Consider the equation below. f(x) = 2x3 + 3x2 − 12x (a) find the interval on which f is increasing. (enter your answer in interval notation.) incorrect: your answer is incorrect. find the interval on which f is decreasing. (enter your answer in interval notation.) incorrect: your answer is incorrect. (b) find the local minimum and maximum values of f. local minimum local maximum (c) find the inflection point. (x, y) = find the interval on which f is concave up. (enter your answer in interval notation.) find the interval on which f is concave down. (enter your answer in interval notation.)

a)

The interval on which f is increasing is (-∞ , -2) ∪ (1 , ∞)

The interval on which f is decreasing is (-2 , 1)

b)

The local minimum value is -7 and the local maximum value is 20

c)

The inflection point is (-0.5 , 6.5)

The interval on f(x) is concave up is (-0.5 , ∞) and the interval on f(x) is concave down is (-∞ ,-0.5)

Step-by-step explanation:

* Lets explain how to solve the problem

Remember that;

# f(x) is increasing when f'(x) > 0

# f(x) is decreasing when f'(x) < 0

# f(x) has minimum or maximum points when f'(x) = 0

# f(x) has inflection point when f''(x) = 0

# When the function y = f(x) has a point of inflection (changes from

  concave up to concave down), the graph of its derivative y = f'(x)

  has a maximum or minimum (and so changes from increasing to

  decreasing or decreasing to increasing respectively

* Lets solve the problem

∵ f(x) = 2x³ + 3x² - 12x

∴ f'(x) = 2(3) x² + 3(2) x - 12

∴ f'(x) = 6x² + 6x - 12

- Factorize it

∵ f(x) = 6(x² + x - 2)

∴ f(x) = 6(x - 1)(x + 2)

- Find the values of x if f'(x) = 0

∵ x - 1 = 0 ⇒ add 1 to both sides

∴ x = 1

∵ x + 2 = 0 ⇒ subtract 2 from both sides

∴ x = -2

a)

- If f(x) is increasing when f'(x) > 0

∴ 6(x - 1)(x + 2) > 0 ⇒ divide both sides by 6

∴ (x - 1)(x + 2) > 0

∴ x > 1 and x < -2

∴ The interval on which f is increasing is (-∞ , -2) ∪ (1 , ∞)

- If f(x) is decreasing when f'(x) < 0

∴ 6(x - 1)(x + 2) < 0 ⇒ divide both sides by 6

∴ (x - 1)(x + 2) < 0

∴ -2 < x < 1

∴ The interval on which f is decreasing is (-2 , 1)

b)

- The Local minimum and maximum values is the values of

  y-coordinates of the maximum and minimum points

∵ f(x) has minimum or/and maximum points when f'(x) = 0

∵ f(x) = 0

∴ x = -2 , x = 1

- To find their y substitute the values of them in f(x)

∵ x = -2

∴ f(-2) = 2(-2)³ + 3(-2)² - 12(-2) = 2(-8) + 3(4) - (-24)

∴ f(-2) = -16 + 12 + 24 = 20

∴ y = 20

∵ x = 1

∴ f(1) = 2(1)³ + 3(1)² - 12(1) = 2(1) + 3(1) - 12(1)

∴ f(1) = 2 + 3 - 12 = -7

∴ y = -7

* The local points are (-2 , 20 ) and (1 , -7)

- To Know which one is minimum and which is maximum calculate

  f"(-2) and f"(1)

∵ f'(x) = 6x² + 6x - 12

∴ f"(x) = 12x + 6

- At x = -2

∴ f"(-2) = 12(-2) + 6 = -24 + 6 = -18 < 0

∴ The point (-2 , 20) is local maximum

- At x = 1

∴ f"(1) = 12(1) + 6 = 18 > 0

∴ The point (1 , -7) is local minimum

∵ The Local minimum and maximum values is the values of

  y-coordinates of the maximum and minimum points

∴ The local minimum value is -7 and the local maximum value is 20

c)

- To find the inflection point put f"(x) = 0

∵ f"(x) = 12x + 6

∵ f"(x) = 0

∴ 12x + 6 = 0 ⇒ subtract 6 from both sides

∴ 12x = -6 ⇒ divide both sides by 12

∴ x = -6/12 = -0.5

- To find the y-coordinate of the inflection point find f(-0.5)

∴ f(-0.5) = 2(-0.5)³ + 3(-0.5)² - 12(-0.5) = 2(-0.125) + 3(0.25) - (-6)

∴ f(-0.5) = -0.25 + 0.75 + 6 = 6.5

∴ y = 6.5

* The inflection point is (-0.5 , 6.5)

∵ The function y = f(x) is concave up, when f"(x) > 0

∵ The function y = f(x) is concave down, when f"(x) < 0

∴ The interval on f(x) is concave up is (-0.5 , ∞)

∴ The interval on f(x) is concave down is (-∞ ,-0.5)

* The interval on f(x) is concave up is (-0.5 , ∞) and the interval on

  f(x) is concave down is (-∞ ,-0.5)

if farmer matt works for 8 hours a day, how many days will it take him to completely plant 12 rows of crops? each batch of cookies needs 3/4 cups of sugar.

step-by-step explanation:

combine all like terms then its just a 2 step equation