Calculate the Intervals of Increase and Decrease for the Following Functions:

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Exercise 1

f(x) = 3x - { x }^{ 3 }

Exercise 2

f(x) = { x }^{ 4 } - 2 { x }^{ 2 } - 8

Exercise 3

f(x) = 4 + 15x + 6 { x }^{ 2 } - { x }^{ 3 }

Exercise 4

f(x) = 3 { x }^{ 4 } - 20 { x }^{ 3 } - 6 { x }^{ 2 } + 60x - 8

Exercise 5

f(x) = x + \frac { 4 }{ x }

Exercise 6

f(x) = \frac { x + 1 }{ { x }^{ 2 } + x - 2}

Exercise 7

f(x) = \frac { { x }^{ 3 } }{ { (x - 1) }^{ 2 } }

Exercise 8

f(x) = \frac { { x }^{ 4 } + 1 }{ { x }^{ 2 } }

Exercise 9

f(x) = \frac { { x }^{ 2 } }{ 2 - x }

Exercise 10

f(x) = \frac { x }{ 1 + { x }^{ 2 } }

Exercise 11

f(x) = x + \sqrt { x }

Exercise 12

f(x) = \sqrt { x + 1 }

Exercise 13

f(x) = { e }^{ - { (x - 1) }^{ 2 } }

Exercise 14

f(x) = { e }^{ \frac { 1 }{ x } }

Exercise 15

f(x) = (x - 1) { e }^{ -x }

Exercise 16

f(x) = \frac { 1 }{ 2 \sqrt { 2 \pi }} { e }^{ - \frac { 1 }{ 2 } { x }^{ 2 } }

Exercise 17

f(x) = x ln (x)

 

Solution of exercise 1

f(x) = 3x - { x }^{ 3 }

f'(x) = 3 - 3 { x }^{ 2 }

 

3 - 3 { x }^{ 2 } = 0

3 { x }^{ 2 } = 3

{ x }^{ 2 } = 1

x = \pm 1

Increasing:

(-1, 1)

Decreasing:

(-\infty, -1 ) U (1, \infty)

 

Solution of exercise 2

f(x) = { x }^{ 4 } - 2 { x }^{ 2 } - 8

f'(x) = 4 { x }^{ 3 } - 4x

4 { x }^{ 3 } - 4x = 0

x ( 4 { x }^{ 2 } - 4 ) = 0

x = 0 \qquad 4 { x }^{ 2 } - 4 = 0

x = 0 \qquad 4 { x }^{ 2 } = 4

x = 0 \qquad  { x }^{ 2 } = 1

x = 0 \qquad x = \pm 1

Increasing: (-1, 0) U (1, \infty)

Decreasing: (- \infty, -1) U (0, 1)

 

Solution of exercise 3

f(x) = 4 + 15x + 6 { x }^{ 2 } - { x }^{ 3 }

f'(x) = 15 + 12x - 3{ x }^{ 2 }

15 + 12x - 3 { x }^{ 2 } = 0

3 { x }^{ 2 } - 12x - 15 = 0

3 { x }^{ 2 } - x (15 - 3) - 15 = 0

3 { x }^{ 2 } - 15x + 3x - 15 = 0

3x (x - 5) + 3 (x - 5) = 0

(3x + 3)(x - 5) = 0

3x + 3 = 0 \qquad x - 5 = 0

3x = -3 \qquad x = 5

x = -1 \qquad x = 5

Increasing: (-1, 5)

Decreasing: (- \infty, -1) U (5, \infty)

 

Solution of exercise 4

f(x) = 3 { x }^{ 4 } - 20 { x }^{ 3 } - 6 { x }^{ 2 } + 60x - 8

f'(x) = 12 { x }^{ 3 } - 60 { x }^{ 2 } - 12 x + 60

f'(x) = 12 ({ x }^{ 3 } - 5 { x }^{ 2 } - x + 5)

12 ({ x }^{ 3 } - 5 { x }^{ 2 } - x + 5) = 0

{ x }^{ 3 } - 5 { x }^{ 2 } - x + 5 = 0

After solving the cubic equation, we will get:

x = \pm 1 \qquad x = 5{}^{}

Increasing: (-1, 1) U (5, \infty)

Decreasing: (- \infty, -1) U (1, 5)

 

Solution of exercise 5

f(x) = x + \frac { 4 }{ x }

D = R - \left \{ 0 \right \}

f'(x) = 1 - \frac { 4 }{ { x }^{ 2 } }

1 - \frac { 4 }{ { x }^{ 2 } } = 0

{ x }^{ 2 } - 4 = 0

x = \pm 2

Increasing: (- \infty, -2) U (2, \infty)

Decreasing: (-2, 0) U (0, 2)

 

Solution of exercise 6

f(x) = \frac { x + 1 }{ { x }^{ 2 } + x - 2 }

{ x }^{ 2 } + x - 2 = 0

{ x }^{ 2 } + x(2 - 1) - 2 = 0

{ x }^{ 2 } + 2x - x - 2 = 0

x (x + 2) - 1 (x + 2) = 0

(x - 1)(x + 2) = 0

x - 1 = 0 \qquad x + 2 = 0

x = 1 \qquad x = -2

 

D = R - \left \{ -2, 1 \right \}

f'(x) = \frac { { x }^{ 2 } + x - 2 - (x + 1)(2x + 1) }{ { ({ x }^{ 2 } + x - 2) }^{ 2 } } = \frac { -({ x }^{ 2 } + 2x + 3) }{ { ({ x }^{ 2 } + x - 2) }^{ 2 } }

\frac { -({ x }^{ 2 } + 2x + 3) }{ { ({ x }^{ 2 } + x - 2) }^{ 2 } } = 0

{ x }^{ 2 } + 2x + 3 = 0

No solutions in .

