Question

Given the function f(x) = x4 – 4r– 3, determine the absolute minimum value of f on the closed interval (-2,4).​

Answer 1

The absolute minimum of the function
`f(x) = x^(4) - 4\cdot x - 3` occurs at
`x = 1` and is
`f(1) = -6`.

Step-by-step explanation:

Statement is incorrect. Correct statement is presented below:

Given the function
`f(x) = x^(4)-4\cdot x - 3`, determine the absolute minimum value of
`f` on the closed interval
`(-2, 4)`. First, we determine the first and second derivatives of the function.

First Derivative

`f'(x) = 4\cdot x^(3)-4` (1)

Second Derivative

`f''(x) = 12\cdot x^(2)` (2)

By equalizing (1) to zero, we solve for
`x`:

`4\cdot x^(3)-4 = 0`

`x^(3)= 1`

`x = 1`

And we evaluated this result in (2):

`f''(1) = 12`

According to criteria of the Second Derivative Test, we conclude that value of
`x` leads to an absolute minimum. The value of the absolute minimum is:

`f(1) = -6`

The absolute minimum of the function
`f(x) = x^(4) - 4\cdot x - 3` occurs at
`x = 1` and is
`f(1) = -6`.