Concavity

We now introduce the concavity of the graph of functions and then define the point of inflection. Next, we describe the second derivative test to find the local maximum and local minimum.
The nature of f ' (x) can determine when the graph of a function is bending up or down. More precisely, when f ' (x) is increasing on (a, b), the slope of the tangent line to the curve increases, and the graph bends upward. Similarly, when f ' (x) is decreasing on (a, b), the slope of the tangent line to the curve decreases, and the graph bends downward.

Definition
Let f be differentiable on an open interval I. The graph of f is
1. concave up on I if f ' is increasing on the interval.
2. concave down on I if f ' is decreasing on the interval.
fig 9.2-1

If f '' (x) > 0 for all x in the interval (a, b), then f '(x) is increasing on (a, b). So the graph of f is concave up on (a, b).
Again, if f '' (x) < 0 for all x in the interval (a, b), then f ' (x) is decreasing on (a, b). So the graph of f is concave down on (a, b).
Therefore we can use the second derivative to decide whether the function is concave up or concave down.

Test for Concavity
Let f be twice differentiable on an open interval I.
1. If f'' (x) > 0 for all x in I, then f is concave up on I.
2. If f''(x) < 0 for all x in I, then f is concave down on I.

Example 11.
Determine the open interval on which the graph of the function f (x) = ¹⁄_{x ² + 1} is concave up or concave down.
Solution
f (x) = ¹⁄_{x ² + 1} ,
f ' (x) = - ^2x⁄_{(x ² + 1)²} .
We calculate that
f '' (x) = ^{-2(x² + 1)² + 2x(2)(x² + 1)(2x)}⁄_{(x² + 1)⁴}
= ^{-2(x² + 1)² + 8x²(x² + 1)}⁄ _{(x² + 1)⁴}
= ^{2(3x² - 1)}⁄_{(x² + 1)³}
f '' (x) is defined for all real number and f '' (x) = 0.
Hence,
^{2(3x² - 1)}⁄_{(x² + 1)³} = 0
2(3x² - 1) = 0
(x + ¹⁄_{√

3}) (x - ¹⁄_{√

3}) = 0.
Therefore x = ± ¹⁄_{√

3}

Interval	x < ^-1⁄_{√ 3}	^-1⁄_{√ 3} < x < ¹⁄_{√ 3}	x > ¹⁄_{√ 3}
x + ¹⁄_{√ 3}	-	+	+
x - ¹⁄_{√ 3}	-	-	+
sign of f '' (x)	+	-	+
Behavior of f	concave up	concave down	concave up

We now describe a behavior of a point at which the concavity changes.
Consider the function f (x) = x³ .
For critical point,
f ' (x) = 0,
3x² = 0
Thus x = 0.
Now f '' (x) = 6x. So f '' (x) < 0 for x < 0 and f '' (x) > 0 for x > 0.
This show that the concavity changes at (0, 0) and the tangent to this curve crosses the curve at (0, 0).
fig 9.2-1

Definition
A point where the graph of a function has a tangent line and where the concavity changes is a point of inflection. At a point of inflection, (c, f (c)), either f '' (c) = 0 or f '' (c) is undefined.
fig 9.2-1

Remark. We have an exceptional case that a point c is not inflection point even if f '' (c) = 0. For example,
f (x) = (x -c)^2k
where the integer k ≥ 2.

Example 12.
Determine the open interval on which the graph of following functions
(a) f (x) = x^{^¹⁄₃}
(b) f (x) = ⁹⁄₈₈ x^{^¹¹⁄₃}
is concave up or concave down. State their inflection points.
Solution
(a) f (x) = x^{^¹⁄₃}
f ' (x) = ¹⁄₃ x^{^{^-2⁄₃}},
f '' (x) = - ²⁄₉ x^{^{^-5⁄
₃}} = - ²⁄_{9x ^{^⁵⁄₃}} .

Interval	x < 0	x = 0	x > 0
sign of f '' (x)	+	undefined	-
Behavior of f	concave up	inflection point	concave down

Therefore on x < 0 the graph of function is concave up and on x > 0 the graph of function is concave down. The point (0, 0) is an inflection point.

