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. . . . . .
Section 4.3Derivatives and the Shapes of
Curves
V63.0121, Calculus I
March 25-26, 2009
.
.Image credit: cobalt123
. . . . . .
Outline
MonotonicityThe Increasing/Decreasing TestFinding intervals of monotonicityThe First Derivative Test
ConcavityDefinitionsTesting for ConcavityThe Second Derivative Test
. . . . . .
The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b), then fis decreasing on (a, b).
Proof.It works the same as the last theorem. Pick two points x and y in(a, b) with x < y. We must show f(x) < f(y). By MVT there exists apoint c in (x, y) such that
f(y) − f(x)y − x
= f′(c) > 0.
Sof(y) − f(x) = f′(c)(y − x) > 0.
. . . . . .
The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b), then fis decreasing on (a, b).
Proof.It works the same as the last theorem. Pick two points x and y in(a, b) with x < y. We must show f(x) < f(y). By MVT there exists apoint c in (x, y) such that
f(y) − f(x)y − x
= f′(c) > 0.
Sof(y) − f(x) = f′(c)(y − x) > 0.
. . . . . .
Finding intervals of monotonicity I
ExampleFind the intervals of monotonicity of f(x) = 2x − 5.
Solutionf′(x) = 2 is always positive, so f is increasing on (−∞,∞).
ExampleDescribe the monotonicity of f(x) = arctan(x).
SolutionSince f′(x) =
11 + x2 is always positive, f(x) is always increasing.
. . . . . .
Finding intervals of monotonicity I
ExampleFind the intervals of monotonicity of f(x) = 2x − 5.
Solutionf′(x) = 2 is always positive, so f is increasing on (−∞,∞).
ExampleDescribe the monotonicity of f(x) = arctan(x).
SolutionSince f′(x) =
11 + x2 is always positive, f(x) is always increasing.
. . . . . .
Finding intervals of monotonicity I
ExampleFind the intervals of monotonicity of f(x) = 2x − 5.
Solutionf′(x) = 2 is always positive, so f is increasing on (−∞,∞).
ExampleDescribe the monotonicity of f(x) = arctan(x).
SolutionSince f′(x) =
11 + x2 is always positive, f(x) is always increasing.
. . . . . .
Finding intervals of monotonicity I
ExampleFind the intervals of monotonicity of f(x) = 2x − 5.
Solutionf′(x) = 2 is always positive, so f is increasing on (−∞,∞).
ExampleDescribe the monotonicity of f(x) = arctan(x).
SolutionSince f′(x) =
11 + x2 is always positive, f(x) is always increasing.
. . . . . .
Finding intervals of monotonicity II
ExampleFind the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative when x is.I We can draw a number line:
. .f′.− ..0.0 .+
.min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).I In fact we can say f is decreasing on (−∞, 0] and increasing on
[0,∞)
. . . . . .
Finding intervals of monotonicity II
ExampleFind the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative when x is.
I We can draw a number line:
. .f′.− ..0.0 .+
.min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).I In fact we can say f is decreasing on (−∞, 0] and increasing on
[0,∞)
. . . . . .
Finding intervals of monotonicity II
ExampleFind the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative when x is.I We can draw a number line:
. .f′.− ..0.0 .+
.min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).I In fact we can say f is decreasing on (−∞, 0] and increasing on
[0,∞)
. . . . . .
Finding intervals of monotonicity II
ExampleFind the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative when x is.I We can draw a number line:
. .f′
.f
.−.↘
..0.0 .+
.↗
.min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).
I In fact we can say f is decreasing on (−∞, 0] and increasing on[0,∞)
. . . . . .
Finding intervals of monotonicity II
ExampleFind the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative when x is.I We can draw a number line:
. .f′
.f
.−.↘
..0.0 .+
.↗
.min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).
I In fact we can say f is decreasing on (−∞, 0] and increasing on[0,∞)
. . . . . .
Finding intervals of monotonicity II
ExampleFind the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative when x is.I We can draw a number line:
. .f′
.f
.−.↘
..0.0 .+
.↗
.min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).I In fact we can say f is decreasing on (−∞, 0] and increasing on
[0,∞)
. . . . . .
Finding intervals of monotonicity IIIExampleFind the intervals of monotonicity of f(x) = x2/3(x + 2).
Solution
f′(x) = 23x−1/3(x + 2) + x2/3 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
. .x−1/3..0
.×.− .+
.5x + 4..−4/5
.0.− .+
.f′(x)
.f(x).
.−4/5
.0 ..0
.×
.+
.↗.−.↘
.+
.↗.max .min
. . . . . .
Finding intervals of monotonicity IIIExampleFind the intervals of monotonicity of f(x) = x2/3(x + 2).
