Maxima and Minima

Let x1,x D f(a) > f(x1) and f(a) > f(x2)

A function is said to have a maximum value at a point 'x=a' in its domain D if f(a) ≥ f(x), x D.

f(a) is called the maximum value of f. "x=a" is called the point of maximum value of f.

Let x1,x D f(a) < f(x1) and f(a) < f(x2)

A function is said to have a minimum value at a point 'a' in its domain D if f(a) ≤ f(x), x D.

f(a) is called the minimum value of f. "x=a" is called the point of minimum value of f.

Extreme value of function f:

Function f(x) is said to have an extreme value in its domain D if there exists a point a  D such

that f (a) is either the maximum value or the minimum value of f in D.

The number f (a) is called an extreme value of f in D, and point 'a' is called an extreme point.

Consider f(x) = 2x, x R

The ordered pairs are (-2,8),(-1,2),(0,0),(1,2) and (2,8).

From the graph, we have f(x) = 0 if x = 0 and f(x) ≥ 0 x R

The minimum value of f is 0, and the point of minimum value of f is x = 0.

From the graph of the function, f has no maximum value, and hence, no point of maximum value of f in R.

Suppose the domain of f is restricted to [- 2, 1].

The maximum value of the function, f(-2) = 2(-2)2 = 8, which is at x = -2

Note: A function has an extreme value at a point even if it is not differentiable at that point.

Every monotonic function assumes its maximum/minimum value at the end points of the domain of definition of the function.

A more general result is

Every continuous function on a closed interval has a maximum and a minimum

value.

N.B.: Every increasing or decreasing function assumes its maximum or minimum value at the end points of the domain of definition of the function.

Or

Every continuous function on a closed interval has a maximum and a minimum value.

Similarly, the function has maximum value in some neighbourhood of points B and D which are at the top of their respective hills. For this reason, the points A and C may be regarded as points of local minimum value (or relative minimum value) and points B and D may be regarded as points of local maximum value (or relative maximum value) for the function. The local maximum value and local minimum value of the function are referred to as local maxima and local minima, respectively, of the function.

Local Maxima and Local Minima

The graph of the function has minimum values in some neighbourhood(nbd) of points Q and S, and has maximum values in some neighbourhood(nbd) of points P, R and T.

The x-coordinates of points P, R and T are called the points of local maximum, and the y-coordinates of points P, R and T, are the local maximum values of f(x).

Similarly, the x-coordinates of points Q and S are called the points of local minimum, and the y-coordinates of points Q and S, are called the local minimum values of f(x).

The local maximum value of a function is said to be the local maxima of that function.

The local minimum value of a function is said to be the local minima of that function.

  • When x = a, if f(x) ≤ f(a) for every x in the domain, then f(x) has an Absolute Maximum value and the point a is the point of the maximum value of f.
  • When x = a, if f(x) ≤ f(a) for every x in some open interval (p, q) then f(x) has a Relative Maximum value.
  • When x= a, if f(x) ≥ f(a) for every x in the domain then f(x) has an Absolute Minimum value and the point a is the point of the minimum value of f.
  • When x = a, if f(x) ≥ f(a) for every x in some open interval (p, q) then f(x) has a Relative Minimum value.

Let the point of local maximum value of f in the graph be x = a. The function f is increasing in

the interval (a - h, a) and decreasing in the interval (a , a+h ), where h > 0.

DefinitionLet 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), x c

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 .

Geometrically, the above definition states that if x = c is a point of local maxima of f,

then the graph of f around c will be as shown in Fig 6.14(a). Note that the function f is

increasing (i.e., f (x) > 0) in the interval (c – h, c) and decreasing (i.e., f (x) < 0) in the

interval (c, c + h).

This suggests that f (c) must be zero.

 

If f is increasing, then f '(x) > 0, and if f is decreasing, then f '(x) < 0.

If f is neither decreasing nor increasing, then f '(x) = 0. i.e. f '(a) = 0.

Let the point of local minimum value of f in the graph be x = a.

Function f is decreasing in the interval (a - h, a) and increasing in the interval (a , a+h ), where h > 0. we have f '(a) = 0

Monotonicity

Functions are said to be monotonic if they are either increasing or decreasing in their entire domain. f(x) = ex, f(x) = nx, f(x) = 2x + 3 are some examples.

Functions which are increasing and decreasing in their domain are said to be non-monotonic

For example: f(x) = sin x , f(x) = x2

Monotonicity Of A function At A Point

A function is said to be monotonically decreasing at x = a if f(x) satisfy;

f(x + h) < f(a) for a small positive h

  • f'(x) will be positive if the function is increasing
  • f'(x) will be negative if the function is decreasing
  • f'(x) will be zero when the function is at its maxima or minima

Theorem:

Let f be a real valued function defined on an open interval I. Suppose point a is any arbitrary point in I. If f has a local maxima or a local minima at x = a, then either f '(a) = 0, or f is not differentiable at a.

