Maxima and Minima of a Function of one variable

MAXIMA AND MINIMA OF A FUNCTION OF ONE VARIABLE

We have already seen the applications of derivative in problems of tangent and normal, in deciding increasing and decreasing nature of a function in an interval. We shall now use it to locate maxima and minima of a function.

In calculus the term "maximum" is used in two senses “absolute” maximum and "relative" maximum.

Similarly, absolute minimum and relative minimum.

Definition 2.21 Let f be the function defined on [a, b] and let c ∈ (a, b). Then

(i) f is said to have a relative maximum (or local maximum) at c, if there is a neighbourhood (c - δ, c + δ) of c such that f(x) < f(c) ∀x ∈ (c - δ, c + δ), x ≠ c.

That is f(c) is the greatest value in a neighbourhood of c.

(ii) f is said to have a relative minimum (or local minimum) at c, if there is a neighbourhood (c - δ, c + δ) of c suchthat  f(x) > f(c) ∀x ∈ (c - δ, c + δ), x ≠ c.

That is f(c) is the least value in a neighbourhood of c.

Note (1) If f(c) is a relative maximum or relative minimum, then f(c) is called an extreme value of ƒ at c or extremum of f at c.

Definition 2.22 Let f be defined on [a, b]. f is said to have an absolute maximum (or global maximum) on [a, b] if there is at least one point c ∈ [a, b] such that f(x) ≤ ƒ(c) ∀x ∈ [a, b].

In otherwords, the largest value of ƒ on [a, b] is called the absolute maximum

Definition 2.23 Let f be defined on [a, b]. f is said to have an absolute minimum (or global minimum) on [a, b] if there is at least one point c ∈ [a, b] such that f(x) ≥ ƒ(c) ∀x ∈ [a, b].

That is the least value of ƒ on [a, b] is called the absolute minimum (or global minimum).

Note

1. Given a function f defined on [a, b], the absolute maximum and the absolute minimum need not exist.

Then ƒ has no absolute maximum on [0, 1].

However, if f is continuous on a closed and bounded interval [a, b], then absolute maximum and absolute minimum exist at some points c, d in [a, b].

2. From the definition of maximum and minimum it is obvious that ƒ(c) is an extreme value of ƒ at c, iff f(x) – f(c) keeps the same sign for all x, other than c, in some neighbourhood of c.

Theorem 2.11 A necessary condition for the existence of an extremum at an interior point

Let f be a function defined on the interval [a, b] and c ∈ (a, b). If f(c) is an extremum at c and if f'(c) exists, then f'(c) = 0.

This theorem is known as Fermat's theorem.

1. Geometrical Meaning

Let y = f(x) be the graph of ƒ on [a, b], then P(c, f(c)) is a point on the curve y = ƒ(x).

If f(c) is a maximum value, then in (c - δ, c] the curve is increasing and so f'(x) > 0 in (c - δ, c) and decreasing in (c, c + δ).

That is f'(x) < 0 in (c, c + δ).

If f'(c) exists, then f'(c) must be zero.

That is the tangent is parallel to the x-axis, because it is increasing up to the point P and momentarily at rest at P and then decreasing.

Similarly, if f(c) is a minimum value, then f'(c) = 0.

Note

1. The points where f'(x) = 0 are called stationary points of f.

2. The converse of the above theorem is not true.

Similarly, if f'(c) = 0, then f(c) is not an extremum.

For example: Consider f(x) = x3.

f'(x) = 3x2     ⸫ ƒ'(0) = 0.

But f(0) is neither a maximum nor a minimum

because there is no neighbourhood of 0 in which ƒ(x) – ƒ(0) keeps the same sign for x ≠ 0.

3. It is possible that f(c) is an extreme value of f even if f'(c) does not exist.

For example: Consider

We know that f'(0) does not exist. 

But f(0) is a minimum value of ƒ(x). 

In fact, ƒ(0) is the absolute minimum.

