Table of Content

• What is Second Derivative Test?
• Steps for Second Derivative Test for Maxima and Minima
• Example of Second Derivative Test
• Uses of Second Derivative Test
• First and Second Derivative Test
• Multivariable Second Derivative Test
• Solved Examples of Second Derivative Test
• Practice Problems on Second Derivative Test

Test	First Derivative Test	Second Derivative Test
Critical Point	Where f'(x) = 0 or f'(x) is undefined	Where f'(x) = 0 or f'(x) is undefined
Local Minima	f'(x) changes from negative to positive	f”(x) > 0 at the critical point
Local Maxima	f'(x) changes from positive to negative	f”(x) < 0 at the critical point
Neither Minima nor Maxima	f'(x) fails to exit.	f”(x) = 0 (inconclusive)

Multivariable Second Derivative Test

Multivariable second derivative test is used in case when the given function has two variable (say x and y). This method makes use of partial differentiation to find the local maxima and local minima. According to this test:

• Let f be a function with two variables x and y.
• Take partial derivatives of function f with respect to each variable and set them equal to 0. Solve the equations obtained to get the values of x and y.
• Let:

[Tex]D (x, y) = \frac{\partial ^ 2f}{\partial x^2} – \frac{\partial ^ 2f}{\partial y^2} – (\frac{\partial ^ 2f}{\partial x \partial y})^2 [/Tex]

• If D(x, y) < 0 then f has a saddle point at x and y
• If D(x, y) = 0 then the test fails
• If D(x, y) > 0 then,
  • If [Tex]\frac{\partial ^ 2}{\partial x^2} \left(f(x, y)\right) > 0 [/Tex], then (x, y) is the point of local minima
  • If [Tex]\frac{\partial ^ 2}{\partial x^2}\left(f(x, y)\right)< 0 [/Tex], then (x, y) is the point of local maxima

Also, Check

• Relative Maxima and Minima
• Higher Order Derivatives
• Derivative Rules

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.

Example 9 : Find and classify the critical points of f(x) = x^3 – 12x using the Second Derivative Test.

Solution :

First derivative: f'(x) = 3x^2 – 12

Critical points: Set f'(x) = 0

3x^2 – 12 = 0

x^2 = 4

x = ±2

Second derivative: f”(x) = 6x

Evaluate f”(x) at critical points:

f”(2) = 12 > 0, so x = 2 is a local minimum

f”(-2) = -12 < 0, so x = -2 is a local maximum

Example 10 : Use the Second Derivative Test to determine the nature of the critical point of f(x) = x^2 e^(-x).

Solution :

First derivative: f'(x) = 2xe^(-x) – x^2e^(-x) = xe^(-x)(2 – x)

Critical points: x = 0 or x = 2

Second derivative: f”(x) = e^(-x)(2 – x) – xe^(-x)(2 – x) – xe^(-x) = e^(-x)(2 – 4x + x^2)

Evaluate f”(x) at critical points:

f”(0) = 2 > 0, so x = 0 is a local minimum

f”(2) = -2e^(-2) < 0, so x = 2 is a local maximum

Practice Problems on Second Derivative Test

1. Use the Second Derivative Test to find and classify the critical points of f(x) = x³ – 3x² – 9x + 5.

2. Determine the nature of the critical points for the function f(x) = x⁴ – 8x² + 16.

3. Apply the Second Derivative Test to classify the critical points of f(x) = x^1/3 (x – 3)²

4. Find and classify the critical points of f(x) = xe^-x2 using the Second Derivative Test.

5. Use the Second Derivative Test to analyze the critical points of f(x) = ln(x² + 1) + 2x.

6. Determine the nature of the critical points for the function f(x) = x³ – 6x² + 12x – 8.

7. Apply the Second Derivative Test to the function f(x) = x² / (x² + 1) and classify its critical points.

8. Find and classify the critical points of f(x) = x⁵ – 5x⁴ + 5x using the Second Derivative Test.

9. Use the Second Derivative Test to analyze the critical points of f(x) = sin(x) + cos(x) on the interval [0, 2π].

10. Determine the nature of the critical points for the function f(x) = x^2/3 – 2x^1/3.

Conclusion

The Second Derivative Test is a powerful tool in calculus for analyzing the behavior of functions at their critical points. By examining the sign of the second derivative at these points, we can determine whether they represent local maxima, local minima, or potentially inflection points. This test complements other techniques like the First Derivative Test and provides valuable insights into the shape and characteristics of functions. Mastering the Second Derivative Test enhances one’s ability to analyze and understand complex mathematical relationships, making it an essential skill for students and professionals in fields such as mathematics, physics, engineering, and economics. Practice with diverse functions and scenarios is key to developing proficiency in applying this test effectively.

FAQs on Second Derivative Test

What is Second Derivative Test for Minimum and Maximum?

Second derivative test is used to find the maxima and minima of any function if the second derivative is negative at a point then it gives the maxima of the function and if it is positive at any point then it gives the minima.

What is the difference between second and first derivative test?

The difference in first and second derivative test lies in the fact that first derivative test only provides us with critical points and does not tell if it is a point of local maximum or local minimum whereas second derivative test tells us if the point is a local maximum or local minimum.

What are the applications of second derivative test?

Applications of second derivative test are as follows: It can also help us to determine the extremities of the curve. It can also help us to know about the orientation of a parabola.

What is a point of Inflection?

A point is said to be point of inflection where the concavity of the curve either changes from concave up to concave down or concave down to concave up.

What is the condition for second derivative test for local Maxima?

For any function f(x) at point x = k, if f”(k) < 0 where f'(k) is either 0 or not defined, then function is said to have local maxima at x = k, and f'(k) is the local maximum value.

What is the condition for second derivative test for local Minima?

For any function f(x) at point x = k, if f”(k) > 0 where f'(k) is either 0 or not defined, then function is said to have local minima at x = k, and f'(k) is the local minimum value.

What do you mean by critical point?

The point at which the first derivative of a function is zero or does not exist is called the critical point.

What is the nature of graph of the function at the point of local maximum?

At the point of local maximum, the graph of the function is concave downward.

What is the nature of graph of the function at the point of local minimum?

At the point of local minimum, the the graph of the function is concave upward.