(b) f (x) = ⁹⁄₈₈ x^{^¹¹⁄₃}
f ' (x) = ³⁄ ₈ x^{^⁸⁄₃},
f '' (x) = x^{^⁵⁄₃}.

Interval	x < 0	x = 0	x > 0
sign of f '' (x)	-	f '' (0) = 0	+
Behavior of f	concave down	inflection point	concave up

Therefore on x < 0 the graph of function is concave down and on x > 0 the graph of function is concave up. The point (0, 0) is an inflection point.

The following example shows that concavity of the curve does not change at (0, 0) even if f '' (0) = 0.

Example 13.
Show that x = 0 is not inflection point of f (x) = x⁶.
Solution
curve: f (x) = x⁶,
f ' (x) = 6x⁵.
fig 9.2-1

For critical points,
f ' (x) = 0, namely x = 0.
Next, we consider f '' (x) = 30x⁴.
At x = 0, f '' (0) = 0.
We observed that
f '' (x) > 0 for x > 0, and f '' (x) > 0 for x < 0,
so the second derivative does not change sign at x = 0.
Therefore, x = 0 is not inflection point of f (x) = x⁶.

Second Derivative Test

We now discuss the use of second derivative to determine whether a function is maximum or minimum without looking for sign changes in f ' at critical points.(see Figure in the previous definition).

Second Derivative Test for Local Maximum and Local Minimum
Suppose f '' is continuous on an open interval that contains x = c.

If f ' (c) = 0 and f '' (c) > 0, then f has a local minimum at x = c.
If f ' (c) = 0 and f '' (c) < 0, then f has a local maximum at x = c.
If f ' (c) = 0 and f '' (c) = 0, then the test fails. The function f may have a local maximum, a local minimum or neither.

Example 14.
Find and classify the critical points of f (x) = x⁴ - 4x³ + 5.
Solution
curve: f(x) = x⁴ - 4x³ + 5
f ' (x) = 4x³ - 12x²
= 4x²(x - 3).
For critical points, f ' (x) = 0, namely x = 0 and x = 3.
When x = 0, f (0) = 5, and when x = 3, f (3) = -22.
Therefore, the critical points are (0, 5) and (3, -22).
Next, f '' (x) = 12 x² - 24x
= 12x(x - 2).
At x = 0, f '' (0) = 0. The second derivative test fails.
It is observed that f ' (x) < 0 when x < 0 and f ' (x) < 0 when 0 < x < 3.
Therefore (0, 5) is a point of inflection.
At x = 3, f '' (3) = 36.
Therefore (3, -22) is a local minimum.

Example 15.
Find the range of f (x) = ln x - x.
Solution
curve: f (x) = ln x - x,
f ' (x) = ¹⁄_x - 1.
f '' (x) = - ¹⁄_x².
For critical point, f ' (x) = 0,
¹⁄_x - 1 = 0.
Therefore x = 1
and
f (1) = ln 1 - 1 = -1.
At x = 1, f '' (1) = -1 < 0.
Thus (1, -1) is a maximum point.
The range of f (x) is {y : y ≤ -1}.
fig 9.2-3

Example 16.
Find the least amount of material needed to build an open cylindrical vessel with a capicity of 400π cm
Solution
Let r = radius and h = height of cylindrical vessel. The volume of vessel is
πr²h = 400 π
h = ⁴⁰⁰⁄_r².
Since vessel is open top, area of material for vessel is
A = πr² + 2πrh + 2πr ⁴⁰⁰⁄_r²
= πr² + ^800π⁄_r, (r > 0).
Differentiating both sides with respect to r, we get
^dA⁄_dr = 2πr - ^800π⁄_r².
For critical point, ^dA⁄_dr = 0
2πr - ^800π⁄_r² = 0
r³ = 400
r = ∛ 400 .
Next, ^{d² A}⁄_dr² = 2π + ^1600π⁄_r³.
When r = ∛ 400 , ^d²A⁄_dr² = 2π + 4π > 0.
Therefore the area of material is the least when r = ∛ 400 .
Therefore the least amount of material
= πr² ^800π⁄_r
= ^1200π⁄_{∛

400} cm³.