Definition 2.24 Critical Points

Let f be a function defined on [a, b]. The points x ∈ (a, b) at which f' (x)=0 or f'(x) does not exist are called critical points of f.

For f(x) = |x|, x = 0 is a critical point but not a stationary point.

2. Tests for Maxima and Minima

(1) Second Derivative Test

Let f be a function defined on [a, b] and let ƒ be twice differentiable in a neighbourhood (c − δ, c + δ) of c ∈ (a, b) and f'(c) = 0. Suppose ƒ"(c) ≠ 0, then

1. f(c) is a maximum if ƒ"(c) < 0 and

2. f(c) is a minimum if ƒ"(c) > 0.

Note If ƒ"(c) = 0, then the second derivative test cannot be applied.

In this case, we use the following general test involving higher derivatives or the first derivative test.

(2) General Test

Let f be differentiable n times and f'(c) = 0, ƒ"(c) = 0,..., f(n-1) (c) = 0 and f(n)((c) ≠ 0.

If n is even, then

1. f(c) is a maximum if f" (c) < 0

2. f(c) is a minimum if f"(c) > 0.

If n is odd, then f(c) is neither a maximum nor a minimum.

(3) First Derivative Test

Let f be defined on [a,b] and c ∈ (a,b).

Let f be differentiable in a neighbourhood (c - δ, c + δ) of c, except possibly at c.

(i) If f'(x) > 0 for x < c and f'(x) < 0 for x > c in the neighbourhood of c, then f(c) is a maximum value.

That is, f'(x) changes from positive to negative in the neighbourhood of c as x increases.

(ii) If f'(x) < 0 for x < c and f'(x) > 0 for x > c in the neighbourhood of c, then f(c) is a minimum value.

That is, f'(x) changes from negative to positive in the neighbourhood of c as x increases.

WORKED EXAMPLES

Example 1 

Find the absolute maximum and minimum of the function

Solution Given

Since f(x) is a polynomial, it is continuous and differentiable in the closed interval (-1/2, 4).

So absolute maximum and minimum will occur at a critical point or at the end points.

Differentiate (1) w. r to x we get

f'(x) = 3x2 - 6x,

Since there is no point where f is not differentiable, all the critical points are given by f'(x) = 0

The largest value is 17 and so it is the absolute maximum and it occurs at the end point x = 4.

The smallest value is -3 and so it is the absolute minimum and it occurs at the critical point x = 2

Example 2 

Find the absolute maximum and minimum of

Solution 

Since the internal (1, 3) is open, absolute maximum or absolute minimum can occur only at the critical points. So, f has an absolute minimum at x = 4/3 There is no absolute maximum

Example 3 

Find the local maximum and minimum of the function f(x) = x + 2 sin x, 0 ≤ x ≤ 2π.

Solution 

Given f(x) = x + 2 sin x, 0 ≤ x ≤ 2π

Example 4 

Investigate the maximum and minimum values of the function

Solution 

Given

⸫ the only critical points x = -1, 1

II Derivative Test

We have

⸫ y is minimum when x = -1

Example 5 

f(x) is a cubic polynomial attaining the maximum and minimum values 10 and -5/2 at x = -3 and x = 2 respectively. Find f(x).

Solution 

Given f(x) is a cubic polynomial

⸫ f'(x) is a quadratic

Given at x = -3, x = 2 maximum and minimum occurs

⸫ they are roots of the quadratic ƒ'(x) = 0

ƒ'(x) = a(x + 3) (x − 2)

= a(x2 + x - 6)

We want to find f(x) and so integrate f'(x)

Example 6 

Find the maxima and minima of the function

10x6 - 24x5 +15x4 − 40x3 + 108.

Solution. 

Let ƒ(x) = 10x6 - 24x5 +15x4 − 40x3 + 108.

Example 7 

Find the maxima and minima of the function x5 - 5x4 + 5x3 + 10

Solution 

Example 8 

Find the area of the largest rectangle with lower base on x-axis and upper vertices on the curve y = 12 - x2

Solution 

Given curve is y = 12 – x2

Since the curve is symmetric about y-axis, the rectangle must be symmetrical about the y-axis

Let S be the area of the rectangle.

