Increasing and Decreasing Functions.
Let f(x) = 4x^3 - 9x^2 -30x +6
A. Find all critical numbers of f(x)
B. Give the open intervals where f (x) is increasing and decreasing and clearly label each.
f' = 12x^2 - 18x - 30
Now use your Algebra I skills.
(A) the critical numbers are where f'=0
(B) f is increasing where f' > 0.
To find the critical numbers of a function, we need to identify the values of x at which the derivative of the function equals zero or is undefined.
A. To find the critical numbers of f(x), we need to find its derivative, f'(x), by taking the derivative of each term:
f'(x) = d/dx (4x^3 - 9x^2 - 30x + 6)
= 12x^2 - 18x - 30
Now we set f'(x) equal to zero and solve for x:
12x^2 - 18x - 30 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 12, b = -18, and c = -30. Plugging these values into the quadratic formula, we get:
x = (-(-18) ± √((-18)^2 - 4 * 12 * (-30))) / (2 * 12)
= (18 ± √(324 + 1440)) / 24
= (18 ± √(1764)) / 24
= (18 ± 42) / 24
This gives two possible values for x: (18 + 42)/24 = 60/24 = 5/2 and (18 - 42)/24 = -24/24 = -1.
Therefore, the critical numbers of f(x) are x = 5/2 and x = -1.
B. To determine the intervals where f(x) is increasing or decreasing, we need to examine the behavior of the derivative f'(x) on different intervals and use the test known as the First Derivative Test.
First, we'll evaluate f'(x) at some test points within the intervals determined by the critical numbers.
For x < -1, we can choose a test point, x = -2:
f'(-2) = 12(-2)^2 - 18(-2) - 30
= 48 + 36 - 30
= 54
Since f'(-2) = 54 > 0, f(x) is increasing on the interval (-∞, -1).
For -1 < x < 5/2, we can choose a test point, x = 0:
f'(0) = 12(0)^2 - 18(0) - 30
= -30
Since f'(0) = -30 < 0, f(x) is decreasing on the interval (-1, 5/2).
For x > 5/2, we can choose a test point, x = 3:
f'(3) = 12(3)^2 - 18(3) - 30
= 108 - 54 - 30
= 24
Since f'(3) = 24 > 0, f(x) is increasing on the interval (5/2, ∞).
Therefore, the open intervals where f(x) is increasing are (-∞, -1) and (5/2, ∞), and the open interval where f(x) is decreasing is (-1, 5/2).