Question

given the function f(x)= x^(4)-4x^(3)-3 determine the absolute minimum value of f on the closed interval [-1,4]

Answer 1

To find the absolute minimum value of f(x) = x⁴ - 4x³ - 3 on the closed interval [-1, 4], we can use the method of finding critical points and evaluating the function at those points and the interval endpoints.

Step-by-step explanation:

Calculation of the absolute minimum value of f(x) = x⁴ - 4x³ - 3 on the closed interval [-1, 4]

Start by finding the critical points by taking the derivative of the function: f'(x) = 4x³ - 12x²

1. Set f'(x) = 0 and solve for x to find the critical points. In this case, the critical points are x = 0 and x = 3.
2. Next, evaluate f(x) at the critical points and the endpoints of the interval: f(-1) = 8, f(0) = -3, f(3) = -18, and f(4) = 53.
3. Compare the values of f(x) at the critical points and endpoints to find the absolute minimum value. In this case, the absolute minimum value is f(3) = -18.

Answer 2

The absolute minimum value of the function over the closed interval [-1,4] is -3.

What the absolute minimum value of a polynomial function?

For us to find the absolute minimum value of the function f(x)= x^(4) - 4x^(3) - 3 on the closed interval [-1,4], we need to find the function at the critical points within the closed interval and at the endpoints.

The critical point is the point where the derivative is equal to zero;

f(x)= x^(4) - 4x^(3) - 3

Derivative of f(x) i.e. f'(x) is

f'(x) = 4x³ - 12x²

If f'(x) = 0

0 = 4x³ - 12x²

4x²(x - 3) = 0

Using the closed interval [-1,4] to evaluate the original function at the critical points and end points;

f(-1) = (-1)⁴ -4(-1)³ - 3

f(-1) = 1 + 4 -3

f(-1) = 2

f(0) = 0⁴ - 4(0)³ - 3

f(0) = -3

f(3) = 3⁴ - 4(3)³ - 3

f(3) = 81 - 108 - 3

f(3) = -30

f(4) = 4⁴ - 4(4)³ - 3

f(4) = 256 - 256 - 3

f(4) = -3

At critical points; f(0) = -3 and f(3) = -30. Thus, the absolute minimum value of f(x) = x⁴ - 4x³ - 3 on the closed interval [-1, 4] is -30 and it occurs at x = 3.