Example 9 

Find two positive numbers such that their sum is 4 and the sum of the square of one and the cube of the other is a maximum

Solution 

Let x and y be numbers such that x + y = 4  ⇒ y = 4 - x

But 6 ∉ (0, 4)

Example 10 

What is the maximum slope of the curve y = -x3 + 3x2 + 9x - 27 and at what points is it?

Solution 

The given curve is y = -x3 + 3x2 + 9x - 27.       (1)

The slope of a curve at a point on it is the same as the slope of the tangent at that point.

Differentiating with respect to x, we get

We have to maximize m.

⸫ when x = 1, m is maximum. Then y = -1 + 3⸱1 + 9⸱1 – 27 = -16

⸫ the point at which maximum slope occurs is (1, −16) and the maximum slope = 12.

Example 11 

Find the maximum and minimum values of f(x) = |4 - x2|, x ∈ [−4, 4]. Also find the absolute maximum and absolute minimum, if they exist.

Solution 

Let f(x) = |4 - x2|, x ∈ [−4, 4]

We know

At x = -2, 2, f'(x) does not exist.

[Since, ƒ is continuous at x = 2 and ƒ'(2−) = −4, ƒ'(2+) = 4

⸫ f'(2) does not exist.

Similarly, f' (-2), does not exist]

The critical points are x = 0, - 2, 2.

When x = 0, ƒ"(0) = −2 < 0.

⸫ f(x) has a maximum at x = 0 and the maximum value = 4.

Since f'(x) does not exist at x = -2, x = 2, we use the first derivative test. In a neighbourhood of -2,

f'(x) < 0 if x < -2 and f'(x) > 0 if x > -2.

i.e., f'(x) changes from -ve to +ve as x increases in the ngd. of -2

So, f(-2) is a minimum and the minimum value = 4.

Similarly, in a neighbourhood of 2,

f'(x) < 0 if x < 2 and f'(x) > 0 if x > 2.

So, ƒ(2) is a minimum and the minimum value is zero.

The least value of f(x) in [−4, 4] is 0.

⸫ the absolute minimum = 0.

Though ƒ(0) = 4 is a relative maximum, it is not the absolute maximum.

Absolute maximum occur at the end points x = 4 or x = -4 and the value is

ƒ(4) = |4 - 42| = |4 - 16| = |-12| = 12.

Example 12 

In a submarine cable the speed of signalling varies as where x is the ratio of the radius of the core to that of the covering. Find the values of x for which the speed of signaling is maximum.

Solution 

Let S be the speed of signalling in a submarine cable.

Then

Example 13 

Find the local maximum and minimum values of using both first and second derivative tests

Solution 

Example 14 

Find the local maxima and minima of the function f(x) = 4x3 + 3x2 - 6x + 10 using first derivative test.

Solution 

Given

f(x) = 4x3 + 3x2 - 6x + 10

f'(x) = 12x2 + 6x - 10

= 6(2x2 + x - 1)

= 6(2x - 1) (x + 1)

The critical points are given by f'(x) = 0

⇒    6(2x − 1) (x + 1) = 0

⇒ x = 1/2  or - 1

Now we find maximum or minimum using first derivative test we study the change of sign of derivative in a neighborhood, of critical points. 

In a neighborhood of x = -1,

f'(x) changes from positive to negative and so, at x = -1, f(x) has a local maximum

The maximum value is ƒ(−1) = 4 (−1)3 + 3(−1)2 − 6(−1) + 10

= 4 + 3 + 6 + 10 = 15

In a neighborhood of x = 1/2,

Example 15 

Find the local maxima and minima of the function

f(x) = 2x3 + 3x2 - 36x + 5, using

(i) first derivative test

(ii) Second derivative test

Solution 

Given f(x) = 2x3 + 3x2 - 36x + 5

ƒ'(x) = 6x2 + 6x – 36           (1)

= 6(x2 + x − 6)

= 6(x + 3) (x − 2)

The critical points are given by f'(x) = 0

⇒ 6(x + 3)(x - 2) = 0

(i) By First derivative test

we have f'(x) = 6(x + 3)(x - 2)

We study change of derivative f'(x), in a neighborhood of the critical points.

