. . . . . .

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