Maxima and Minima

In this section, we find the method to calculate the maximum and the minimum values of a function in a given domain. i.e. we will find the turning points of the graph of a function at which the graph reaches its highest or lowest. We use these points is for sketching the graph of a given function.

 Again, we will also find the absolute maximum and absolute minimum of a function that are used for the solution of many applied problems.

 There are some definition of maxima and minima which are going to discuss one by one:

 

Definition 1: Let f be a function defined on an interval I. Then

(a) f is said to have a maximum value in I, if there exists a point c in I such that f(c) ≥ f(x), for all x Є I.

The number f(c) is called the maximum value of f in I and the point c is called a point of maximum value of f in I.

(b) f is said to have a minimum value in I, if there exists a point c in I such that f(c) ≤ f(x), for all x Є I.

The number f(c), in this case, is called the minimum value of f in I and the point c, in this case, is called a point of minimum value of f in I.

 (c) f is said to have an extreme value in I if there exists a point c in I such that f(c) is either a maximum value or a minimum value of f in I.

The number f(c), in this case, is called an extreme value of f in I and the point c is called an extreme point.

Definition 2: Let f be a real valued function and let c be an interior point in the domain of f. Then

 

(a) c is called a point of local maxima if there is an h > 0 such that f(c) ≥ f(x), for all x in (c – h, c + h)

 The value f(c) is called the local maximum value of f.

 (b) c is called a point of local minima if there is an h > 0 such that f(c) ≤ f(x), for all x in (c – h, c + h)

 The value f(c) is called the local minimum value of f.

The above definition leads to the following theorem:

Theorem: Let f be a function defined on an open interval I. Suppose c ∈ I be any point. If f has a local maxima or a local minima at x = c, then either f ′(c) = 0 or f is not differentiable at c.

Now, using only the first order derivatives, we calculate for finding points of local maxima or points of local minima.

Theorem (First Derivative Test): Let f be a function defined on an open interval I. Again let f be continuous at a critical point c in I. Then

(i) If f ′(x) changes sign from positive to negative as x increases through c, i.e. if f ′(x) > 0 at every point sufficiently close to and to the left of c, and f ′(x) < 0 at every point sufficiently close to and to the right of c, then c is a point of local maxima.

(ii) If f ′(x) changes sign from negative to positive as x increases through c, i.e. if f ′(x) < 0 at every point sufficiently close to and to the left of c, and f ′(x) > 0 at every point sufficiently close to and to the right of c, then c is a point of local minima.

(iii) If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Such a point is called point of inflection as shown in the figure.

    

 

 

 

Theorem (Second Derivative Test): Let f be a function defined on an interval I and c ∈ I. Let f be twice

differentiable at c. Then

(i) x = c is a point of local maxima if f ′(c) = 0 and f ″(c) < 0

The value f (c) is local maximum value of f.

(ii) x = c is a point of local minima if f′(c) = 0 and f ″(c) > 0

In this case, f (c) is local minimum value of f.

(iii) The test fails if f ′(c) = 0 and f ″(c) = 0.

In this case, we go back to the first derivative test and find whether c is a point of local maxima, local minima or a point of inflexion.

Problem: Find the local maxima and local minima, if any, of the following functions. Also find the local maximum and the local minimum values, as the case may be:

(i) f(x) = x²                                                                  (ii) g(x) = x³ − 3x           

Solution:

(i) Given, f(x) = x²

So f’(x) = 2x

Now, f’(x) = 0

=> 2x = 0

=> x = 0

Thus, x = 0 is the only critical point which could possibly be the point of local maxima or local minima of f.

We have, f”(0) = 2 which is positive.

Therefore, by second derivative test, x = 0 is a point of local minima and local minimum value

of f at x = 0 is f(0) = 0.

