Maxima and minima: In linear algebra and game theory, finding maxima or minima is crucial. They are the maximum and minimum extrema of a function. The maximum and smallest values of a function within a certain set of ranges are known as maxima and minima. The largest value of the function under the full range is known as the absolute maxima, while the least value is known as the absolute minima. There are various applications in real life. The goal of piping system design is to minimize pressure drop, which reduces the size of required pumps and saves money. Steel beam forms are designed to maximize strength.

Let’s take a closer look at local maxima and minima, absolute maxima and minima, and how to calculate the maximum and minima of a function.

Turning Points

The notion of derivatives is used to find maxima and minima in calculus. We locate the points where the gradient is zero. These points are called turning points or stationary points, as we know the idea of derivatives offers us information about the gradient or slope of the function. These are the points corresponding to the largest and smallest values of the function.

Learn about Derivative of a Function here

What are Maximum and Minimum Values?

Maximum Value

Let $f\left(x\right)$ be a real function defined on an interval $\left[a,b\right].$ Then, $f\left(x\right)$ is said to have the maximum value in $\left[a,b\right],$ If there exists a point $c$ in $\left[a,b\right]$ such that $f\left(x\right)⩽f\left(c\right)$ for all $x\in \left[a,b\right].$
In such a case, the number $f\left(c\right)$ is called the maximum value of $f\left(x\right)$ in the interval $\left[a,b\right],$ and the point $c$ is called the point of a maximum value of $f$ in the interval $\left[a,b\right].$

Consider the function $f$ given by $f\left(x\right)=–{\left(x–1\right)}^2+10.$ Here, the domain $\left(f\right)=R=\left(–\mathrm{\infty },\mathrm{\infty }\right).$
We observe that
$–{\left(x–1\right)}^2⩽0$ for all $x\in R$
$\Rightarrow –{\left(x–1\right)}^2+10⩽10$ for all $x\in R$
$\Rightarrow f\left(x\right)⩽10$ for all $x\in R$
$\Rightarrow f\left(x\right)⩽f\left(1\right)$ for all $x\in R$ $\left[\because f\left(1\right)=–{\left(1–1\right)}^2+10=10\right]$
Here,
$f\left(1\right)=10$ is the maximum value of function $f$
point of the maximum value of $f$ is $x=1$

Minimum Value

Let $f\left(x\right)$ be a real function defined on an interval $\left[a,b\right].$ Then, $f\left(x\right)$ is said to have a minimum value in the interval $\left[a,b\right],$ if there exists a point $c\in \left[a,b\right]$ such that $f\left(x\right)⩾f\left(c\right)$ for all $x\in \left[a,b\right].$
In such a case, the number $f\left(c\right)$ is called the minimum value of $f\left(x\right)$ in the interval $\left[a,b\right].$ The point $c$ is called a point of minimum value of $f$ in the interval $\left[a,b\right].$

Consider $f\left(x\right)=x^2+5.$ Here, domain $\left(f\right)=R=\left(–\mathrm{\infty },\mathrm{\infty }\right).$
We know that
$x^2⩾0$ for all $x\in R$
$\Rightarrow x^2+5⩾5$ for all $x\in R$
$\Rightarrow f\left(x\right)⩾5$ for all $x\in R$
$\Rightarrow f\left(x\right)⩾f\left(0\right)$ for all $x\in R$
It is clear that the minimum value of function $f\left(x\right)=x^2+5$ defined on $R$ is $5.$
Point of minimum value of $f$ is $x=0.$

