ′′(x) = − 2
f ′′(−1) = − 2
Relative maximum at (−1, 4)
Joseph Lee Second Derivative Test
Example 3.
Use the second derivative test to identify any extrema.
f (x) = −x2 − 2x + 3
Solution.f ′(x) =
− 2x − 2
0 = 2(x + 1)
x = − 1
f ′′(x) = − 2
f ′′(−1) = − 2
Relative maximum at (−1, 4)
Joseph Lee Second Derivative Test
Example 3.
Use the second derivative test to identify any extrema.
f (x) = −x2 − 2x + 3
Solution.f ′(x) = − 2x − 2
0 = 2(x + 1)
x = − 1
f ′′(x) = − 2
f ′′(−1) = − 2
Relative maximum at (−1, 4)
Joseph Lee Second Derivative Test
Example 3.
Use the second derivative test to identify any extrema.
f (x) = −x2 − 2x + 3
Solution.f ′(x) = − 2x − 2
0 = 2(x + 1)
x =
− 1
f ′′(x) = − 2
f ′′(−1) = − 2
Relative maximum at (−1, 4)
Joseph Lee Second Derivative Test
Example 3.
Use the second derivative test to identify any extrema.
f (x) = −x2 − 2x + 3
Solution.f ′(x) = − 2x − 2
0 = 2(x + 1)
x = − 1
f ′′(x) =
− 2
f ′′(−1) = − 2
Relative maximum at (−1, 4)
Joseph Lee Second Derivative Test
Example 3.
Use the second derivative test to identify any extrema.
f (x) = −x2 − 2x + 3
Solution.f ′(x) = − 2x − 2
0 = 2(x + 1)
x = − 1
f ′′(x) = − 2
f ′′(−1) =
− 2
Relative maximum at (−1, 4)
Joseph Lee Second Derivative Test
Example 3.
Use the second derivative test to identify any extrema.
f (x) = −x2 − 2x + 3
Solution.f ′(x) = − 2x − 2
0 = 2(x + 1)
x = − 1
f ′′(x) = − 2
f ′′(−1) = − 2
Relative maximum at (−1, 4)
Joseph Lee Second Derivative Test
Example 3.
Use the second derivative test to identify any extrema.
f (x) = −x2 − 2x + 3
Solution.f ′(x) = − 2x − 2
0 = 2(x + 1)
x = − 1
f ′′(x) = − 2
f ′′(−1) = − 2
Relative maximum at (−1, 4)
Joseph Lee Second Derivative Test
Example 4.
Use the second derivative test to identify any extrema.
f (x) =x2 + 1
x
Solution.
f (x) = x +1
x
f ′(x) = 1− 1
x2
Critical numbers: −1, 1
f ′′(x) =2
x3
f ′′(−1) = −2f ′′(1) = 2
Relative maximum at (−1,−2)Relative minimum at (1, 2)
Joseph Lee Second Derivative Test
Example 4.
Use the second derivative test to identify any extrema.
f (x) =x2 + 1
x
Solution.
f (x) =
x +1
x
f ′(x) = 1− 1
x2
Critical numbers: −1, 1
f ′′(x) =2
x3
f ′′(−1) = −2f ′′(1) = 2
Relative maximum at (−1,−2)Relative minimum at (1, 2)
Joseph Lee Second Derivative Test
Example 4.
Use the second derivative test to identify any extrema.
f (x) =x2 + 1
x
Solution.
f (x) = x +1
x
f ′(x) =
1− 1
x2
Critical numbers: −1, 1
f ′′(x) =2
x3
f ′′(−1) = −2f ′′(1) = 2
Relative maximum at (−1,−2)Relative minimum at (1, 2)
Joseph Lee Second Derivative Test
Example 4.
Use the second derivative test to identify any extrema.
f (x) =x2 + 1
x
Solution.
f (x) = x +1
x
f ′(x) = 1− 1
x2
Critical numbers: −1, 1
f ′′(x) =2
x3
f ′′(−1) = −2f ′′(1) = 2
Relative maximum at (−1,−2)Relative minimum at (1, 2)
Joseph Lee Second Derivative Test
Example 4.
Use the second derivative test to identify any extrema.
f (x) =x2 + 1
x
Solution.
f (x) = x +1
x
f ′(x) = 1− 1
x2
Critical numbers: −1, 1
f ′′(x) =2
x3
f ′′(−1) = −2f ′′(1) = 2
Relative maximum at (−1,−2)Relative minimum at (1, 2)
Joseph Lee Second Derivative Test
Example 4.
Use the second derivative test to identify any extrema.
f (x) =x2 + 1
x
Solution.
f (x) = x +1
x
f ′(x) = 1− 1
x2
Critical numbers: −1, 1
f ′′(x) =
2
x3
f ′′(−1) = −2f ′′(1) = 2
Relative maximum at (−1,−2)Relative minimum at (1, 2)
Joseph Lee Second Derivative Test
Example 4.
Use the second derivative test to identify any extrema.
f (x) =x2 + 1
x
Solution.
f (x) = x +1
x
f ′(x) = 1− 1
x2
Critical numbers: −1, 1
f ′′(x) =2
x3
f ′′(−1) = −2f ′′(1) = 2
Relative maximum at (−1,−2)Relative minimum at (1, 2)
Joseph Lee Second Derivative Test
Example 4.