Exercise 9.3
1. Determine the open intervals on which the graph of the following functions are concave up or concave down. State their points of inflection.
(a) f (x) = (x - 1)⁵,
Solution
f (x) = (x - 1)⁵
f '(x) = 5(x - 1)⁴
f "(x) = 20(x - 1)³
f" is defined for all real numbers and
f" (x) = 0
20(x - 1)³ = 0
∴ x = 1

Interval	x < 1	x = 1	x > 1
x	-	f "(1) = 0	+
Behavior of f	Concave down	inflection point	concave up

(b) g (x) = x⁴ - 2x³
Solution
g (x) = x⁴ - 2x³
g' (x) = 4x³ - 6x²
g" (x) = 12x² - 12x
g " (x) is defined for all real numbers and g " (x) = 0.
12x² - 12x = 0
x² - x = 0
x(x - 1) = 0
x = 0 or x = 1

Interval	x < 0	x = 0	0 < x < 1	x = 1	x > 1
x	-			+
x - 1	-
sign of g "	+	g " (0) = 0	-	g " (1) = 0	+
Behavior of g	concave up		concave down		concave up

(c) h (x) = sin x [π, -π]
Solution
h (x) = sin x [π, -π]
h ' (x) = cos x
h " (x) = -sin x
h " (x) is defined on [-π , π] and h "(x) = 0
-sin x = 0
x = -π , 0, π

Interval	-π < x < 0	x = 0	0 < x < π
sin x	-	0	+
sign of g "	+	0	-
Behavior of g	Concave up	inflection point	concave down

2. Find and classify the critical points of the following functions.
(a) f (x) = x⁴ - 4x³ + 5
Solution
curve: f(x) = x⁴ - 4x³ + 5
f ' (x) = 4x³ - 12x²
= 4x²(x - 3).
For critical points, f ' (x) = 0, namely x = 0 and x = 3.
When x = 0, f (0) = 5, and when x = 3, f (3) = -22.
Therefore, the critical points are (0, 5) and (3, -22).
Next, f '' (x) = 12 x² - 24x
= 12x(x - 2).
At x = 0, f '' (0) = 0. The second derivative test fails.
It is observed that f ' (x) < 0 when x < 0 and f ' (x) < 0 when 0 < x < 3.
Therefore (0, 5) is a point of inflection.
At x = 3, f '' (3) = 36.
Therefore (3, -22) is a local minimum.

(b) g (x) = x⁵ - 5x + 5
Solution
g (x) = x⁵ - 5x + 5
g ' (x) = 5x⁴ - 5
For critical points, g ' (x) = 0
5x⁴ - 5 = 0
x⁴ = 1
x = -1 or x = 1
g(-1) = 9
g (1) = 1
The critical points are (-1, 9) and (1, 1).
g " (x) = 20x³
At x = -1, g " (-1) = -20 < 0
∴ (-1, 9) is a local maximum.
At x = 1, g " (1) = 20 > 0
∴ (1, 1) is a local minimum.

(c) h (x) = xe^-x
Solution
h (x) = xe^-x
h ' (x) = xe^-x (-1) + e^-x
= -xe^-x + e^-x
For critical points, h ' (x) = 0,
-xe^-x + e^-x = 0
e^-x (-x + 1) = 0
e^-x ≠ 0,
-x + 1 = 0
x = 1
h(1) = e^-1 = ¹⁄_e
The critical point is (1, ¹⁄_e).
h " (x) = -x e^-x (-1) + e^-x (-1) + e^-x (-1)
= xe^-x - 2e^-x
At (1, ¹⁄_e),
h " (1) = e^-1 - 2e^--1
= - ¹⁄_e < 0
∴ (1, ¹⁄_e) is a maximum point.