The function $f\left(x\right)=–{\left(x–1\right)}^2+10,x\in R$ has the maximum value, but it does not attain the minimum value, because $–{\left(x–1\right)}^2+10$ can be made as small as possible.
The function $f\left(x\right)=x^2+5$ attains the minimum value $5$ at $x=0,$ but it does not attain the maximum value at any point in its domain.
Example:
Consider the function $f\left(x\right)=\sin x,$ defined on the interval $\left[0,2\pi \right].$
Since $–1⩽\sin x⩽1$ for all $x\in \left[0,2\pi \right]$
$\Rightarrow –1⩽f\left(x\right)⩽1$ for all $x\in \left[0,2\pi \right]$
Since $f\left(\frac{\pi }{2}\right)=1$ and $f\left(\frac{3\pi }{2}\right)=–1$
$\therefore f\left(\frac{3\pi }{2}\right)⩽f\left(x\right)⩽f\left(\frac{\pi }{2}\right),$ for all $x\in \left[0,2\pi \right].$

Thus, $f\left(x\right)$ attains both the maximum value $1$ and the minimum value $–1$ in the interval $\left[0,2\pi \right].$ Points $x=\frac{\pi }{2}$ and $x=\frac{3\pi }{2}$ are respectively the points maximum and minimum values of $f$ in the interval $\left[0,2\pi \right].$
Example:
Consider the function $f\left(x\right)=x^3$ defined on $\left(–2,2\right).$ It is an increasing function in the given interval. So, it should have the minimum value at a point closest right of $–2$ and the maximum value at a point closest left of $2.$ In fact, it is not possible to locate such points. Therefore, $f\left(x\right)=x^3$ has neither the maximum value nor the minimum value in $\left(–2,2\right).$

From these examples, we can say that $f$ defined on an interval $I$ may:

1. attain the maximum at a point in $I$ but not the minimum value at any point in $I.$
2. attain the minimum at a point in $I$ but not the maximum value at any point in $I.$
3. attain both the maximum and minimum values at some points in $I.$
4. not attain both the maximum and minimum values at any point in $I.$

Local Maxima and Local Minima

There may be points in the domain of a function where it does not attain the greatest (or the least) value, but the values at these points are greater than or less than the values of the function at the neighbouring points. Such points are known as the points of local minima or local maxima.
Local Maxima: A function $f\left(x\right)$ is said to attain a local maximum at $x=a$ if there exists a neighbourhood $\left(a–\delta ,a+\delta \right)$ of a such that $f\left(x\right)<f\left(a\right)$ for all $x\in \left(a–\delta ,a+\delta \right),x\ne a.$
Here, $f\left(a\right)$ is called the local maximum value of $f\left(x\right)$ at $x=a.$
Local Minima: A function $f\left(x\right)$ is said to attain a local minimum at $x=a$ if there exists a neighbourhood $\left(a–\delta ,a+\delta \right)$ of a such that $f\left(x\right)>f\left(a\right)$ for all $x\in \left(a–\delta ,a+\delta \right),x\ne a.$
Here, $f\left(a\right)$ is called the local minimum value of $f\left(x\right)$ at $x=a.$
Consider the following figure

In the above figure, observe that the $x-$coordinates of $A,C$ and $E$ are the points of the local maximum. The values at the points,i.e. their $y-$coordinates are the local maximum values of $f\left(x\right).$ Similarly, $x-$coordinates of $B$ and $D$ are points of local minimum, and their $y-$coordinates are the local minimum values of $f\left(x\right).$
Note: A function can have any number of local maximum (or minimum) points, and a local minimum value can even be bigger than a local maximum value.
In simpler words, a local maximum may not be the highest value of the function in its domain, and a local minimum may not be the lowest value in its domain.

First Derivative Test for Local Maxima and Local Minima

Let $f$ be a differentiable function defined in an interval $I$ and let $a\in I.$ Then,

1. $x=a$ is a point of local maximum, if $f^{\prime }\left(a\right)=0$ and $f^{\prime }\left(x\right)$ changes sign from positive to negative as $x$ increases through $a.$
2. $x=a$ is a point of local minimum, if $f^{\prime }\left(a\right)=0$ and $f^{\prime }\left(x\right)$ changes sign from negative to positive as $x$ increases through $a.$
3. If $f^{\prime }\left(a\right)=0$ and $f^{\prime }\left(x\right)$ does not change the sign as it increases through $a,$ i.e.$f^{\prime }\left(x\right)$ has the same sign in the neighbourhood of $a,$ then $a$ is neither a point of a local maximum value nor a point of local minimum value. Such a point is called as the point of inflection.

