1. For the function f(x)=4x^3-30x^2+4:
a) Find the critical numbers of f(if any);
b) Find the open intervals where the function is increasing or decreasing; and
c) Apply the First Derivative Test to identify all relative extrema
2. For the function f(x)=(x-1)^(8/11)
a) Find the critical numbers of f(if any);
b) Find the open intervals where the function is increasing or decreasing; and
c) Apply the First Derivative Test to identify all relative extrema
f' = 12x^2 - 60x = 12x(x-5)
as you know
extrema where f'=0
increasing where f' > 0
To answer these questions, we need to find the derivative of the given functions and analyze its properties. Let's address each question step by step:
1. For the function f(x) = 4x^3 - 30x^2 + 4:
a) To find the critical numbers, we set the derivative equal to zero and solve for x. So, let's find the derivative of f(x): f'(x) = 12x^2 - 60x. Now we set f'(x) = 0: 12x^2 - 60x = 0. Factoring out 12x, we get: 12x(x - 5) = 0. This equation holds true when x = 0 or x = 5. Thus, the critical numbers of f(x) are x = 0 and x = 5.
b) To determine the open intervals where the function is increasing or decreasing, we need to examine the sign of the derivative in these intervals. On a number line, we mark the critical numbers (0 and 5) and choose test points in the intervals on either side.
For x < 0, we can use x = -1 as a test point. Plugging -1 into f'(x) = 12x^2 - 60x, we get f'(-1) = 72, which is positive. Therefore, f(x) is increasing on the interval (-∞, 0).
For 0 < x < 5, we can use x = 3 as a test point. Plugging 3 into f'(x) = 12x^2 - 60x, we get f'(3) = -36, which is negative. Therefore, f(x) is decreasing on the interval (0, 5).
For x > 5, we can use x = 6 as a test point. Plugging 6 into f'(x) = 12x^2 - 60x, we get f'(6) = 72, which is positive. Therefore, f(x) is increasing on the interval (5, ∞).
c) To apply the First Derivative Test, we examine the sign of the derivative around the critical numbers to determine relative extrema.
At x = 0, we see that f'(x) changes sign from positive to negative, indicating a relative maximum at x = 0.
At x = 5, f'(x) changes sign from negative to positive, indicating a relative minimum at x = 5.
2. For the function f(x) = (x - 1)^(8/11):
a) To find the critical numbers, we will again find the derivative of f(x) and set it equal to zero. The derivative of f(x) is f'(x) = (8/11) * (x - 1)^(8/11 - 1) = (8/11) * (x - 1)^(-3/11). Setting f'(x) = 0, we get (8/11) * (x - 1)^(-3/11) = 0. Since a fraction is only zero if the numerator is zero, we have (x - 1)^(-3/11) = 0. However, any real number to the power of -3/11 cannot be zero. Hence, there are no critical numbers for f(x).
b) Without critical numbers, we cannot determine open intervals of increasing or decreasing.
c) Since there are no critical numbers, we cannot apply the First Derivative Test to identify relative extrema.
In summary:
1. For the function f(x) = 4x^3 - 30x^2 + 4:
a) The critical numbers are x = 0 and x = 5.
b) The function is increasing on (-∞, 0) and (5, ∞), and decreasing on (0, 5).
c) There is a relative maximum at x = 0 and a relative minimum at x = 5.
2. For the function f(x) = (x - 1)^(8/11):
a) There are no critical numbers.
b) N/A (No critical numbers).
c) N/A (No critical numbers).