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)

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

step-by-step explanation: