Consider the function f (x)=x1=5(x��4). This function

has two critical numbers A < B
Find A
and B
For each of the following intervals, tell whether f (x) is increasing
(type in INC) or decreasing (type in DEC).
(-inf;A]:
[A;B]:
[B;inf)
The critical number A is a relative (type in MAX,
MIN or NEITHER) and the critical number B is a relative
(type in MAX, MIN or NEITHER)
There are two numbers C < D where either f 00(x) = 0 or f 00(x)
is undefined.
Find C and D
Finally for each of the following intervals, tell whether f (x) is
concave up (type in CU) or concave down (type in CD).
(-inf;C):
(C;D)
(D;inf)

To find the critical numbers A and B, we need to find where the derivative of the function f(x) is equal to zero or undefined.

Step 1: Find the derivative of f(x)
f'(x) = 5(4x^3 - 20x^2)

Step 2: Set the derivative equal to zero and solve for x
5(4x^3 - 20x^2) = 0
4x^3 - 20x^2 = 0
4x^2(x - 5) = 0

The critical numbers are x = 0 and x = 5. So A = 0 and B = 5.

Now let's determine the intervals where f(x) is increasing or decreasing.

Step 1: Choose a value less than A, between A and B, and greater than B to test the sign of the derivative.

Let's choose -1, 2, and 6 as test values.

Step 2: Plug the test values into f'(x) = 5(4x^3 - 20x^2) and determine the sign.

For x = -1:
f'(-1) = 5(4(-1)^3 - 20(-1)^2) = -60
Since f'(-1) is negative, f(x) is decreasing on (-inf;A].

For x = 2:
f'(2) = 5(4(2)^3 - 20(2)^2) = 40
Since f'(2) is positive, f(x) is increasing on [A;B].

For x = 6:
f'(6) = 5(4(6)^3 - 20(6)^2) = -480
Since f'(6) is negative, f(x) is decreasing on [B;inf).

The critical number A is a relative minimum, and the critical number B is a relative maximum.

To find the numbers C and D where f''(x) = 0 or f''(x) is undefined, we need to find the second derivative of f(x).

Step 1: Find the second derivative of f(x)
f''(x) = 20(12x^2 - 40x)

Step 2: Set the second derivative equal to zero and solve for x
20(12x^2 - 40x) = 0
12x^2 - 40x = 0
4x(3x - 10) = 0

The numbers C = 0 and D = 10 are where f''(x) = 0. At these points, the second derivative is undefined.

Finally, let's determine the intervals where f(x) is concave up or concave down.

Step 1: Choose a value less than C, between C and D, and greater than D to test the sign of the second derivative.

Let's choose -1, 5, and 12 as test values.

Step 2: Plug the test values into f''(x) = 20(12x^2 - 40x) and determine the sign.

For x = -1:
f''(-1) = 20(12(-1)^2 - 40(-1)) = 20(12 - (-40)) = 1200
Since f''(-1) is positive, f(x) is concave up on (-inf;C).

For x = 5:
f''(5) = 20(12(5)^2 - 40(5)) = 20(300 - 200) = 2000
Since f''(5) is positive, f(x) is concave up on (C;D).

For x = 12:
f''(12) = 20(12(12)^2 - 40(12)) = 20(1728 - 480) = 20(1248) = 24960
Since f''(12) is positive, f(x) is concave up on (D;inf).

To summarize:
(-inf;A]: f(x) is decreasing
[A;B]: f(x) is increasing
[B;inf): f(x) is decreasing

The critical number A is a relative minimum, and the critical number B is a relative maximum.

C = 0 and D = 10 are where f''(x) = 0 or f''(x) is undefined.

(-inf;C): f(x) is concave up
(C;D): f(x) is concave up
(D;inf): f(x) is concave up

To find the critical numbers of the function f(x) = x^5 - 4x, we need to find the values of x where the derivative of the function is either 0 or undefined.

Step 1: Find the derivative of f(x).
f'(x) = 5x^4 - 4

Step 2: Set f'(x) = 0 to find the values where the derivative is 0.
5x^4 - 4 = 0

Step 3: Solve the equation for x.
5x^4 = 4
x^4 = 4/5
x = ±(4/5)^(1/4)

The critical numbers are the values of x where the derivative is either 0 or undefined. In this case, there are no values where the derivative is undefined.

Critical number A: x = -(4/5)^(1/4) (approximately -0.8706)
Critical number B: x = (4/5)^(1/4) (approximately 0.8706)

Now, let's analyze the intervals and determine whether f(x) is increasing or decreasing and whether it is concave up or concave down.

Interval (-∞, A]:
To determine if f(x) is increasing or decreasing in this interval, we can choose any value less than A. Let's choose x = -1. Substitute x = -1 into the derivative f'(x).
f'(-1) = 5(-1)^4 - 4 = 5 - 4 = 1
Since the derivative is positive (1), f(x) is increasing in the interval (-∞, A]. Therefore, f(x) is INC (increasing).

Interval [A, B]:
To determine if f(x) is increasing or decreasing in this interval, we can choose any value between A and B. Let's choose x = 0. Substitute x = 0 into the derivative f'(x).
f'(0) = 5(0)^4 - 4 = -4
Since the derivative is negative (-4), f(x) is decreasing in the interval [A, B]. Therefore, f(x) is DEC (decreasing).

Interval [B, ∞):
To determine if f(x) is increasing or decreasing in this interval, we can choose any value greater than B. Let's choose x = 1. Substitute x = 1 into the derivative f'(x).
f'(1) = 5(1)^4 - 4 = 5 - 4 = 1
Since the derivative is positive (1), f(x) is increasing in the interval [B, ∞). Therefore, f(x) is INC (increasing).

The critical number A is a relative MIN (minimum) because f'(x) changes from negative to positive at x = A. The critical number B is a relative MIN (minimum) because f'(x) changes from positive to negative at x = B.

To find the values of x where the second derivative f''(x) is either 0 or undefined, we need to find the derivative of f'(x) obtained earlier.

f''(x) = d/dx (5x^4 - 4)
= 20x^3

Set f''(x) = 0 to find the values where the second derivative is 0.
20x^3 = 0
x^3 = 0
x = 0

The second derivative is defined for all real numbers, so there are no values where it is undefined.

There are no values where either f''(x) = 0 or f''(x) is undefined.

Finally, let's analyze the intervals and determine whether f(x) is concave up or concave down.

Interval (-∞, C):
To determine if f(x) is concave up or concave down in this interval, we can choose any value less than C. Since there are no values for C, we do not have an interval here.

Interval (C, D):
To determine if f(x) is concave up or concave down in this interval, we can choose any value between C and D. Since there are no values for C and D, we do not have an interval here.

Interval (D, ∞):
To determine if f(x) is concave up or concave down in this interval, we can choose any value greater than D. Since there are no values for D, we do not have an interval here.

In conclusion, there are no values for C and D, and therefore, we cannot determine the concavity of the function.