Answer:
x = (0, -√(3/5), √(3/5)) are the critical points.
Local maximum at √(3/5), and local minimum is at -√(3/5).
The function is increasing in the interval
(−∞, -√(3/5)) U (√(3/5), ∞)
And decreasing in the interval
(-√(3/5), 0) U (0, √(3/5))
Step-by-step explanation:
Given the function
f(x) = 9x^5 - 3x³ + 6
First of all, take the first derivative of this, to have
f'(x) = 45x^4 - 27x²
The critical point are the points where the first derivative vanishes, that is
f'(x) = 0
Now, solve the equation
45x^4 - 27x² = 0
9x²(5x² - 3) = 0
x = 0 twice
Or
5x² - 3 = 0
5x² = 3
x² = 3/5
x = ±√(3/5)
So, x = (0, -√(3/5), √(3/5)) are the critical points.
Local maximum is when f'(x) > 0, this is √(3/5) in this case,
and local minimum is when f'(x) < 0, this is -√(3/5) in this case.
Now, we need to test for the various intervals to determine where the function increases and decreases.
(−∞, -√(3/5)):
f'(-√(4/5)) = 45(-√(4/5))^4 - 27(-√(4/5))²
= 36/5 > 0. Increasing
(-√(3/5), 0):
f'(-√(2/5)) = 45(-√(2/5))^4 - 9(-√(2/5))²
= -18/5 < 0. Decreasing
(0, √(3/5)):
f'(√(1/5)) = 45(√(1/5))^4 - 9(√(1/5))² = -18/5 < 0. Decreasing
(√(3/5), ∞): f'(1) = 45(1)^4 - 9(1)² =
36 > 0. Increasing.
Therefore, the function is increasing in the interval
(−∞, -√(3/5)) U (√(3/5), ∞)
And decreasing in the interval
(-√(3/5), 0) U (0, √(3/5))
