Derivatives - Second derivative test

• The second derivative test is a method for finding the concavity of a function and determining whether a critical point is a relative maximum or minimum.
• It involves the use of the second derivative of a function.
• Let’s understand the steps to perform the second derivative test.
• Step 1: Find the first derivative of the function.
• Step 2: Find the second derivative of the function.
• Step 3: Set the second derivative equal to zero and solve for x to find any possible inflection points.
• Step 4: Use the sign of the second derivative to determine the concavity of the graph.
• Step 5: Use the critical points and inflection points to determine whether they are relative maxima or minima. Example: Consider the function f(x) = x^3 - 9x^2 + 24x - 20. Let’s perform the second derivative test on this function. Equations:
• First derivative: f’(x) = 3x^2 - 18x + 24
• Second derivative: f’’(x) = 6x - 18 Step 1: Find the first derivative: f’(x) = 3x^2 - 18x + 24 Step 2: Find the second derivative: f’’(x) = 6x - 18 Step 3: Set f’’(x) = 0 and solve:

6x - 18 = 0 x = 3 Step 4: Analyze the sign of the second derivative:

• For x < 3, f’’(x) < 0, so the graph is concave down.
• For x > 3, f’’(x) > 0, so the graph is concave up. Step 5: Determine the nature of critical and inflection points:
• At x = 3, f’’(x) changes sign from negative to positive, so there is a relative minimum at x = 3.

11. Second Derivative Test - Example 1

• Let’s apply the second derivative test to the function f(x) = x^3 - 3x^2 - 9x + 5.
• First, find the first derivative: f’(x) = 3x^2 - 6x - 9.
• Next, find the second derivative: f’’(x) = 6x - 6.
• Find the critical points by setting f’(x) = 0: 3x^2 - 6x - 9 = 0.
• Solve this quadratic equation to get x = -1 and x = 3.
• Analyze the sign of f’’(x) for x < -1, between -1 and 3, and for x > 3.
• Determine the nature of the critical points using the sign changes of f’’(x).

12. Second Derivative Test - Example 2

• Consider the function f(x) = x^4 - 4x^3 + 3x^2.
• Calculate the first derivative: f’(x) = 4x^3 - 12x^2 + 6x.
• Next, find the second derivative: f’’(x) = 12x^2 - 24x + 6.
• Set f’’(x) = 0 to find the critical points.
• Solve the quadratic equation 12x^2 - 24x + 6 = 0.
• Find the nature of the critical points based on the sign changes of f’’(x).

13. Second Derivative Test - Example 3

• Let’s consider the function f(x) = x^3 - 3x + 2.
• Calculate the first derivative: f’(x) = 3x^2 - 3.
• Next, find the second derivative: f’’(x) = 6x.
• Set f’’(x) = 0 to find the critical points.
• Solve the equation 6x = 0 to get x = 0.
• Analyze the sign of f’’(x) to determine the concavity of the graph.
• Determine the nature of the critical point using the sign change.

14. Second Derivative Test - Example 4

• Consider the function f(x) = -x^3 + 6x^2 - 9x + 4.
• Calculate the first derivative: f’(x) = -3x^2 + 12x - 9.
• Next, find the second derivative: f’’(x) = -6x + 12.
• Set f’’(x) = 0 to find the critical point.
• Solve the equation -6x + 12 = 0 to get x = 2.
• Analyze the sign of f’’(x) to determine the concavity.
• Determine the nature of the critical point by using the sign change.

15. Second Derivative Test - Example 5

• Let’s consider the function f(x) = x^3 - 6x^2 + 12x - 8.
• Calculate the first derivative: f’(x) = 3x^2 - 12x + 12.
• Next, find the second derivative: f’’(x) = 6x - 12.
• Set f’’(x) = 0 to find the critical point.
• Solve the equation 6x - 12 = 0 to get x = 2.
• Analyze the sign of f’’(x) to determine the concavity.
• Determine the nature of the critical point using the sign change.

16. Inflection Points

• An inflection point is a point on the graph of a function where the curve changes its concavity.
• A function may have multiple inflection points or none at all.
• To find the inflection points, we need to analyze the second derivative of the function.
• If the concavity changes at a particular x-value, then it is an inflection point.
• Inflection points can occur when the second derivative is zero or when it does not exist.
• Sometimes, inflection points may also occur at vertical asymptotes or in isolated regions of the graph.

17. Inflection Points - Example 1

• Consider the function f(x) = x^3 - 3x.
• Find the first derivative: f’(x) = 3x^2 - 3.
• Next, calculate the second derivative: f’’(x) = 6x.
• Set f’’(x) = 0 to find the critical point.
• Solve the equation 6x = 0 to get x = 0.
• Analyze the sign of f’’(x) to determine the concavity.
• Determine whether the critical point is an inflection point based on the concavity change.