Algorithm to find the Local Maxima or Local Minima

Step 1: Put $y=f\left(x\right)$ and find $\frac{dy}{dx}.$
Step 2: Put $\frac{dy}{dx}=0$
Step 3: Solve the equation for $x.$
Let $c_1,c_2,c_3.,\dots .,c_n$ be the roots of this equation. These points are called the critical points and these are the possible points where the function can attain a local maximum or a local minimum. So, we test the function at each of these points.
Step 4: Consider $x=c_1$

Change in sign of $\frac{dy}{dx}$ as $x$ increases through $c_1$	Function attains
Positive to negative	Local maximum at $x=c_1$
Negative to positive	Local minimum at $x=c_1$

If $\frac{dy}{dx}$ does not change the sign as $x$ increases through $c_1,$ then $x=c_1$ is not a point of local maximum, nor is it a point of local minimum. In this case $x=c_1$ is a point of inflection.

Second-Order Derivative Test:

Theorem: Let $f$ be a differentiable function defined on an interval $I$ and let $c$ be an interior point of $I$ such that
$f^{\prime }\left(c\right)=0$ and $f”\left(c\right)$ exists and is non-zero.
Then,
If $f”\left(c\right)<0\Rightarrow x=c$ is a point of local maximum, and the value $f\left(c\right)$ is the local maximum value of $f.$
If $f”\left(c\right)>0\Rightarrow x=c$ is a point of local minimum, and $f\left(c\right)$ is the local minimum value of $f.$
The test fails if $f^{\prime }\left(c\right)=0$ and $f”\left(c\right)=0.$
Algorithm:
Step 1: Put $y=f\left(x\right)$ and find $f^{\prime }\left(x\right).$
Step 2: Put $f^{\prime }\left(x\right)=0$ and solve the equation for $x.$ Let $c_1,c_2,c_3,\dots ,c_n$ be the roots of this equation. Points $c_1,c_2,c_3,\dots ,c_n$ are critical points and these are the possible points where the function can attain a local maximum or a local minimum. So, we test the function at each of these points.
Step 3: Find $f”\left(x\right).$ Consider $x=c_1,$
If $f”\left(c_1\right)<0$ then $x=c_1$ is a point of local maximum.
If $f”\left(c_1\right)>0$ then $x=c_1$ is a point of local minimum.
If $f”\left(c_1\right)=0,$ then the test fails, and then we can apply the first derivative test.
Similarly, we can check for all critical points.

Absolute Maxima and Minima

An absolute maximum point is a point where the function obtains its greatest possible value. Similarly, an absolute minimum point is a point where the function obtains its least possible value.
Algorithm to find absolute maxima and minima:
Step 1: Find $f^{\prime }\left(x\right).$
Step 2: Put $f^{\prime }\left(x\right)=0$ and find values of $x.$ Let $c_1,c_2,\dots .,c_n$ be the values of $x.$
Step 3: Take the maximum and minimum values out of the values $f\left(a\right),f\left(c_1\right),f\left(c_2\right),\dots ,f\left(c_n\right),f\left(b\right).$
These maximum and minimum values are the function’s absolute maximum and absolute minimum values, respectively.

Solved Examples-Maxima and Minima

Q.1. Find the maximum and the minimum values of the following function.
$f\left(x\right)=3x^2+6x+8,x\in R$
Ans: Given:
$f\left(x\right)=3x^2+6x+8,x\in R$
$\Rightarrow f\left(x\right)=3\left(x^2+2x+1\right)+5$
$=3{\left(x+1\right)}^2+5$
Clearly,
$3{\left(x+1\right)}^2⩾0$ for all $x\in R$
$\Rightarrow 3{\left(x+1\right)}^2+5⩾5$ for all $x\in R$
$\Rightarrow f\left(x\right)⩾f\left(–1\right)$ for all $x\in R.$ $\left[\because f\left(–1\right)=3{\left(–1\right)}^2+6\left(–1\right)+8=5\right]$
Thus, $5$ is the minimum value of $f\left(x\right)$ at $x=–1.$

Note that $f\left(x\right)$ can be made as large as we please. Therefore, the maximum value does not exist, as seen in the graph.