(ii) g(x) = x³ − 3x

So, g’(x) = 3x² – 3

Now, g’(x) = 0

=> 3x² – 3 = 0

=> 3(x² – 1) = 0

=> x² – 1 = 0

=> x² = 1

=> x = ±1

Now, g”(x) = 6x

g”(1) = 6 * 1 = 6 > 0

g”(-1) = 6 * (-1) = -6 < 0

By second derivative test, x = 1 is a point of local minima and local minimum value of g at x = 1

is g(1) = 1³ − 3 = 1 − 3 = −2.

However, x = −1 is a point of local maxima and local maximum value of g at x = −1 is

g(1) = (−1)³ − 3 (− 1) = − 1 + 3 = 2.

There are two theorems to find the absolute maximum and absolute minimum values of a function on a closed interval I.

Theorem: Let f be a continuous function on an interval I = [a, b]. Then f has the absolute maximum value and f attains it at least once in I. Also, f has the absolute minimum value and attains it at least once in I.

Theorem: Let f be a differentiable function on a closed interval I and let c be any interior point of I. Then

(i) f′(c) = 0 if f attains its absolute maximum value at c.

(ii) f′(c) = 0 if f attains its absolute minimum value at c.

The working rule for finding absolute maximum and/or absolute minimum values of a function in a given

closed interval [a, b] as given below:

Working Rule:

Step 1: Find all critical points of f in the interval, i.e., find points x where either f′(x) =0 or f is not differentiable.

Step 2: Take the end points of the interval.

Step 3: At all these points from Step 1 and 2 and calculate the values of f.

Step 4: Identify the maximum and minimum values of f out of the values calculated in Step 3. This maximum value will be the absolute maximum (greatest) value of f and the minimum value will be the absolute minimum (least) value of f.

Problem: Find both the maximum value and the minimum value of 3x⁴ − 8x³ + 12x² − 48x + 25 on the interval [0, 3].

Solution:

Given, f(x) = 3x⁴ − 8x³ + 12x² − 48x + 25

f′(x) = 12x³ − 24x² + 24x − 48

       =12(x³ −2x² +2x −4)

       =12[x²(x − 2) + 2(x − 2)]

       =12(x² + 2)(x − 2)

For maxima and minima, f′(x) = 0

=> 12(x² + 2)(x − 2)=0

=> x = 2, x² = -2

Since x² = -2 is not possible

So, x = 2 ∈ [0, 3]

Now we evaluate the value of f at critical point x = 2 and at the end points of the interval [0, 3]

f(0) = 25

f(2) = 3 * 2⁴ – 8 * 2³ + 12 * 2² – 48 * 2 + 25

        = 48 – 64 + 48 – 96 + 25

        = −39

f(3) = 3 * 3⁴ – 8 * 3³ + 12 * 3² – 48 * 3 + 25

       = 243 – 216 + 108 −144 + 25

       = 16

Hence, at x = 0, Maximum value = 25

At x = 2, Minimum value = -39

Problem: Find the maximum and minimum values of x + sin 2x on [0, 2π].

Solution:

Let f(x) = x + sin 2x

So, f’(x) = 1 + 2 cos 2x

Now, f’(x) = 0

=> 1 + 2 cos 2x = 0

=> cos 2x = -1/2

=> cos 2x = -cos π/3

=> cos 2x = cos (π - π/3)

=> cos 2x = cos 2π/3

=> 2x = 2nπ ± 2π/3, n є Z

=> x = nπ ± π/3, n є Z

=> x = π/3, 2π/3, 4π/3, 5π/3 є [0, 2π]

Now,

f(π/3) = π/3 + sin 2π/3 = π/3 + √3/2

f(2π/3) = 2π/3 + sin 4π/3 = π/3 - √3/2

f(4π/3) = 4π/3 + sin 8π/3 = 4π/3 + √3/2

f(5π/3) = 5π/3 + sin 10π/3 = 5π/3 - √3/2

f(0) = 0 + sin 0 = 0

f(2π) = 2π + sin 2π = 2π + 0 = 2π

Hence, we can conclude that the absolute maximum value of f(x) in the interval [0, 2π] is 2π occurring at x = 2π and the absolute minimum value of f(x) in the interval [0, 2π] is 0 occurring at x = 0.