Answer:
The intervals on which f is increasing are (- ∞, 0) ∪ (2, ∞)
and the interval on which f is decreasing is (0, 2)
P(0, 15) is local maximum
P(2, 3) is a local minimum
The intervals on which f is increasing are (- ∞, 0) ∪ (2, ∞)
and the interval on which f is decreasing is (0, 2)
P(1, 9) is the inflection point
Explanation:
Let
f(x) = 15−9x²+3x³
then we can apply
f'(x) = 0 ⇒ (15 − 9x² + 3x³)' = -18x + 9x² = 0
⇒ 9x*(x - 2) = 0
⇒ x₁ = 0 ∧ x₂ = 2
When - ∞ < x < 0
Example: x = -1
f'(-1) = -18*(-1) + 9*(-1)² = 18 + 9 = 27 > 0
⇒ f'(x) > 0
When 0 < x < 2
Example: x = 1
f'(1) = -18*(1) + 9*(1)² = -18 + 9 = -9 < 0
⇒ f'(x) < 0
When 2 < x < ∞
Example: x = 3
f'(3) = -18*(3) + 9*(3)² = -54 + 81 = 27 > 0
⇒ f'(x) > 0
The intervals on which f is increasing are (- ∞, 0) ∪ (2, ∞)
and the interval on which f is decreasing is (0, 2)
We can find f(x₁) and f(x₂) as follows
f(x₁) = f(0) = 15−9(0)²+3(0)³ = 15
f(x₂) = f(2) = 15−9(2)²+3(2)³ = 15 - 36 + 24 = 3
P(0, 15) is local maximum
P(2, 3) is a local minimum
Now, we can apply
f"(x) = 0 ⇒ (-18x + 9x²)' = -18 + 18x = 0
⇒ 18*(x- 1) = 0
⇒ x = 1
When - ∞ < x < 1
Example: x = 0
f"(0) = 18*(0- 1) = -18 < 0
⇒ f"(x) < 0
When 1 < x < ∞
Example: x = 2
f"(0) = 18*(2- 1) = 18 > 0
⇒ f"(x) > 0
then
the interval on which f is concave up is (1, ∞) and
the interval on which f is concave down is (- ∞, 1)
We can find f(1) as follows
f(1) = 15−9(1)²+3(1)³ = 9
P(1, 9) is the inflection point