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# Applications of Second Derivative

Second derivative of a function is used to determine the concavity, convexity, the points of inflection, and local extrema of functions.

Applications of Second Derivative

Second derivative of a function is used to determine the concavity, convexity, the points of inflection, and local extrema of functions.

## Concavity, Convexity, and Points of Inflection

A graph is said to be concave down (convex up) at a point if the tangent line lies above the graph in the vicinity of the point. It is said to be concave up (convex down) at a point if the tangent line to the graph at that point lies below the graph in the vicinity of the point. This may be easily observed from the adjoining graph. ### Definition 7.8

Let f ( x) be a function whose second derivative exists in an open interval I = (a, b) . Then the function f ( x) is said to be

(i) If f ŌĆ▓( x) is strictly increasing on I , then the function is concave up on an open interval I

(ii) If f ŌĆ▓( x) is strictly decreasing on I , then the function is concave down on an open interval I.

Analytically, given a differentiable function whose graph y = f ( x) , then the concavity is given by the following result.

### Theorem 7.11 (Test of Concavity)

(i) If f ŌĆ▓ŌĆ▓(x) > 0 on an open interval I , then f ( x) is concave up on I .

(ii) If f ŌĆ▓ŌĆ▓(x) < 0 on an open interval I , then f ( x) is concave down on I .

Remark

(1) Any local maximum of a convex upward function defined on the interval [a , b] is also its absolute maximum on this interval.

(2) Any local minimum of a convex downward function defined on the interval [a , b] is also its absolute minimum on this interval.

(3) There is only one absolute maximum (and one absolute minimum) but there can be more than one local maximum or minimum.

## Points of Inflection

### Definition 7.9

The points where the graph of the function changes from ŌĆ£concave up to concave downŌĆØ or ŌĆ£concave down to concave upŌĆØ are called the points of inflection of f ( x) .

## Theorem 7.12 (Test for Points of Inflection)

(i) If f ŌĆ▓ŌĆ▓(c) exists and f ŌĆ▓ŌĆ▓(c) changes sign when passing through x = c , then the point (c , f (c)) is a point of inflection of the graph of f .

(ii) If f ŌĆ▓ŌĆ▓(c) exists at the point of inflection, then f ŌĆ▓ŌĆ▓(c) = 0 .

Remark

To determine the position of points of inflexion on the curve y = f ( x) it is necessary to find the points where f ŌĆ▓ŌĆ▓( x) changes sign. For ŌĆśsmoothŌĆÖ curves (no sharp corners), this may happen when either

(i) f ŌĆ▓ŌĆ▓(x) = 0 or

(ii) f ŌĆ▓ŌĆ▓( x) does not exist at the point.

Remark

(1) It is also possible that f ŌĆ▓ŌĆ▓(c) may not exist, but (c , f (c)) could be a point of inflection. For instance, f (x) = x1/3 at c = 0 .

(2) It is possible that f ŌĆ▓ŌĆ▓(c) = 0 at a point but (c , f (c)) need not be a point of inflection. For instance, f (x) = x4 at c = 0 .

(3) A point of inflection need not be a stationary point. For instance, if f (x) = sin x then, f  ŌĆ▓ (x) = cos x and f  ŌĆ▓ŌĆ▓(x) = ŌłÆ sin x and hence (ŽĆ, 0) is a point of inflection but not a stationary point for f (x) .

Example 7.57

Determine the intervals of concavity of the curve f (x) = (x ŌłÆ1)3 Ōŗģ ( x ŌłÆ 5), x Ōłł ŌäØ and, points of inflection if any.

Solution

The given function is a polynomial of degree 4. Now,

f ŌĆ▓(x) = (x ŌłÆ 1)3 + 3(x ŌłÆ 1)2 Ōŗģ (x ŌłÆ 5)

= 4(x ŌłÆ 1)2 Ōŗģ (x ŌłÆ 4)

f ŌĆ▓ŌĆ▓ (x)= 4((x ŌłÆ 1)2 + 2(x ŌłÆ 1) Ōŗģ (x ŌłÆ 4))

= 12(x ŌłÆ 1) Ōŗģ (x ŌłÆ 3) Now,

f ŌĆ▓ŌĆ▓( x) = 0 ŌćÆ x =1, x = 3 .

The intervals of concavity are tabulated in Table 7.7. The curve is concave upwards on (ŌłÆŌł×,1) and (3, Ōł×) .

The curve is concave downwards on (1, 3) .

As f ŌĆ▓ŌĆ▓(x) changes its sign when it passes through x = 1 and x = 3, (1, f (1)) = (1, 0) and (3, f (3)) = (3, ŌłÆ16) are points of inflection for the graph y = f ( x) . The sign change may be observed from the adjoining figure of the curve f ŌĆ▓ŌĆ▓( x) .

### Example 7.58

Determine the intervals of concavity of the curve y = 3 + sin x .