Use the second derivative test to identify any extrema.
f (x) =x2 + 1
x
Solution.
f (x) = x +1
x
f ′(x) = 1− 1
x2
Critical numbers: −1, 1
f ′′(x) =2
x3
f ′′(−1) = −2f ′′(1) = 2
Relative maximum at (−1,−2)Relative minimum at (1, 2)
Joseph Lee Second Derivative Test
Example 4.
Use the second derivative test to identify any extrema.
f (x) =x2 + 1
x
Solution.
f (x) = x +1
x
f ′(x) = 1− 1
x2
Critical numbers: −1, 1
f ′′(x) =2
x3
f ′′(−1) = −2f ′′(1) = 2
Relative maximum at (−1,−2)Relative minimum at (1, 2)
Joseph Lee Second Derivative Test
Example 5.
Use the second derivative test to identify any extrema.
g(x) = sin x + cos x , 0 < x < 2π
Solution.g ′(x) = cos x − sin x
sin x = cos x
x =π
4,
5π
4
g ′′(x) = − sin x − cos x
g ′′ (π4
)= −
√2
g ′′ (5π4
)=√
2
Relative maximum at(π4 ,√
2)
Relative minimum at(5π4 ,−√
2)
Joseph Lee Second Derivative Test
Example 5.
Use the second derivative test to identify any extrema.
g(x) = sin x + cos x , 0 < x < 2π
Solution.g ′(x) =
cos x − sin x
sin x = cos x
x =π
4,
5π
4
g ′′(x) = − sin x − cos x
g ′′ (π4
)= −
√2
g ′′ (5π4
)=√
2
Relative maximum at(π4 ,√
2)
Relative minimum at(5π4 ,−√
2)
Joseph Lee Second Derivative Test
Example 5.
Use the second derivative test to identify any extrema.
g(x) = sin x + cos x , 0 < x < 2π
Solution.g ′(x) = cos x − sin x
sin x = cos x
x =π
4,
5π
4
g ′′(x) = − sin x − cos x
g ′′ (π4
)= −
√2
g ′′ (5π4
)=√
2
Relative maximum at(π4 ,√
2)
Relative minimum at(5π4 ,−√
2)
Joseph Lee Second Derivative Test
Example 5.
Use the second derivative test to identify any extrema.
g(x) = sin x + cos x , 0 < x < 2π
Solution.g ′(x) = cos x − sin x
sin x = cos x
x =
π
4,
5π
4
g ′′(x) = − sin x − cos x
g ′′ (π4
)= −
√2
g ′′ (5π4
)=√
2
Relative maximum at(π4 ,√
2)
Relative minimum at(5π4 ,−√
2)
Joseph Lee Second Derivative Test
Example 5.
Use the second derivative test to identify any extrema.
g(x) = sin x + cos x , 0 < x < 2π
Solution.g ′(x) = cos x − sin x
sin x = cos x
x =π
4,
5π
4
g ′′(x) =
− sin x − cos x
g ′′ (π4
)= −
√2
g ′′ (5π4
)=√
2
Relative maximum at(π4 ,√
2)
Relative minimum at(5π4 ,−√
2)
Joseph Lee Second Derivative Test
Example 5.
Use the second derivative test to identify any extrema.
g(x) = sin x + cos x , 0 < x < 2π
Solution.g ′(x) = cos x − sin x
sin x = cos x
x =π
4,
5π
4
g ′′(x) = − sin x − cos x
g ′′ (π4
)=
−√
2
g ′′ (5π4
)=√
2
Relative maximum at(π4 ,√
2)
Relative minimum at(5π4 ,−√
2)
Joseph Lee Second Derivative Test
Example 5.
Use the second derivative test to identify any extrema.
g(x) = sin x + cos x , 0 < x < 2π
Solution.g ′(x) = cos x − sin x
sin x = cos x
x =π
4,
5π
4
g ′′(x) = − sin x − cos x
g ′′ (π4
)= −
√2
g ′′ (5π4
)=
√2
Relative maximum at(π4 ,√
2)
Relative minimum at(5π4 ,−√
2)
Joseph Lee Second Derivative Test
Example 5.
Use the second derivative test to identify any extrema.
g(x) = sin x + cos x , 0 < x < 2π
Solution.g ′(x) = cos x − sin x
sin x = cos x
x =π
4,
5π
4
g ′′(x) = − sin x − cos x
g ′′ (π4
)= −
√2
g ′′ (5π4
)=√
2
Relative maximum at(π4 ,√
2)
Relative minimum at(5π4 ,−√
2)
Joseph Lee Second Derivative Test
Example 5.
Use the second derivative test to identify any extrema.
g(x) = sin x + cos x , 0 < x < 2π
Solution.g ′(x) = cos x − sin x
sin x = cos x
x =π
4,
5π
4
g ′′(x) = − sin x − cos x
g ′′ (π4
)= −
√2
g ′′ (5π4
)=√
2
Relative maximum at(π4 ,√
2)
Relative minimum at(5π4 ,−√
2)
Joseph Lee Second Derivative Test