Q.2. Find all points of local maxima and minima of the function $f\left(x\right)=x^3–6x^2+9x–8.$
Ans: Let $y=f\left(x\right)=x^3–6x^2+9x–8.$ Then,
$\frac{dy}{dx}=f^{\prime }\left(x\right)=3x^2–12x+9=3\left(x^2–4x+3\right)$
The critical points of $f\left(x\right)$ are given by $f^{\prime }\left(x\right)=0$ or $\frac{dy}{dx}=0.$
Now, $\frac{dy}{dx}=0\Rightarrow 3\left(x^2–4x+3\right)=0\Rightarrow x=1,3.$
We have to examine whether these points are points of local maxima or local minima or neither of them.
We have, $\frac{dy}{dx}=3\left(x–1\right)\left(x–3\right)$
The changes in signs of $\frac{dy}{dx}$ for different values of $x$ are shown in the figure given below:

Clearly, $\frac{dy}{dx}$ changes sign from positive to negative asx increases through $1.$
So, $x=1$ is a point of the local maximum.
Also, $\frac{dy}{dx}$ changes sign from negative to positive as $x$ increases through $3.$
So, $x=3$ is a point of local minima.

Q.3. Show that the function $f\left(x\right)=4x^3–18x^2+27x–7$ has neither maxima nor minima.
Ans: Given: $f\left(x\right)=4x^3–18x^2+27x–7$
$f^{\prime }\left(x\right)=12x^2–36x+27=3\left(4x^2–12x+9\right)=3{\left(2x–3\right)}^2$
The critical points of $y=f\left(x\right)$ are given by $\frac{dy}{dx}=0.$
Now, $\frac{dy}{dx}=0\Rightarrow 3{\left(2x–3\right)}^2=0\Rightarrow 2x–3=0\Rightarrow x=\frac{3}{2}$
Clearly, $\frac{dy}{dx}=3{\left(2x–3\right)}^2>0$ for all $x\ne \frac{3}{2}$
Thus, $\frac{dy}{dx}$ does not change its sign as $x$ increases through $x=\frac{3}{2}.$ Hence, $x=\frac{3}{2}$ is not a point of local maximum, nor is it a point of local minimum. This is a point of inflection.

Q.4.Find the points of local maxima and local minima, if any, of the following function.
$f\left(x\right)=\sin 2x–x,$ where $–\frac{\pi }{2}<x<\frac{\pi }{2}$
Ans: Given,
$f\left(x\right)=\sin 2x–x$
$\Rightarrow f^{\prime }\left(x\right)=2\cos 2x–1$
The critical points of $f\left(x\right)$ are given by $f^{\prime }\left(x\right)=0.$
$\therefore f^{\prime }\left(x\right)=0$
$\Rightarrow 2\cos 2x–1=0$
$\Rightarrow \cos 2x=\frac{1}{2}$
$\Rightarrow 2x=–\frac{\pi }{3}$ or, $2x=\frac{\pi }{3}$
$\Rightarrow x=–\frac{\pi }{6}$ or $x=\frac{\pi }{6}$
Thus, $x=–\frac{\pi }{6}$ and $x=\frac{\pi }{6}$ are possible points of local maxima or minima.
Now, we test the function at each of these points.
Clearly, $f”\left(x\right)=–4\sin 2x$
At $x=–\frac{\pi }{6},$ we have
$f”\left(–\frac{\pi }{6}\right)=–4\sin \left(–\frac{\pi }{3}\right)=–4\times \frac{–\sqrt{3}}{2}=2\sqrt{3}>0.$
So, $x=–\frac{\pi }{6},$ is a point of local minimum.
At $x=\frac{\pi }{6},$ we have
$f”\left(\frac{\pi }{6}\right)=–4\sin \frac{\pi }{3}=–4\left(\frac{\sqrt{3}}{2}\right)=–2\sqrt{3}<0$
So, $x=\frac{\pi }{6}$ is a point of local maximum.
Hence, $x=\frac{\pi }{6}$ and $x=–\frac{\pi }{6}$ are the points of local maxima and local minima, respectively.

