Applications of Second Derivative

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 ′( x) is strictly increasing on I , then the function is concave up on an open interval I

(ii) If f ′( 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 ′′(x) > 0 on an open interval I , then f ( x) is concave up on I .

(ii) If f ′′(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 “concave up to concave down” or “concave down to concave up” are called the points of inflection of f ( x) .

Theorem 7.12 (Test for Points of Inflection)

(i) If f ′′(c) exists and f ′′(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 ′′(c) exists at the point of inflection, then f ′′(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 ′′( x) changes sign. For ‘smooth’ curves (no sharp corners), this may happen when either

(i) f ′′(x) = 0 or

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

Remark

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

(2) It is possible that f ′′(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  ′ (x) = cos x and f  ′′(x) = − sin x and hence (π, 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 −1)³ ⋅ ( x − 5), x ∈ ℝ and, points of inflection if any.

Solution

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

f ′(x) = (x − 1)³ + 3(x − 1)² ⋅ (x − 5)

= 4(x − 1)² ⋅ (x − 4)

f ′′ (x)= 4((x − 1)² + 2(x − 1) ⋅ (x − 4))

 = 12(x − 1) ⋅ (x − 3)

Now,

f ′′( x) = 0 ⇒ x =1, x = 3 .

The intervals of concavity are tabulated in Table 7.7.

The curve is concave upwards on (−∞,1) and (3, ∞) .

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

As f ′′(x) changes its sign when it passes through x = 1 and x = 3, (1, f (1)) = (1, 0) and (3, f (3)) = (3, −16) are points of inflection for the graph y = f ( x) . The sign change may be observed from the adjoining figure of the curve f ′′( 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 2π and hence there will be stationary points and points of inflections in each period interval. We have,

We now consider an interval, (−π , π ) by splitting into two sub intervals (−π , 0) and ( 0, π) .

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

In the interval (0,π ), 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 (nπ , (n +1)π ) , where n is any integer which can be discussed as before to conclude that (nπ, 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 ′(c) = 0 , that f ′( x) exists in a neighborhood of c , and that f ′′(c) exists. Then f has a relative maximum value at c if f ′′ (c) < 0 and a relative minimum value at c if f ′′ (c ) > 0 . If f ′′ (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 ′( x) = 4x³ + 32 = 0 gives x³ = −8

x = −2

and f ′′( x) = 12x².

As f′′ (−2) > 0, the function has local minimum at x = −2 . The local minimum value is f (− 2) = −48 . Therefore, the extreme point is (−2, −48).

Example 7.60

Find the local extrema of the function f ( x ) = 4x⁶ −6x⁴ .

Solution

Differentiating with respect to x, we get

  f ′( x) ₌ 24x⁵ -24x³

 = 24x³ ( x² −1)

 = 24x³ ( x +1)( x −1)

 f  ′( x ) = 0 ⇒ x = −1, 0, 1. Hence the critical numbers are x = −1, 0,1.

Now, f ′′( x ) = 120x⁴ −72x² = 24x² (5 x² −3).

⇒ f ′′(−1) = 48 , f ′′( 0) = 0 , f ′′(1) = 48 .

As f ′′(−1) and f ′′(1) are positive by the second derivative test, the function f ( x) has local minimum. But at x = 0,  f ′′(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 = −1, its local minimum value is −2 . 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 −2 .

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 − x)². Now,

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

Therefore, f ′(x) = 4x³ − 60x² + 200x = 4x(x² −15x + 50)

 f ′(x)  = 4x (x² − 15x + 50) = 0 ⇒ x = 0, 5, 10

and f′′ (x) = 12x² − 120x + 200

The stationary numbers of f ( x) are x = 0, 5, 10 at these points the values of f ′′ ( x) are respectively 200, −100 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 .