Solution
f′(x) = 23x−1/3(x + 2) + x2/3 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
. .x−1/3..0
.×.− .+
.5x + 4..−4/5
.0.− .+
.f′(x)
.f(x).
.−4/5
.0 ..0
.×
.+
.↗.−.↘
.+
.↗.max .min
. . . . . .
Finding intervals of monotonicity IIIExampleFind the intervals of monotonicity of f(x) = x2/3(x + 2).
Solution
f′(x) = 23x−1/3(x + 2) + x2/3 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
. .x−1/3..0
.×.− .+
.5x + 4..−4/5
.0.− .+
.f′(x)
.f(x).
.−4/5
.0 ..0
.×
.+
.↗.−.↘
.+
.↗.max .min
. . . . . .
Finding intervals of monotonicity IIIExampleFind the intervals of monotonicity of f(x) = x2/3(x + 2).
Solution
f′(x) = 23x−1/3(x + 2) + x2/3 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
. .x−1/3..0
.×.− .+
.5x + 4..−4/5
.0.− .+
.f′(x)
.f(x).
.−4/5
.0 ..0
.×.+
.↗.−.↘
.+
.↗
.max .min
. . . . . .
The First Derivative Test
Theorem (The First Derivative Test)Let f be continuous on [a, b] and c a critical point of f in (a, b).
I If f′(x) > 0 on (a, c) and f′(x) < 0 on (c, b), then c is a localmaximum.
I If f′(x) < 0 on (a, c) and f′(x) > 0 on (c, b), then c is a localminimum.
I If f′(x) has the same sign on (a, c) and (c, b), then c is not a localextremum.
. . . . . .
Finding intervals of monotonicity II
ExampleFind the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative when x is.I We can draw a number line:
. .f′
.f
.−.↘
..0.0 .+
.↗
.min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).I In fact we can say f is decreasing on (−∞, 0] and increasing on
[0,∞)
. . . . . .
Finding intervals of monotonicity II
ExampleFind the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative when x is.I We can draw a number line:
. .f′
.f
.−.↘
..0.0 .+
.↗.min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).I In fact we can say f is decreasing on (−∞, 0] and increasing on
[0,∞)
. . . . . .
Finding intervals of monotonicity IIIExampleFind the intervals of monotonicity of f(x) = x2/3(x + 2).
Solution
f′(x) = 23x−1/3(x + 2) + x2/3 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
. .x−1/3..0
.×.− .+
.5x + 4..−4/5
.0.− .+
.f′(x)
.f(x).
.−4/5
.0 ..0
.×.+
.↗.−.↘
.+
.↗
.max .min
. . . . . .
Finding intervals of monotonicity IIIExampleFind the intervals of monotonicity of f(x) = x2/3(x + 2).
Solution
f′(x) = 23x−1/3(x + 2) + x2/3 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
. .x−1/3..0
.×.− .+
.5x + 4..−4/5
.0.− .+
.f′(x)
.f(x).
.−4/5
.0 ..0
.×.+
.↗.−.↘
.+
.↗.max .min
. . . . . .
Outline
MonotonicityThe Increasing/Decreasing TestFinding intervals of monotonicityThe First Derivative Test
ConcavityDefinitionsTesting for ConcavityThe Second Derivative Test
. . . . . .
DefinitionThe graph of f is called concave up on and interval I if it lies aboveall its tangents on I. The graph of f is called concave down on I if itlies below all its tangents on I.
.
concave up
.
concave downWe sometimes say a concave up graph “holds water” and a concavedown graph “spills water”.
. . . . . .
DefinitionA point P on a curve y = f(x) is called an inflection point if f iscontinuous there and the curve changes from concave upward toconcave downward at P (or vice versa).
..concavedown
. concave up
..inflection point
. . . . . .
Theorem (Concavity Test)
I If f′′(x) > 0 for all x in I, then the graph of f is concave upward on II If f′′(x) < 0 for all x in I, then the graph of f is concave downward on I
Proof.Suppose f′′(x) > 0 on I. This means f′ is increasing on I. Let a and xbe in I. The tangent line through (a, f(a)) is the graph of
L(x) = f(a) + f′(a)(x − a)
By MVT, there exists a b between a and x withf(x) − f(a)
x − a= f′(b). So
f(x) = f(a) + f′(b)(x − a) ≥ f(a) + f′(a)(x − a) = L(x)
. . . . . .