Q.5.Find the points of local maxima and local minima, if any, of the following function. $f\left(x\right)=\sin x–\cos x,$ where $0<x<2\pi .$
Ans: Given,$f\left(x\right)=\sin x–\cos x,$ where $0<x<2\pi .$
$\Rightarrow f^{\prime }\left(x\right)=\cos x+\sin x$
At points of local maximum and local minimum, we must have
$f^{\prime }\left(x\right)=0$
$\Rightarrow \cos x+\sin x=0$
$\Rightarrow \sin x=–\cos x$
$\Rightarrow \tan x=–1$
$\Rightarrow x=\frac{3\pi }{4}$ or $x=\frac{7\pi }{4}.$ $\left[\because 0<x<2\pi \right]$
Thus, $x=\frac{3\pi }{4}$ or $x=\frac{7\pi }{4}$ are the possible points of local maximum or minimum.
Now, we test the function at each of these points.
Clearly, $f”\left(x\right)=–\sin x+\cos x$
At $x=\frac{3\pi }{4},$ we have
$f”\left(\frac{3\pi }{4}\right)=–\sin \frac{3\pi }{4}+\cos \frac{3\pi }{4}=–\frac{1}{\sqrt{2}}–\frac{1}{\sqrt{2}}=–\frac{2}{\sqrt{2}}<0.$
Thus, $x=\frac{3\pi }{4}$ is a point of local maximum.
At $x=\frac{7\pi }{4},$ we have
$f”\left(\frac{7\pi }{4}\right)=–\sin \frac{7\pi }{4}+\cos \frac{7\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}>0.$
Thus, $x=\frac{7\pi }{4}$ is a point of local minimum.

Summary

Let $f\left(x\right)$ be a real function defined on an interval $\left[a,b\right].$ Then, $f\left(x\right)$ is said to have the maximum and minimum value in $\left[a,b\right],$ If there exists a point $c$ in $\left[a,b\right]$ such that $f\left(x\right)⩽f\left(c\right)$ and $f\left(x\right)⩾f\left(c\right)$ for all $x\in \left[a,b\right]$ respectively. In first derivative test, we observe the change in the sign of $f^{\prime }\left(x\right)$ at all critical points. And, in second-order derivative test we check the sign of the second-order derivatives at critical points to find the points of local maximum and minimum. Based on the various methods we have provided the solved examples, which can help in understanding all concepts in a better way.

Frequently Asked Questions (FAQs)

Q.1. What is maxima and minima in Math?
Ans: The maxima and minima of a function are the function’s largest and smallest values, either within a specific range or on the entire domain.

Q.2. How do you find the maxima and minima?
Ans: Observing the graph of a function can reveal the local maxima and minima. A local maxima is the point that is greater than the points directly adjacent to it on both sides. A local minimum, on the other hand, is any point that is smaller than the points directly adjacent to it on both sides.

Q.3. What is the use of maxima and minima in real life?
Ans: Maxima and minima are used in Economics, Business, and Engineering. For example, profit can usually be expressed as a function of the number of units sold because of which we can find the maximum and minimum profit using the first-order derivative or second-order derivative test.

Q.4. What is a critical point in maxima and minima?
Ans: A critical point is where the derivative is either zero or undefined. These critical points are locations on the graph where the function’s slope is zero. A function can have its local maximum and local minimum values at critical points.

Q.5. What is the point of inflection in maxima and minima?
Ans: A point of inflection is where a curve changes from concave upward to concave downward, or vice-versa.

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