Maxima and Minima: Definition, Types and Examples

Maxima and Minima: In simple terms, maxima and minima are the highest and lowest points of a function respectively. They appear within a set of ranges. While the largest value of the function under a given range is called absolute maxima, the minimum value of the function under a given range is called the absolute minima.  

It is interesting to note that apart from absolute maxima and minima, we also have what are called local maxima and local minima, which are defined as per particular intervals. 

Maxima and Minima: Meaning 

If we take any curve, it has what we call a peak and a valley. While a peak corresponds to maxima, minima corresponds to a valley. It is to be noted that a curve can have more than one such peak or valley. When they are combined, they can be referred to as extrema. 

Maxima is the plural of maximum while minima is the plural of minimum and extrema is the plural of extremum. All these terms form a part of calculus. 

Absolute Maxima and Absolute Minima 

The highest point of a function in a given domain is called the absolute maxima. On the other hand, absolute minima is the lowest point of a function in a given domain. Since these points pertain to highest and lowest, there can be only one absolute maxima and only one absolute minima for a given function, for an entire domain. In other words, absolute maxima and absolute minima can also be called global maxima and global minima of a function. 

Local Maxima and Local Minima

Local maxima and local minima are terms that we use in relation to a particular interval. Local maxima of a function would be the highest point at that particular interval. It should be noted that the values of the function near that point are always lesser. 

On the other hand, local minima of a function is the least point at that particular interval. The values of the function near that point are always greater. 

Maxima and Minima: How To Find Them?

Now that we know the meanings of maxima, minima, their absolute values and least values, let’s know how to find them. In simple words, they can be identified using what’s called derivatives. A derivative is the rate of change of a function with respect to a variable. 

Also Read: Maxima, minima: Their Identification  

First Order Derivative

We know that the slope of a function increases as it nears the maxima. At maxima, it becomes still. Later on, the slope begins to decrease as we move away from the maxima. Similarly, as the slope nears minima, it starts decreasing. At minima, it becomes still and then starts increasing. In order to find out maxima and minima, we use this inflection point precisely. 

Local maxima: If f’(x) goes from positive to negative when x increases over a point c, the maximum value of the function of that particular range can be given by f(c).

Local minima: If f’(x) goes from negative to positive when x increases over a point c, the minimum value of the function of that particular range can be given by f(c).

Inflection Point: If f’(x) remains without switching signs when x increases over c, and c is not the maxima or minima of that particular function, then c is the inflection point.

Second Order Derivative

In the second order derivative, we first use the first derivative and find out the value of the function at the critical point. If it gives any value at the critical point, then we find the second order derivative 

Maxima: f(c) <0

Minima f(C)>0

Also Read: More on Derivatives Test 

Solved Examples

Let us look at some of the solved examples below:

Q.1. Find the maximum and the minimum values of the following function.

f(x)=3×2+6x+8,x∈R

Ans: Given:

f(x)=3×2+6x+8,x∈R

⇒f(x)=3(x2+2x+1)+5

=3(x+1)2+5

Clearly,

3(x+1)2⩾0 for all x∈R

⇒3(x+1)2+5⩾5 for all x∈R

⇒f(x)⩾f(–1) for all x∈R. [∵f(–1)=3(–1)2+6(–1)+8=5]

Thus, 5 is the minimum value of f(x) at x=–1.

Note that f(x) 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(x)=x3–6×2+9x–8.

Ans: Let y=f(x)=x3–6×2+9x–8. Then,

dydx=f′(x)=3×2–12x+9=3(x2–4x+3)

The critical points of f(x) are given by f′(x)=0 or dydx=0.

Now, dydx=0⇒3(x2–4x+3)=0⇒x=1,3.

We have to examine whether these points are points of local maxima or local minima or neither of them.

We have, dydx=3(x–1)(x–3)

The changes in signs of dydx for different values of x are shown in the figure given below:

Clearly, dydx changes sign from positive to negative asx increases through 1.

So, x=1 is a point of the local maximum.

Also, dydx 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(x)=4×3–18×2+27x–7 has neither maxima nor minima.

Ans: Given: f(x)=4×3–18×2+27x–7

f′(x)=12×2–36x+27=3(4×2–12x+9)=3(2x–3)2

The critical points of y=f(x) are given by dydx=0.

Now, dydx=0⇒3(2x–3)2=0⇒2x–3=0⇒x=32

Clearly, dydx=3(2x–3)2>0 for all x≠32

Thus, dydx does not change its sign as x increases through x=32. Hence, x=32 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(x)=sin2x–x, where –π2<x<π2

Ans: Given,

f(x)=sin2x–x

⇒f′(x)=2cos2x–1

The critical points of f(x) are given by f′(x)=0.

∴f′(x)=0

⇒2cos2x–1=0

⇒cos2x=12

⇒2x=–π3 or, 2x=π3

⇒x=–π6 or x=π6

Thus, x=–π6 and x=π6 are possible points of local maxima or minima.

Now, we test the function at each of these points.

Clearly, f”(x)=–4sin2x

At x=–π6, we have

f”(–π6)=–4sin(–π3)=–4×–3√2=23–√>0.

So, x=–π6, is a point of local minimum.

At x=π6, we have

f”(π6)=–4sinπ3=–4(3√2)=–23–√<0

So, x=π6 is a point of local maximum.

Hence, x=π6 and x=–π6 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(x)=sinx–cosx, where 0<x<2π.

Ans: Given,f(x)=sinx–cosx, where 0<x<2π.

⇒f′(x)=cosx+sinx

At points of local maximum and local minimum, we must have

f′(x)=0

⇒cosx+sinx=0

⇒sinx=–cosx

⇒tanx=–1

⇒x=3π4 or x=7π4. [∵0<x<2π]

Thus, x=3π4 or x=7π4 are the possible points of local maximum or minimum.

Now, we test the function at each of these points.

Clearly, f”(x)=–sinx+cosx

At x=3π4, we have

f”(3π4)=–sin3π4+cos3π4=–12√–12√=–22√<0.

Thus, x=3π4 is a point of local maximum.

At x=7π4, we have

f”(7π4)=–sin7π4+cos7π4=12√+12√=22√>0.

Thus, x=7π4 is a point of local minimum.

We hope this article on maxima and minima helps you understand the concept better. For more such articles, keep visiting Embibe.