Theorem (Concavity Test)
I If f′′(x) > 0 for all x in I, then the graph of f is concave upward on II If f′′(x) < 0 for all x in I, then the graph of f is concave downward on I
Proof.Suppose f′′(x) > 0 on I. This means f′ is increasing on I. Let a and xbe in I. The tangent line through (a, f(a)) is the graph of
L(x) = f(a) + f′(a)(x − a)
By MVT, there exists a b between a and x withf(x) − f(a)
x − a= f′(b). So
f(x) = f(a) + f′(b)(x − a) ≥ f(a) + f′(a)(x − a) = L(x)
. . . . . .
ExampleFind the intervals of concavity for the graph of f(x) = x3 + x2.
Solution
I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x + 2.I This is negative when x < −1/3, positive when x > −1/3, and 0 when
x = −1/3
I So f is concave down on (−∞,−1/3), concave up on (1/3,∞), andhas an inflection point at (−1/3, 2/27)
. . . . . .
ExampleFind the intervals of concavity for the graph of f(x) = x3 + x2.
Solution
I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x + 2.
I This is negative when x < −1/3, positive when x > −1/3, and 0 whenx = −1/3
I So f is concave down on (−∞,−1/3), concave up on (1/3,∞), andhas an inflection point at (−1/3, 2/27)
. . . . . .
ExampleFind the intervals of concavity for the graph of f(x) = x3 + x2.
Solution
I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x + 2.I This is negative when x < −1/3, positive when x > −1/3, and 0 when
x = −1/3
I So f is concave down on (−∞,−1/3), concave up on (1/3,∞), andhas an inflection point at (−1/3, 2/27)
. . . . . .
ExampleFind the intervals of concavity for the graph of f(x) = x3 + x2.
Solution
I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x + 2.I This is negative when x < −1/3, positive when x > −1/3, and 0 when
x = −1/3
I So f is concave down on (−∞,−1/3), concave up on (1/3,∞), andhas an inflection point at (−1/3, 2/27)
. . . . . .
ExampleFind the intervals of concavity of the graph of f(x) = x2/3(x + 2).
Solution
I f′′(x) =109
x−1/3 − 49
x−4/3 =29
x−4/3(5x − 2)
I x−4/3 is always positive, so the concavity is determined by the 5x − 2factor
I So f is concave down on (−∞, 2/5), concave up on (2/5,∞), and hasan inflection point when x = 2/5
. . . . . .
ExampleFind the intervals of concavity of the graph of f(x) = x2/3(x + 2).
Solution
I f′′(x) =109
x−1/3 − 49
x−4/3 =29
x−4/3(5x − 2)
I x−4/3 is always positive, so the concavity is determined by the 5x − 2factor
I So f is concave down on (−∞, 2/5), concave up on (2/5,∞), and hasan inflection point when x = 2/5
. . . . . .
ExampleFind the intervals of concavity of the graph of f(x) = x2/3(x + 2).
Solution
I f′′(x) =109
x−1/3 − 49
x−4/3 =29
x−4/3(5x − 2)
I x−4/3 is always positive, so the concavity is determined by the 5x − 2factor
I So f is concave down on (−∞, 2/5), concave up on (2/5,∞), and hasan inflection point when x = 2/5
. . . . . .
ExampleFind the intervals of concavity of the graph of f(x) = x2/3(x + 2).
Solution
I f′′(x) =109
x−1/3 − 49
x−4/3 =29
x−4/3(5x − 2)
I x−4/3 is always positive, so the concavity is determined by the 5x − 2factor
I So f is concave down on (−∞, 2/5), concave up on (2/5,∞), and hasan inflection point when x = 2/5
. . . . . .
The Second Derivative Test
Theorem (The Second Derivative Test)Let f, f′, and f′′ be continuous on [a, b]. Let c be be a point in (a, b) withf′(c) = 0.
I If f′′(c) < 0, then f(c) is a local maximum.I If f′′(c) > 0, then f(c) is a local minimum.
If f′′(c) = 0, the second derivative test is inconclusive (this does notmean c is neither; we just don’t know yet).
. . . . . .
ExampleFind the local extrema of f(x) = x3 + x2.
Solution
I f′(x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.I Remember f′′(x) = 6x + 2I Since f′′(−2/3) = −2 < 0, −2/3 is a local maximum.I Since f′′(0) = 2 > 0, 0 is a local minimum.
. . . . . .
ExampleFind the local extrema of f(x) = x3 + x2.
Solution
I f′(x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
I Remember f′′(x) = 6x + 2I Since f′′(−2/3) = −2 < 0, −2/3 is a local maximum.I Since f′′(0) = 2 > 0, 0 is a local minimum.
. . . . . .
ExampleFind the local extrema of f(x) = x3 + x2.
Solution
I f′(x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.I Remember f′′(x) = 6x + 2
I Since f′′(−2/3) = −2 < 0, −2/3 is a local maximum.I Since f′′(0) = 2 > 0, 0 is a local minimum.