3. Find the range of f (x) = x - e^x.
Solution
f (x) = x - e^x
f ' (x) = 1 - e^x
f " (x) = - e^x
For critical points,
f ' (x) = 0
1 - e^x = 0
e^x = 1
e^x = 1⁰
x = 0
f (0) = 0 - e⁰
= -1
The critical point is (0, -1).
At (0, -1), f " (0) = -1 < 0
∴ (0, -1) is a local maximum.
Range of f = {y: y ≤ -1}

4. If a piece of string of fixed length is made to enclose a rectangle, show that the enclosed area is the greatest when the rectangle is a square.
Solution
Let a fixed length of string = ℓ cm.
Let the length of a rectangle = x cm
Let the breadth of a rectangle = y cm
∴ 23x + 2y = ℓ
2y = ℓ - 2x
y = ^{ℓ - 2x}⁄₂
y = ^ℓ⁄₂ - x
Area = xy
A = x(^ℓ⁄₂ - x)
= ^ℓ⁄₂ x - x²
^dA⁄_dx = ^ℓ⁄₂ - 2x
^d²A⁄_dx² = -2
^dA⁄_dx = 0,
^ℓ⁄₂ - 2x = 0
x = ^ℓ⁄₄ cm
y = ^ℓ⁄₂ - ^ℓ⁄₄
= ^ℓ⁄₄ cm
∴ The rectangle is a square.
^d²A⁄_dx² = -2 < 0
∴ The enclosed area is the greatest.

5. Find the minimum value of the sum of a positive number and its reciprocal.
Solution
Let x be a positive number.
reciprocal of x = ¹⁄_x
sum = s = x + ¹⁄_x
^ds⁄_dx = 1 - ¹⁄_x²
^ds²⁄_dx² = ²⁄_x³
^ds⁄_dx = 0 ⇒ 1 - ¹⁄_x² = 0
- ¹⁄_x² = -1
x² = 1
x = 1 (∵ x is a positive)
when x = 1,
^ds²⁄_dx² = ²⁄₁ = 2 > 0
Required minimum value = 1 + 1 = 2

6. A rectangular field is surrounded by a fence on three of its sides and a straight hedge on the fourth side. If the length of the fence is 320 meters, find the maximum area of the field enclosed.
Solution
Let the length of a rectangle = x m
the breadth = y m.

2y + x = 320
2y = 320 - x
y = 160 - ^x⁄₂
Area A = xy
= x(160 - ^x⁄₂)
= 160x - ^x²⁄₂
^dA⁄_dx = 160 - x
^d²A⁄_dx² = -1
^dA⁄_dx = 0 ⇒ 160 - x = 0
x = 160 m
y = 160 - 80 = 80 m
when x = 160,
^d²A⁄_dx² = -1 < 0
∴ Maximum area of the field = 160 × 80
= 12800 m²

7. A rectangular box has a square base of side x cm. If the sum of one side of the square and height is 15 cm, express the volume of the box in terms of x. Use this expression to determine the maximum volume of the box.
Solution

Volume of the box V = x ⋅ x ⋅ (15 - x)
= 15 x² - x³
^dV⁄_dx = 30x - 3x²
^d²V⁄_dx² = 30 - 6x
When ^dV⁄_dx = 0,
30x - 3x² = 0
3x (10 - x) = 0
10 - x = 0 (∵ x > 0)
x = 10 cm
When x = 10,
^d²V⁄_dx² = 30 - 6(10)
= -30 < 0
∴ The volume is maximum.
Maximum volume of the box = 10 × 10 × (15 - 10) = 500 cm³.

8. Two sides of triangle have a lengths x and y, and the angle between them is θ .
What volume of θ will maximize the triangle's area?
Solution

Area of triangle A = ¹⁄₂ bh
A = ¹⁄₂ xy sin θ
^dA⁄_dθ = ¹⁄₂ xy cos θ
^d²A⁄_dθ² = - ¹⁄₂ xy sin θ
When ^dA⁄_dθ = 0,
¹⁄₂ xy cos θ = 0
cos θ = 0 (∵ x, y > 0)
θ = ^π⁄₂
When θ = ^π⁄₂ ,
^d²A⁄_dθ² = - ¹⁄₂ xy sin ^π⁄₂
= - ¹⁄₂ xy < 0 (∵ x, y > 0)
∴ The triangle area is maximum.
Hence, θ = ^π⁄₂ will maximize the triangle's area.

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