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Second Derivative Test Joseph Lee Metropolitan Community College Joseph Lee Second Derivative Test
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