⸫ f'(x) changes from positive to negative and so at x = -3, f(x)  has a local maximum

The maximum value is f(−3) = 2 (−3)+ + 3(−3)2 − 36(−3) + 5 = 86

In a neighborhood of 2,

⸫  f'(x) changes from negative to positive and so at x = 2, f(x) has a local minimum

The minimum value is f(2) = 2⸱23 +3⸱22 - 36·2 + 5 = -39

(ii) By second derivative test

we have by (1), ƒ'(x) = 6x2 + 6x – 36

⸫ ƒ"(x) = 12x + 6

when x = -3, f" (-3)= 12(-3) + 6 = −30 < 0

⸫  f(x) is a maximum at x = -3

and maximum value is f(-3) = 86

when x = 2, ƒ"(2) = 12(3) + 6 = 42 > 0

⸫  f(x) is a minimum at x = 2

and the minimum value is f(2) = −39

Example 16 

Find the local maxima and minima of the function f(x) = x2 – x -In x, x > 0, by using the first derivative test.

Solution 

Given f(x) = x2 – x - ln x; x > 0 is the domain

To find critical points, solve ƒ'(x) = 0

⸫ f'(x) changes from negative to positive and so at x = 1, f(x) has a minimum 

The minimum value is f(1) = 12 - 1 – ln l = 0

Example 17 

Suppose the derivative of a function ƒ is ƒ'(x) = (x + 1)2 (x − 2)3 (x − 4)4, find

(i) the intervals in which f is increasing or decreasing. and (ii) the points where f(x) is a maximum or minimum

Solution 

Given

ƒ'(x) = (x + 1)2 (x − 2)3 (x − 4)4 

To find critical points, solve f'(x) = 0 

⇒  (x + 1)2 (x − 2)3 (x − 4)4 = 0 ⇒ x = -1, 2 or 4

The critical points divided the domain of the function, (-∞, ∞) into subintervals (-∞, -1), (1, 2), (2, 4), (4, ∞) where the function is either increasing or decreasing.

Note

We know maxima or minima occurs only at the critical points. But it is not necessary that at a critical point maximum or minimum should occur. In example 17 above we find at the critical points x = -1, x = 4, there is no maximum or minimum. The situation can be visualized from the diagram given below.

EXERCISE 2.11

1. Find the maximum and minimum values of x5 -5x + 5x3 −1. 

2. Find maxima and minima of 2x3 - 3x2 - 36x + 10.

3. Find the maximum and minimum values of a2 sin2 x + b2 cos2x 

4. If xy = 4, find the maximum and minimum values of 4x + 9y. 

5. Find the maximum and minimum values of | x - x2 | on [2,2].

6. Show that maximum value of

7. A cubic function f(x) has turning points x = 1, x = -3/2. It vanishes when x = 0, and its value is 4 when x = 2, find the function 

8. Find maxima and minima of

9. Find maxima and minima of the function

10. f(x) = a/x + bx has an extremum at x = 2 and f(2) = 1. Determine the values of a and b. Is the point (2, 1) a point of maximum or minimum? 

11. Find the maximum possible slope for a tangent line to the curve

12. If 2x + y = 5 find the maximum value of x2 + 3xy + y2.

13. Divide 64 into two numbers such that the sum of their cubes is a minimum

14. Show that the area of a rectangle of given perimeter 20 m is maximum when it is a square?

ANSWERS TO EXERCISE 2.11

1. Maximum at x = 2, maximum value = -9 

Minimum at x = 3, minimum value = -28 

x = 0, is neither maximum or minimum