Decreasing: R - \left \{ -2, 1 \right \}

 

Solution of exercise 7

f(x) = \frac { { x }^{ 3 } }{ { (x - 1) }^{ 2 } }

f'(x) = \frac { { x }^{ 3 } - 3 { x }^{ 2 } }{ { (x - 1) }^{ 3 } }

\frac { { x }^{ 3 } - 3 { x }^{ 2 } }{ { (x - 1) }^{ 3 } } = 0

{ x }^{ 3 } - 3 { x }^{ 2 } = 0

x = 0 \qquad x = 3

Increasing: (- \infty, 0) U (0, 1) U (3, \infty)

Decreasing: (1, 3)

 

Solution of exercise 8

f(x) = \frac { { x }^{ 4 } + 1 }{ { x }^{ 2 } }

f'(x) = \frac { 2({ x }^{ 4 } - 1) }{ { x }^{ 3 } }

x = \pm 1

Increasing: (-1, 0) U (1, \infty)

Decreasing: ( -\infty, -1) U (0, 1)

 

Solution of exercise 9

f(x) = \frac { { x }^{ 2 } }{ 2 - x }

f'(x) = \frac { 4x - { x }^{ 2 } }{ { (2 - x) }^{ 2 } }

\frac { 4x - { x }^{ 2 } }{ { (2 - x) }^{ 2 } } = 0

x = 0 \qquad x = 4

Increasing: (0, 4)

Decreasing: ( -\infty, 0) U (4, \infty)

 

Solution of exercise 10

f(x) = \frac { x }{ 1 + { x }^{ 2 } }

f'(x) = \frac { 1 - { x }^{ 2 } }{ { (1 + { x }^{ 2 }) }^{ 2 } }

\frac { 1 - { x }^{ 2 } }{ { (1 + { x }^{ 2 }) }^{ 2 } } = 0

1 - { x }^{ 2 } = 0

{ x }^{ 2 } = 1

x = \pm 1

Increasing: (-1, 1)

Decreasing: ( -\infty, -1) U (1, \infty)

 

Solution of exercise 11

f(x) = x + \sqrt { x }

f'(x) = 1 + \frac { 1 }{ 2 \sqrt { x }}

1 + \frac { 1 }{ 2 \sqrt { x }} = 0

2 \sqrt { x } + 1 = 0

\sqrt { x } = - \frac { 1 }{ 2 }

Hence we got no solution

Increasing: (0, \infty)

 

Solution of exercise 12

f(x) = \sqrt { x + 1 }

x + 1 \geq 0

x \geq -1

D = [ -1, \infty )

f'(x) = \frac { 1 }{ 2 \sqrt { x + 1 } }

Increasing: (-1, \infty)

 

Solution of exercise 13

f(x) = { e }^{ -{ (x - 1) }^{ 2 } }

f'(x) = -2 (x - 1) { e }^{ - { (x - 1) }^{ 2 } }

-2 (x - 1) { e }^{ - { (x - 1) }^{ 2 } } = 0

-2 (x - 1) = 0 \qquad { e }^{ - { (x - 1) }^{ 2 } } = 0

-2(x - 1) = 0 \qquad { e }^{ - { (x - 1) }^{ 2 } } = 0

x = 1 \qquad No solution

Increasing: (-\infty, 1)

Decreasing: ( 1, \infty)

 

Solution of exercise 14

f(x) = { e }^{ \frac { 1 }{ x } }

f'(x) = - \frac { 1 }{ { x }^{ 2 } } { e }^{ \frac { 1 }{ x } }

-\frac { 1 }{ { x }^{ 2 } } { e }^{ \frac { 1 }{ x } } = 0

-\frac { 1 }{ { x }^{ 2 } } = 0 \qquad { e }^{ \frac { 1 }{ x } } = 0

Both solutions can't be solved

Decreasing: ( - \infty, 0) U (0, \infty)

 

Solution of exercise 15

f(x) = (x - 1) { e }^{ -x }

f'(x) = { e }^{ -x } (2 - x)

{ e }^{ -x } (2 - x) = 0

{ e }^{ -x } = 0 \qquad (2 - x) = 0

x = 2 is the only possible solution

Increasing: (-\infty, 2)

Decreasing: ( 2, \infty)

 

Solution of exercise 16

f(x) = \frac { 1 }{ 2 \sqrt { 2 \pi } } { e }^{ -\frac { 1 }{ 2 } { x }^{ 2 } }

f'(x) = - \frac { 1 }{ \sqrt { 2 \pi } } x { e }^{ -\frac { 1 }{ 2 } { x }^{ 2 } } = 0

x = 0 is the only possible solution

Increasing: (-\infty, 0)

Decreasing: ( 0, \infty)

 

Solution of exercise 17

f(x) = x . ln (x)

x > 0 \qquad D = (0, \infty)

f'(x) = ln (x) + x \frac { 1 }{ x } = ln (x) + 1

ln(x) + 1 = 0

ln(x) = -1

x = { e }^{ -1 }

Increasing: ({ e }^{ -1 }, \infty)

Decreasing: ( 0, { e }^{ -1 })

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Emma

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