The word ‘maxima’ is a plural of ‘maximum’ and the word ‘minima’ is the plural of minimum. Before we proceed further let us first discuss what a function is. A function is relation, subject to certain conditions, describes the output for a given input. In simple terms it is a ‘mathematical machine’ which converts the input and the type of conversion depends on the nature of the relation described in the function.

In any case, the output of a function varies with the variation of input except in case of constant functions. The output may increase or decrease or may undergo both. In many cases the phenomenon may repeat. When a function is increasing in an interval and starts decreasing subsequently, the changeover point represents the maximum value in the given interval. Hence we call that point as ‘local’ or ‘relative’ maximum. We use such a prefix because a similar situation may occur also in some other intervals in the domain of the function. In other words, a function can have a number of peak points which is collectively called as ‘maxima’ of the function. As the reverse case, a function can also have a number of local minimum points.

## Formula

As said earlier a function could have maxima and minima at several points. At those points, if we draw tangents to the graph of the function, all of them would be horizontal. Means the slope of the tangent at such places is zero. As one of the applications of derivative in differential calculus, we can identify the points of maxima and minima of a function without actually graphing the function. We shall come back to this study a little later.

However, a formula can be derived in case of quadratic functions for maximum or minimum points and values without the help of calculus concepts. The general form of a quadratic function is $f(x)$ = $ax^2 + bx + c$ and it can have either a minimum value (if $a > 0$) or a maximum value (if $x < 0$). The maximum or the minimum occurs at $x$ = $h$ = $\frac{-b}{2a}$ and the maximum or minimum value of the function would be $f(h)$.

## Maxima and Minima of Functions

The graph of a typical continuous function is shown above. Consider the interval $(A, C)$. The function is increasing till it reaches B and starts decreasing immediately after that. Obviously the point B is a maximum point local or relative to the interval $(A, C)$. The case is reverse in the interval $(B, D)$. The function is decreasing till the point $C$ and starts increasing immediately after that. So, the point $C$ is a local or relative minimum. Like this, we notice another local maximum and minimum at points $D$ and $E$ respectively. It may be noted that the maximum and minimum values are different compared to the previous ones.

The tangents drawn at all local maxima or minima points are horizontal meaning their slope are $0$. But the slope of a tangent drawn anywhere in the increasing intervals is positive and the slope of the tangent drawn anywhere in the decreasing intervals is negative. Now as per the concept of calculus the slope of a tangent drawn at any point on the graph of a function is equal to the value of the derivative of the function at that point. Therefore, conversely we can say that if the derivative of a function is positive at any point, then the function is increasing at that point and vice-versa. If the derivative is $0$, then that point could be a local maximum or local minimum. This is a necessary condition for maxima and minima of a function but not a sufficient condition. Look at the situation at point $G$. There also the slope of the tangent and hence the derivative of the function is $0$ but it is neither a local maximum nor a local minimum. What is that then? On either side of that point we see the slopes of the tangent are positive because the function is in increasing status on both sides of point $G$. Such a point is called as point of inflection. It is also possible that in a point of inflection the function may continue to decrease.

It may be noted that a local minimum point in a given interval need not be the point of the lowest value. It only shows up to what minimum value the function decreases at that point before it starts increasing. It may be possible some other point in the given interval may have a lesser value but at that point the function may be still decreasing and not turning. Such a point is called as point of Global (or Absolute) Minimum and the value at that point is called as Global (or Absolute) minimum value. Same way we can define a global maximum.

## Rules

The theory discussed in the previous section helps us to frame the rules for determining and identifying the points of maxima and minima of a function.

1) Find the derivative of the function and equate it to $0$  . Solve for the values of the variable which satisfy the equation.

2) For each value obtained, find the sign of the derivative with two test points, one on the lower side and one on the higher side. If the sign changes from positive to negative, then the point correspond to the obtained value is a local maximum. If the case is reverse, then the point is a local minimum. If there is no change in sign, then it is a point of inflection.

3) In a given interval, find all the values of local maxima and minima. Also evaluate the function at both ends of the interval. The least of all these values is the global minimum and the largest of all these values is the global maximum.

## Examples

Example 1:

Find the vertex of the quadratic function $f(x)$ = $2x^2 – 8x + 3$. Identify whether it is a point of maximum or minimum of the function and also its value.

Solution:

Since the given function is a quadratic, the maximum occurs at $x$ = $\frac{-b}{2a}$ = $-$ $\frac{(-8)}{(2 \times 2)}$ = $2$. Since the leading coefficient of the function is positive, the vertex represents only the minimum point. And the minimum value of the function is $f(2)$ = $2(2^2) – 8(2) + 3$ = $-5$.
Example 2:

Find the local maximum and minimum values of the function $f(x)$ = $x^3 - 6x^2 + 9x - 3$ and also find the global maximum and minimum in the interval $[0, 5]$

Solution:

The derivative $f’(x)$ = $3x^2 – 12x + 9$ = $3(x^2 – 4x + 3)$ = $3(x -1)(x - 3)$

$f’(x)$ = $0$ at $x$ = $1$ and at $x$ = $3$.

For $x$ = $1$, consider test points $x$ = $0$ and $x$ = $2$

$f’(0)$ = $9$ and $f’(2)$ = $-3$.

Since the first derivative changes from positive to negative at $x$ = $1$, it is a point of local maximum. And its value is $f(1)$ = $1$

For $x$ = $3$, consider test points $x$ = $2$ and $x$ = $4$

$f’(2)$ = $-3$ and $f’(4)$ = $1$

Since the first derivative changes from negative to positive at $x$ = $3$, it is a point of local minimum. And its value is $f(3)$ = $-3$

Evaluating the function at the end points of $[0, 5],\ f(0)$ = $-3$ and $f(5)$ = $17$.

Therefore, the global maximum value is $17$ at $x$ = $5$ and the global minimum value is $-3$ at $x$ = $0$ in the interval $[0, 5]$