Solved Examples of Second Derivative Test

Example 1: Find the point of local maxima and local minima of the function x³ – 12x using second derivative test.

Solution:

Given f(x) = x³-12x

In order to calculate the local maxima and local minima, differentiate f(x) w.r.t x

f'(x) = 3x² – 12

Equate f'(x) to 0

3x² – 12 = 0

⇒ 3x² = 12

⇒ x = 2 or -2

Now calculate f”(x)

f”(x) = 6x

At x = 2, f”(2) = 12 > 0. This means that x = 2 is the point of local minima.

At x = -2, f”(-2) = -12 < 0. This means that x = -2 is the point of local maxima.

Example 2: Find the point of local maxima and local minima of the function x³ – x² – 5x using second derivative test.

Solution:

Given f(x) = x³-x²– 5x

In order to calculate the local maxima and local minima, differentiate f(x) w.r.t x

f'(x) = 3x² – 2x – 5

Equate f'(x) to 0

3x² – 2x – 5 = 0

⇒ [Tex]x = \frac{2\pm \sqrt{4-4(3)(-5)}}{2(3)} \\ \implies x = \frac{2\pm \sqrt{64}}{6} \\ \implies x = \frac{5}{3}~or~-1 [/Tex]

Now calculate f”(x)

f”(x) = 6x – 2

At x = 5/3, f”(5/3) = 8 > 0. This means that x = 5/3 is the point of local minima.

At x = -1, f”(-1) = -8 < 0. This means that x = -1 is the point of local maxima.

Example 3: Find the point of local maxima and local minima of the function x⁴– x² using second derivative test.

Solution:

Given f(x) = x⁴-x²

In order to calculate the local maxima and local minima, differentiate f(x) w.r.t x

f'(x) = 4x³ – 2x

Equate f'(x) to 0

4x³ – 2x = 0

⇒ 4x³ = 2x

⇒ 2x² = 1

x = 1/√2 or -1√2

Now calculate f”(x)

f”(x) = 12x² – 2

At x = 1/√2, f”(1/√2) = 4 > 0. This means that x = 1/√2 is the point of local minima.

At x = -1/√2, f”(-1/√2) = -8 < 0. This means that x = -1/√2 is the point of local maxima.

Example 4: A bus is moving along the curve 2y = 2x²+10. A man standing at point (3,5) wants to find the nearest distance between him and bus. Calculate the nearest distance.

Solution:

Given 2y = 2x² + 10

⇒ y = x² + 10

f(x) = x² + 5

We need to calculate the nearest distance which means minimum distance. Let the distance be minimum when bus is at a point (x, y)

Distance of point (3, 5) from the point (x, y) = D = √(x-3)²+(y-5)²

D² = (x-3)² + (y-5)²

Substituting the value of y = x² + 5

D² = (x-3)² + (x²)² ……………….(1)

D is minimum when D² is minimum. Thus we will calculate D’

D’ = 2(x-3) + 4x³

D’ = 2x – 6 + 4x³

D’ = (x−1)(4x²+4x+6) = 2(x−1)(2x²+2x+3)

Equating D’ = 0

x-1 = 0 or 2x²+2x+3 = 0

⇒x = 1 OR

[Tex]⇒ x = \frac{-2\pm \sqrt{4-4(2)(3)}}{2(2)} \\ \implies x = \frac{-2\pm \sqrt{-20}}{4} [/Tex]

This is not possible as x does not assume real value

Thus x = 1

Now D” = 2+12x²

At x =1, D” = 14 > 0. Thus x = 1 is the point of local minima.

Thus minimum distance is at x = 1 which is calculated by substituting x = 1 in equation (1)

D² = 4 + 1 = 5

D = √5

Thus the minimum distance is √5

Example 5: Calculate the maximum height of a cricket ball if its height is given by the function h(x) = 2x³-12x²+1 using second derivative test.

Solution:

Given h(x) = 2x³– 12x²+1

We need to calculate the maximum height.

In order to calculate the maximum height, differentiate h(x) w.r.t x

h'(x) = 6x²– 24x

Equate h'(x) to 0

6x²– 24x = 0

⇒ 6x(x-4) = 0

⇒ x = 0 or 4

Now calculate f”(x)

h”(x) = 12x – 24

At x = 0, h”(0) = -24 < 0. This means that x = 1/√2 is the point of local maxima.

At x = 4, h”(4) = 24 > 0. This means that x = -1/√2 is the point of local minima.

Thus the height is maximum at a x = 0.

Maximum height = h(0) = 2(0)³-12(0) + 1 = 1 unit

Example 6: Find the point of local maxima and local minima of the function x⁴ – 12x³ using second derivative test.

Solution:

Given f(x) = x⁴ – 12x³

In order to calculate the local maxima and local minima, differentiate f(x) w.r.t x

f'(x) = 4x³ – 36x²

Equate f'(x) to 0

4x³ – 36x² = 0

⇒ 4x²(x-9) = 0

⇒ x = 0 or 9

Now calculate f”(x)

f”(x) = 12x² – 72x

At x = 0, f”(0) = 0 > 0. This means that x = 0 is the point of inflection.

At x = 9, f”(9) = 12(9)²-72(9) = 324 > 0. This means that x = 9 is the point of local minima.

Example 7: Find the point of local maxima or local minima of the function x² – 1 using second derivative test.

Solution:

Given f(x) = x² – 1

In order to calculate the local maxima and local minima, differentiate f(x) w.r.t x

f'(x) = 2x

Equate f'(x) to 0

2x = 0

⇒ x = 0

Now calculate f”(x)

f”(x) = 2

At x = 0, f”(0) = 2 > 0. This means that x = 0 is the point of local minima.

Example 8: Find the point of local maxima and local minima of the function 2x² – 3x using second derivative test.

Solution:

Given f(x) = 2x² – 3x

In order to calculate the local maxima and local minima, differentiate f(x) w.r.t x

f'(x) = 4x – 3

Equate f'(x) to 0

4x – 3 = 0

⇒ x = 3/4

Now calculate f”(x)

f”(x) = 4

At x = 3/4, f”(3/4) = 4 > 0. This means that x = 3/4 is the point of local minima.

Second Derivative Test is one of the methods in calculus to find the maxima and minima of a curve. Other than, the second derivative test there is also a first derivative, which can be referred to as a rudimentary version of the second derivative test.

First derivative test helps us find critical points for a given function but does not tell us about the nature of the function at these points. We also come across cases where we cannot get critical points as the first derivative test fails. Second derivative test is used in these cases. The second derivative test clearly tells us if the critical point obtained is a point of local maximum or local minimum. Second derivative test is also helpful in solving various problems in different fields such as science, physics, and engineering. In this article, we shall discuss the second derivative test in detail.