Solution

The given function is a periodic function with period 2ŽĆ and hence there will be stationary points and points of inflections in each period interval. We have, We now consider an interval, (ŌłÆŽĆ , ŽĆ ) by splitting into two sub intervals (ŌłÆŽĆ , 0) and ( 0, ŽĆ) .

In the interval (ŌłÆŽĆ , 0) , d2y/dx2 > 0 and hence the function is concave upward.

In the interval (0,ŽĆ ), d2y/dx2 < 0 and hence the function is concave downward. Therefore (0, 3) is a point of inflection (see Fig. 7.25). The general intervals need to be considered to discuss the concavity of the curve are (nŽĆ , (n +1)ŽĆ ) , where n is any integer which can be discussed as before to conclude that (nŽĆ, 3) are also points of inflection.

## Extrema using Second Derivative Test

The Second Derivative Test: The Second Derivative Test relates the concepts of critical points, extreme values, and concavity to give a very useful tool for determining whether a critical point on the graph of a function is a relative minimum or maximum.

## Theorem 7.13 (The Second Derivative Test)

Suppose that c is a critical point at which f ŌĆ▓(c) = 0 , that f ŌĆ▓( x) exists in a neighborhood of c , and that f ŌĆ▓ŌĆ▓(c) exists. Then f has a relative maximum value at c if f ŌĆ▓ŌĆ▓ (c) < 0 and a relative minimum value at c if f ŌĆ▓ŌĆ▓ (c ) > 0 . If f ŌĆ▓ŌĆ▓ (c) = 0 , the test is not informative.

Example 7.59

Find the local extremum of the function f (x) = x4 + 32x .

Solution

We have,

f ŌĆ▓( x) = 4x3 + 32 = 0 gives x3 = ŌłÆ8

x = ŌłÆ2

and f ŌĆ▓ŌĆ▓( x) = 12x2.

As fŌĆ▓ŌĆ▓ (ŌłÆ2) > 0, the function has local minimum at x = ŌłÆ2 . The local minimum value is f (ŌłÆ 2) = ŌłÆ48 . Therefore, the extreme point is (ŌłÆ2, ŌłÆ48).

### Example 7.60

Find the local extrema of the function f ( x ) = 4x6 ŌłÆ6x4 .

### Solution Differentiating with respect to x, we get

f ŌĆ▓( x) = 24x5 -24x3

= 24x3 ( x2 ŌłÆ1)

= 24x3 ( x +1)( x ŌłÆ1)

f  ŌĆ▓( x ) = 0 ŌćÆ x = ŌłÆ1, 0, 1. Hence the critical numbers are x = ŌłÆ1, 0,1.

Now, f ŌĆ▓ŌĆ▓( x ) = 120x4 ŌłÆ72x2 = 24x2 (5 x2 ŌłÆ3).

ŌćÆ f ŌĆ▓ŌĆ▓(ŌłÆ1) = 48 , f ŌĆ▓ŌĆ▓( 0) = 0 , f ŌĆ▓ŌĆ▓(1) = 48 .

As f ŌĆ▓ŌĆ▓(ŌłÆ1) and f ŌĆ▓ŌĆ▓(1) are positive by the second derivative test, the function f ( x) has local minimum. But at x = 0,  f ŌĆ▓ŌĆ▓(0) = 0 . That is the second derivative test does not give any information about local extrema at x = 0 . Therefore, we need to go back to the first derivative test. The intervals of monotonicity is tabulated in Table 7.8. By the first derivative test f ( x) has local minimum at x = ŌłÆ1, its local minimum value is ŌłÆ2 . At x = 0 , the function f ( x) has local maximum at x = 0 , and its local maximum value is 0. At x = 1, the function f ( x) has local minimum at x = 1, and its local minimum value is ŌłÆ2 .

### Remark

When the second derivative vanishes, we have no information about extrema. We have used the first derivative test to find out the extrema of the function!

### Example 7.61

Find the local maximum and minimum of the function x 2 y2 on the line x + y = 10 .

Solution

Let the given function be written as f (x) = x2 (10 ŌłÆ x)2. Now,

f (x) = x2 (100 ŌłÆ 20x + x2 ) = x4 ŌłÆ 20x3 +100x2

Therefore, f ŌĆ▓(x) = 4x3 ŌłÆ 60x2 + 200x = 4x(x2 ŌłÆ15x + 50)

f ŌĆ▓(x)  = 4x (x2 ŌłÆ 15x + 50) = 0 ŌćÆ x = 0, 5, 10

and fŌĆ▓ŌĆ▓ (x) = 12x2 ŌłÆ 120x + 200

The stationary numbers of f ( x) are x = 0, 5, 10 at these points the values of f ŌĆ▓ŌĆ▓ ( x) are respectively 200, ŌłÆ100 and 200 . At x = 0 , it has local minimum and its value is f (0) = 0 . At x = 5 , it has local maximum and its value is f (5) = 625 . At x = 10 , it has local minimum and its value is f (10) = 0 .

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12th Maths : UNIT 7 : Applications of Differential Calculus : Applications of Second Derivative | Applications of Differential Calculus | Mathematics