. . . . . .
ExampleFind the local extrema of f(x) = x3 + x2.
Solution
I f′(x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.I Remember f′′(x) = 6x + 2I Since f′′(−2/3) = −2 < 0, −2/3 is a local maximum.
I Since f′′(0) = 2 > 0, 0 is a local minimum.
. . . . . .
ExampleFind the local extrema of f(x) = x3 + x2.
Solution
I f′(x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.I Remember f′′(x) = 6x + 2I Since f′′(−2/3) = −2 < 0, −2/3 is a local maximum.I Since f′′(0) = 2 > 0, 0 is a local minimum.
. . . . . .
ExampleFind the local extrema of f(x) = x2/3(x + 2)
Solution
I Remember f′(x) =13
x−1/3(5x + 4) which is zero when x = −4/5
I Remember f′′(x) =109
x−4/3(5x − 2), which is negative when
x = −4/5
I So x = −4/5 is a local maximum.I Notice the Second Derivative Test doesn’t catch the local minimum
x = 0 since f is not differentiable there.
. . . . . .
ExampleFind the local extrema of f(x) = x2/3(x + 2)
Solution
I Remember f′(x) =13
x−1/3(5x + 4) which is zero when x = −4/5
I Remember f′′(x) =109
x−4/3(5x − 2), which is negative when
x = −4/5
I So x = −4/5 is a local maximum.I Notice the Second Derivative Test doesn’t catch the local minimum
x = 0 since f is not differentiable there.
. . . . . .
ExampleFind the local extrema of f(x) = x2/3(x + 2)
Solution
I Remember f′(x) =13
x−1/3(5x + 4) which is zero when x = −4/5
I Remember f′′(x) =109
x−4/3(5x − 2), which is negative when
x = −4/5
I So x = −4/5 is a local maximum.I Notice the Second Derivative Test doesn’t catch the local minimum
x = 0 since f is not differentiable there.
. . . . . .
ExampleFind the local extrema of f(x) = x2/3(x + 2)
Solution
I Remember f′(x) =13
x−1/3(5x + 4) which is zero when x = −4/5
I Remember f′′(x) =109
x−4/3(5x − 2), which is negative when
x = −4/5
I So x = −4/5 is a local maximum.
I Notice the Second Derivative Test doesn’t catch the local minimumx = 0 since f is not differentiable there.
. . . . . .
ExampleFind the local extrema of f(x) = x2/3(x + 2)
Solution
I Remember f′(x) =13
x−1/3(5x + 4) which is zero when x = −4/5
I Remember f′′(x) =109
x−4/3(5x − 2), which is negative when
x = −4/5
I So x = −4/5 is a local maximum.I Notice the Second Derivative Test doesn’t catch the local minimum
x = 0 since f is not differentiable there.
. . . . . .
Graph
Graph of f(x) = x2/3(x + 2):
. .x
.y
..(−4/5, 1.03413)
..(0, 0)
..(2/5, 1.30292)
..(−2, 0)
. . . . . .
When the second derivative is zero
I At inflection points c, if f′ is differentiable at c, then f′′(c) = 0I Is it necessarily true, though?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
All of them have f′′(0) = 0. But the first has a local min at 0, thesecond has a local max at 0, and the third has an inflection point at 0.This is why we say 2DT has nothing to say when f′′(c) = 0.
. . . . . .
When the second derivative is zero
I At inflection points c, if f′ is differentiable at c, then f′′(c) = 0I Is it necessarily true, though?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
All of them have f′′(0) = 0. But the first has a local min at 0, thesecond has a local max at 0, and the third has an inflection point at 0.This is why we say 2DT has nothing to say when f′′(c) = 0.
. . . . . .
When the second derivative is zero
I At inflection points c, if f′ is differentiable at c, then f′′(c) = 0I Is it necessarily true, though?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
All of them have f′′(0) = 0. But the first has a local min at 0, thesecond has a local max at 0, and the third has an inflection point at 0.This is why we say 2DT has nothing to say when f′′(c) = 0.
. . . . . .
Summary
I Concepts: Mean Value Theorem, monotonicity, concavityI Facts: derivatives can detect monotonicity and concavityI Techniques for drawing curves: the Increasing/Decreasing Test
and the Concavity TestI Techniques for finding extrema: the First Derivative Test and the
Second Derivative Test
Next week: Graphing functions
. . . . . .
Summary
I Concepts: Mean Value Theorem, monotonicity, concavityI Facts: derivatives can detect monotonicity and concavityI Techniques for drawing curves: the Increasing/Decreasing Test
and the Concavity TestI Techniques for finding extrema: the First Derivative Test and the
Second Derivative Test
Next week: Graphing functions