# Meaning of relative maxima and minima

The relative minima and maxima for a given function g(x) are the smallest and largest value respectively that a function can gain in selected range of x.

The figure given below depicts the picture of the calculus maxima and minima of a function.

## Meaning of absolute maxima and minima

Since we have gone through relative maxima and minima, we will be learning other types of minima and maxima. If g(x) is a any function defined on some fixed range [x1,x2], then absolute maxima and minima is defined as maximum point at x0 if g(x0) ≥ g(x) for all x in [x1,x2] and minimum point at x0 if g(x0) ≤ g(x) for all x in [x1,x2].

Some of the common facts for absolute max and min of a function are given below.

• A function can have both max and min or either of these or none of these absolute points.

• For a continuous function there is always absolute max and min on a closed interval.

Now let us discuss about maxima and minima calculus. With the help of calculus method, we can find maxima and minima very easily. Suppose g(x) is a function of x, the value of x at which g'(x) (derivative of g(x)) have zero value is either a maximum or minimum point of a function. If g''(x) (second derivative of g(x)) is positive, then function has a minimum value and if g''(x) is negative, then function has a positive value for that particular point.

## How we can get absolute maximum of particular function?

To find absolute maximum of particular function g(x), find g'(x) (derivative of g(x)) and find x for which g'(x) become zero. Then find g''(x) (second derivative of g(x)), and put values of x found in step 1, the value for which g''(x) become negative is called maximum of function. Let see an example to get proper understanding of the subject.

(1) Find the maximum value for g(x) = 2x3 – 9x2 -24x + 2 for the interval 0 ≦ x ≦ 5:

Finding the derivative of given function, g'(x) = 6x2 – 18x -24 = 6(x2 – 3x – 4) = 6(x – 4)(x + 1). So g'(x) = 0 if x = 4 or x = -1. But the range is between 0 and 5. Now g''(x) = 12x – 18. For x = 0, g''(x) is negative. Also g(0) = 2. So the function has maximum value at x = 0. The process for finding maximum is explain in below graph.

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