Respuesta :
Answer:
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)
