- First, we have to check whether a function is continuous or not in an interval [a, b].
- Then, we have to find out all the critical points of function f(x) in the interval [a, b]. Here, we only observe the critical points in a given interval, where the function is continuous for any other interval.
- Then, we have to find out the function where the critical points are found.
- Finally, we have to identify the absolute extrema.

To understand this more deeply, we just take an example of any function.

k(x) = 2x

^{3}+ 3x

^{2}- 12x + 4 on [-4, 2]

Here, we have to find out the absolute extrema for a given function and interval. We can easily see that, this is a polynomial equation. So, it is continuous in each interval, especially in [-4, 2]. Now, we have to find the critical points of the function. Thatswhy, we find the derivative of the above function which is,

k'(x) = 6x

^{2}+ 6x - 12

Then, we separate the equation in factors as follows:

k'(x) = 6 (x + 2) (x - 1)

So, after solving further, we find that x = -2 and x = 1 as the two critical points. Now, we have to calculate the end points of interval and the function at the critical points. So,

K(-2) = 24k(1) = -3

K(-4) = -28k(2) = 8

There are four points, where the calculus absolute extrema can occur. But, according to absolute extrema, it is the largest and the smallest of function. So,

- Absolute maximum k(x) = 24 occurs at x = -2.
- Absolute minimum k(x) = -28 occurs at x = -4.