f(x)=4x3−18x2−480x−2
is decreasing on what interval?
It is increasing on what interval(s) ?
The function has a local maximum at ?
To determine if a function is increasing or decreasing on a given interval, we need to analyze the sign of its derivative.
To find the derivative of the given function f(x) = 4x^3 - 18x^2 - 480x - 2, we differentiate it with respect to x using the power rule of differentiation.
f'(x) = 12x^2 - 36x - 480
To find where the function is decreasing, we need to find the interval(s) where the derivative is negative. Set f'(x) < 0 and solve for x.
12x^2 - 36x - 480 < 0
Since this is a quadratic inequality, we can either factor it or use the quadratic formula to solve it.
Factoring, we have:
12(x - 16)(x + 5) < 0
We find that the critical points are x = -5 and x = 16.
To determine the sign intervals, we create a number line and test the intervals between the critical points and beyond:
For x < -5:
Pick a test point x = -6
Substitute it into the inequality:
12(-6 - 16)(-6 + 5) < 0
(-22)(-1) < 0
22 > 0
This interval satisfies the inequality.
For -5 < x < 16:
Pick a test point x = 0
Substitute it into the inequality:
12(0 - 16)(0 + 5) < 0
(-16)(5) < 0
-80 < 0
This interval satisfies the inequality.
For x > 16:
Pick a test point x = 17
Substitute it into the inequality:
12(17 - 16)(17 + 5) < 0
(1)(22) < 0
22 > 0
This interval does not satisfy the inequality.
Thus, the function is decreasing on the interval (-∞, -5) and (16, ∞).
To find where the function is increasing, we need to find the interval(s) where the derivative is positive. Set f'(x) > 0 and solve for x.
12x^2 - 36x - 480 > 0
Using the same critical points, x = -5 and x = 16, we can create the sign intervals as follows:
For x < -5:
Pick a test point x = -6
Substitute it into the inequality:
12(-6 - 16)(-6 + 5) > 0
(-22)(-1) > 0
22 > 0
This interval does not satisfy the inequality.
For -5 < x < 16:
Pick a test point x = 0
Substitute it into the inequality:
12(0 - 16)(0 + 5) > 0
(-16)(5) > 0
-80 > 0
This interval does not satisfy the inequality.
For x > 16:
Pick a test point x = 17
Substitute it into the inequality:
12(17 - 16)(17 + 5) > 0
(1)(22) > 0
22 > 0
This interval satisfies the inequality.
Thus, the function is increasing on the interval (-5, 16).
To find the local maximum, we need to find the critical points of the function. These occur where the derivative is equal to zero or does not exist.
From the derivative, we already found the critical points at x = -5 and x = 16.
To determine if these points are local maximum or minimum, we can use the second derivative test. Let's find the second derivative of the function:
f''(x) = 24x - 36
Now substitute the critical points x = -5 and x = 16 into the second derivative:
For x = -5:
f''(-5) = 24(-5) - 36 = -120 - 36 = -156
Since f''(-5) is negative, the function has a local maximum at x = -5.
For x = 16:
f''(16) = 24(16) - 36 = 384 - 36 = 348
Since f''(16) is positive, the function does not have a local maximum or minimum at x = 16.
Thus, the function f(x) = 4x^3 - 18x^2 - 480x - 2 is decreasing on the interval (-∞, -5), increasing on the interval (-5, 16), and has a local maximum at x = -5.
f'(x) = 12x^2 - 36x - 480
=0 for max/mins
divide by 12
x^2 - 3x - 40=0
(x-8)(x+5) = 0
x = 8 or x = -5
You should know the general behaviour of cubic curve with a positive x^3 term and critical values at -5 and 8
I will give you the decreasing interval, it is
between -5 and 8 or
-5 < x < 8