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One of the major applications of derivatives is determination of the maximum and minimum values of a function. We know that the derivative provides the information about the gradient of the graph of a function which can be used to identify the points where the gradient is zero. These points are usually related to the largest and the smallest values of the function.

The maximum and minimum values of a function are together termed as extrema. Pierre de Fermat was one of the mathematicians to propose a general technique for finding the maxima & minima.Here, we shall throw some light on some interesting facts about maxima and minima including its sub-topics. And that’s not all…. we shall go into the intricacies and discuss the tips to master this topic and the practical applications of maxima and minima in real life as well.

The various topics that form a part of maxima and minima include

  • Stationary points
  • Turning points
  • Local maxima and  minima
  • Ways to identify local maxima and minima
  • Global maxima and minima

We shall give a brief outline of these topics in addition to the tips to master them.

Local Maxima and Minima

  • Mathematically, we can define the local maxima of a function f to be a point ‘a’ such that f(a) > f(a-h) and f(a) > f(a+h), where h > 0 is a small quantity.
  • Similarly, a function f(x) is said to have a local minimum at the point ‘a’ if the value of the function at the point x = a is less than the value at the neighboring points.

Global Maxima and Minima

Global maxima or minima of f(x) in [a, b] refers to the greatest or least value of f(x) in [a, b]. Mathematically, it can be explained as:

The function f(x) has a global maximum at the point ‘a’ in the interval I if

Similarly, f(x) has a global minimum at the point ‘a’ if f (a) ≤f (x), for all x ∈ I.

From the figure given below, it becomes clear that while global maxima denotes the maxima of the function in the entire interval, local maxima denotes the poitn at which the value of the function is maximum as compare to the neighbouring points.

What are the various ways of testing the local maximum and minimum of a function?

Generally speaking, there are three ways to test a function for local maxima/ minima:

 

Maxima & Minima is an extremely important topic of differential calculus. One cannot dare to leave this topic as it often fetches good number of questions in JEE. The differential calculus portion accounts for 13% of the JEE screening and maxima & minima is an important component of application of derivatives.

Illustrations

Illustration 1:

If f(x) = (x2-1)/(x2+1), for every real number x, then the minimum value of f   (IIT JEE 1998)

A. does not exist because f is unbounded                     B. is not attained even though f is bounded

C. is equal to 1                                                                 D. is equal to -1.

Solution:

Given f(x) = (x2-1)/(x2+1) = 1- 2/(x2+1)

This shows that f(x) will be minimum when 2/(x2+1) is maximum

i.e. when (x2+1) is maximum i.e. at the point x = 0

Hence, the minimum value of f(x) is f(0) = -1.

—————————————————————————–

Finding Maxima and Minima using Derivatives

Where is a function at a high or low point? Calculus can help!

A maximum is a high point and a minimum is a low point:

function local minimum and maximum

In a smoothly changing function a maximum or minimum is always where the function flattens out  (except for a saddle point).

Where does it flatten out?  Where the slope is zero.

Where is the slope zero?  The Derivative tells us!

Let’s dive right in with an example:

quadratic graph

Example: A ball is thrown in the air. Its height at any time t is given by:

h = 3 + 14t − 5t2

What is its maximum height?

 

Using derivatives we can find the slope of that function:

d/dth = 0 + 14 − 5(2t)
= 14 − 10t

(See below this example for how we found that derivative.)

 

quadratic graph

Now find when the slope is zero:

14 − 10t = 0
10t = 14
t = 14 / 10 = 1.4

The slope is zero at t = 1.4 seconds

And the height at that time is:

h = 3 + 14×1.4 − 5×1.42
h = 3 + 19.6 − 9.8 = 12.8

And so:

The maximum height is 12.8 m (at t = 1.4 s)

 

A Quick Refresher on Derivatives

A derivative basically finds the slope of a function.

In the previous example we took this:

h = 3 + 14t − 5t2

and came up with this derivative:

d/dth = 0 + 14 − 5(2t)
= 14 − 10t

Which tells us the slope of the function at any time t

slope examples: y=3, slope=0; y=2x, slope=2

 

We used these Derivative Rules:

  • The slope of a constant value (like 3) is 0
  • The slope of a line like 2x is 2, so 14t has a slope of 14
  • A square function like t2 has a slope of 2t, so 5t2 has a slope of 5(2t) = 10t
  • And then we added them up

 

How Do We Know it is a Maximum (or Minimum)?

We saw it on the graph! But otherwise … derivatives come to the rescue again.

Take the derivative of the slope (the second derivative of the original function):

The Derivative of 14 − 10t is −10

This means the slope is continually getting smaller (−10): traveling from left to right the slope starts out positive (the function rises), goes through zero (the flat point), and then the slope becomes negative (the function falls):

slope positive then zero then negative
A slope that gets smaller (and goes though 0) means a maximum.

This is called the Second Derivative Test

On the graph above I showed the slope before and after, but in practice we do the test at the point where the slope is zero:

Second Derivative Test

When a function’s slope is zero at x, and the second derivative at x is:

  • less than 0, it is a local maximum
  • greater than 0, it is a local minimum
  • equal to 0, then the test fails (there may be other ways of finding out though)

“Second Derivative: less than 0 is a maximum, greater than 0 is a minimum”

 

Example: Find the maxima and minima for:

y = 5x3 + 2x2 − 3x

The derivative (slope) is:

d/dxy = 15x2 + 4x − 3

Which is quadratic with zeros at:

  • x = −3/5
  • x = +1/3

 

Could they be maxima or minima? (Don’t look at the graph yet!)

 

The second derivative is y” = 30x + 4

At x = −3/5:

y” = 30(−3/5) + 4 = −14
it is less than 0, so −3/5 is a local maximum

At x = +1/3:

y” = 30(+1/3) + 4 = +14
it is greater than 0, so +1/3 is a local minimum

(Now you can look at the graph.)

5x^3 2x^2 3x

Words

A high point is called a maximum (plural maxima).

A low point is called a minimum (plural minima).

The general word for maximum or minimum is extremum (plural extrema).

We say local maximum (or minimum) when there may be higher (or lower) points elsewhere but not nearby.

One More Example

Example: Find the maxima and minima for:

y = x3 − 6x2 + 12x − 5

The derivative is:

d/dxy = 3x2 − 12x + 12

Which is quadratic with only one zero at x = 2

Is it a maximum or minimum?

 

The second derivative is y” = 6x − 12

At x = 2:

y” = 6(2) − 12 = 0
it is 0, so the test fails

And here is why:

x^3 6x^2 12x 5

It is a saddle point … the slope does become zero, but it is neither a maximum or minimum.

 

Must Be Differentiable.

And there is an important technical point:

The function must be differentiable (the derivative must exist at each point in its domain).

Example: How about the function f(x) = |x| (absolute value) ?

|x| looks like this: Absolute Value function

At x=0 it has a very pointy change!

In fact it is not differentiable there (as shown on the differentiable page).

So we can’t use this method for the absolute value function.

Section 1Introduction
Section 2Stationary Points
Section 3Turning Points
Section 4Distinguishing maximum points from minimum points
Section 5Using the first derivative to distinguish maxima from minima
Section 6Examples