6.
f(x)= 12x^5 + 30x^4 - 160x^3 +4
For this function there are four important intervals:(-inf, A] ,
[A,B] ,[B,C],[C, inf) where A ,B ,C are the critical numbers.
Find A B AND C and at those points do they have a local min, max or neither??
To find the critical points of the function f(x), we need to find the values of x where the derivative of f(x) is equal to zero or undefined. The derivative of f(x) can be found by taking the derivative of each term separately:
f'(x) = 60x^4 + 120x^3 - 480x^2
Setting f'(x) equal to zero and solving for x:
60x^4 + 120x^3 - 480x^2 = 0
Factoring out the common term x^2:
60x^2(x^2 + 2x - 8) = 0
Using the zero-product property, we have two cases:
1. 60x^2 = 0
This gives us x = 0 as a critical point.
2. x^2 + 2x - 8 = 0
This equation can be factored as (x + 4)(x - 2) = 0
So we have two more critical points: x = -4 and x = 2.
Now, we need to determine the nature of these critical points by analyzing the concavity of the function.
Taking the second derivative of f(x):
f''(x) = 240x^3 + 360x^2 - 960x
To determine the concavity of the function, we evaluate f''(x) at each critical point.
For x = 0:
f''(0) = 0
For x = -4:
f''(-4) = (-4)^3 * 240 + (-4)^2 * 360 + (-4) * (-960)
= -3840 + 5760 + 3840
= 5760
For x = 2:
f''(2) = (2)^3 * 240 + (2)^2 * 360 + (2) * (-960)
= 960 + 720 - 1920
= -240
Now we can analyze the concavity based on the second derivatives:
At x = 0, f''(0) = 0, so the concavity is unknown.
At x = -4, f''(-4) = 5760 > 0, so we have a local minimum.
At x = 2, f''(2) = -240 < 0, so we have a local maximum.
Therefore, the critical points A, B, and C are:
A = -4 (local minimum)
B = 0 (unknown concavity)
C = 2 (local maximum)