However, the converse need not be true, i.e. a point at which the derivative vanishes need not be the point of local maxima or local minima.

Every continuous function on a closed interval has a maximum and a minimum value.

Point of Inflection:

For continuous function f(x), if f'(x0) = 0 or f’”(x0) does not exist at points where f'(x0) exists and if f”(x) changes sign when passing through x = xthen x0 is called the point of inflection.

(a) If f”(x) < 0, x ∈ (a, b) then the curve y = f(x) in concave downward

(b) if f” (x) > 0, x ∈ (a, b) then the curve y = f(x) is concave upwards in (a, b)

For example: f(x) = sin x

Solution: f'(x) = cos x

f”(x) = sinx = 0 x = nπ, n ∈ z

Critical point:

A point c in the domain of a function f at which either f (c) = 0 or f is not

differentiable is called a critical point of f.

 

Theorem  (First Derivative Test) : Let f be a function defined on an open interval I.

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. Infact, such a point is called point of inflection .

Example : Find all points of local maxima and local minima of the function f

given by

f (x) = x3 – 3x + 3.

Solution We have

f (x) = x3 – 3x + 3.

or f (x) = 3x2 – 3 = 3(x – 1) (x + 1)

or f (x) = 0 at x = 1 and x = – 1

Thus, x = ± 1 are the only critical points which could possibly be the points of local

maxima and/or local minima of f . Let us first examine the point x = 1.

Example ; Find all the points of local maxima and local minima of the function f

given by

f (x) = 2x3 – 6x2 + 6x +5.

Solution We have

f (x) = 2x3 – 6x2 + 6x + 5

or f (x) = 6x2 – 12x + 6 = 6(x – 1)2

or f (x) = 0 at x = 1

Thus, x = 1 is the only critical point of f . We shall now examine this point for local maxima and/or local minima of f. Observe that f (x) 0, for all x R and in particular f (x) > 0, for values close to 1 and to the left and to the right of 1. Therefore, by first derivative test, the point x = 1 is neither a point of local maxima nor a point of local minima. Hence x = 1 is a point of inflexion.

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.

Example : Find local maximum and local minimum values of the function f given by

f (x) = 3x4 + 4x3 – 12x2 + 12

Solution We have

f (x) = 3x4 + 4x3 – 12x2 + 12

or f (x) = 12x3 + 12x2 – 24x = 12x (x – 1) (x + 2)

or f (x) = 0 at x = 0, x = 1 and x = – 2.

Now f (x) = 36x2 + 24x – 24 = 12(3x2 + 2x – 2)

  f ‘’(0)=-24 <0

  f ‘’(1)= 36 >0

  f ‘’(-2)= 72 >0

Therefore, by second derivative test, x = 0 is a point of local maxima and local maximum value of f at x = 0 is f (0) = 12 while x = 1 and x = – 2 are the points of local minima and local minimum values of f at x = – 1 and – 2 are f (1) = 7 and f (–2) = –20, respectively.

Example : If length of three sides of a trapezium other than base are equal to 10cm, then find the area of the trapezium when it is maximum.

Solution The required trapezium is as given in Figure 6.19.

Draw perpendiculars DP and

CQ on AB. Let AP = x cm. Note that APD ~ BQC. Therefore, QB = x cm. Also, by

Pythagoras theorem, DP = QC = √(100-x 2). Let A be the area of the trapezium. Then

AA(x) = (sum of parallel sides) (height)

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.

In view of the above results, we have the following working rule for finding absolute maximum and/or absolute minimum values of a function in a given closed interval [a, b].

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 (listed in Step 1 and 2), 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 .

Example: Find the absolute maximum and minimum values of a function f given by

f (x) = 2x3 – 15x2 + 36x +1 on the interval [1, 5].

Solution We have  f (x) = 2x3 – 15x2 + 36x +1

or f (x) = 6x2 – 30x + 36 = 6(x – 3) (x – 2)

Note that f (x) = 0 gives x = 2 and x = 3.

We shall now evaluate the value of f at these points and at the end points of the

interval [1, 5], i.e., at x = 1, x = 2, x = 3 and at x = 5. So

f (1) = 2(13) – 15(12) + 36(1) + 1 = 24

f (2) = 2(23) – 15(22) + 36(2) + 1 = 29

f (3) = 2(33) – 15(32) + 36(3) + 1 = 28

f (5) = 2(53) – 15(52) + 36(5) + 1 = 56

Thus, we conclude that absolute maximum value of f on [1, 5] is 56, occurring at x =5, and absolute minimum value of f on [1, 5] is 24 which occurs at x = 1.