18. Inflection Points - Example 2

• Let’s consider the function f(x) = x^4 - 4x^2.
• Calculate the first derivative: f’(x) = 4x^3 - 8x.
• Next, find the second derivative: f’’(x) = 12x^2 - 8.
• Set f’’(x) = 0 to find the critical points.
• Solve the quadratic equation 12x^2 - 8 = 0.
• Analyze the sign of f’’(x) to determine the concavity.
• Determine if the critical points are inflection points based on the concavity change.

19. Summary

• The second derivative test helps determine the concavity of a graph and the nature of critical points.
• To perform the second derivative test:
  • Find the first derivative and the second derivative of the function.
  • Set the second derivative equal to zero to find any possible inflection points.
  • Analyze the sign of the second derivative to determine the concavity of the graph.
  • Use the critical points and inflection points to determine whether they are relative maxima or minima.
• Inflection points indicate a change in concavity, while critical points can be either maxima, minima, or points of inflection.

20. Conclusion

• The second derivative test is a useful tool for analyzing the concavity and nature of critical points of a function.
• By finding the first and second derivatives of a function, we can determine the concavity of the graph and identify inflection points.
• The sign changes of the second derivative at critical points help us classify those points as relative maxima, minima, or points of inflection.
• Practice using the second derivative test with different examples to enhance your understanding and problem-solving skills.

Examples of Second Derivative Test

• Example 1:
  • Function: f(x) = x^3 - 9x^2 + 24x - 20
  • First derivative: f’(x) = 3x^2 - 18x + 24
  • Second derivative: f’’(x) = 6x - 18
  • Critical points: x = 3 (relative minimum)
• Example 2:
  • Function: f(x) = x^4 - 4x^3 + 3x^2
  • First derivative: f’(x) = 4x^3 - 12x^2 + 6x
  • Second derivative: f’’(x) = 12x^2 - 24x + 6
  • Critical points: x = -1 (relative maximum), x = 3 (relative minimum)

Examples of Second Derivative Test (Contd.)

• Example 3:
  • Function: f(x) = x^3 - 3x + 2
  • First derivative: f’(x) = 3x^2 - 3
  • Second derivative: f’’(x) = 6x
  • Critical points: x = 0 (point of inflection)
• Example 4:
  • Function: f(x) = -x^3 + 6x^2 - 9x + 4
  • First derivative: f’(x) = -3x^2 + 12x - 9
  • Second derivative: f’’(x) = -6x + 12
  • Critical point: x = 2 (point of inflection)

Examples of Second Derivative Test (Contd.)

• Example 5:
  • Function: f(x) = x^3 - 6x^2 + 12x - 8
  • First derivative: f’(x) = 3x^2 - 12x + 12
  • Second derivative: f’’(x) = 6x - 12
  • Critical point: x = 2 (point of inflection)
• Example 6:
  • Function: f(x) = 4x^3 - 12x^2 - 3x + 5
  • First derivative: f’(x) = 12x^2 - 24x - 3
  • Second derivative: f’’(x) = 24x - 24
  • Critical point: None (no inflection)

Inflection Points - Example 1

• Function: f(x) = x^3 - 3x
• First derivative: f’(x) = 3x^2 - 3
• Second derivative: f’’(x) = 6x
• Critical point: x = 0 (point of inflection)

Inflection Points - Example 2

• Function: f(x) = x^4 - 4x^2
• First derivative: f’(x) = 4x^3 - 8x
• Second derivative: f’’(x) = 12x^2 - 8
• Critical points: x = -√(2) (point of inflection), x = √(2) (point of inflection)

Summary

• Second derivative test helps determine the concavity and nature of critical points.
• Inflection points occur at places where the curve changes its concavity.
• Critical points can be maxima, minima, or points of inflection.
• Examples covered: f(x) = x^3 - 3x^2 - 9x + 5, f(x) = x^4 - 4x^3 + 3x^2, f(x) = x^3 - 3x + 2, f(x) = -x^3 + 6x^2 - 9x + 4, f(x) = x^3 - 6x^2 + 12x - 8.
• Inflection point examples: f(x) = x^3 - 3x, f(x) = x^4 - 4x^2.

Conclusion

• Second derivative test is a valuable tool for analyzing functions and their critical points.
• It helps determine the concavity of a graph and identify inflection points.
• Practice using the second derivative test with more examples to strengthen your understanding.
• Mastering this test will enable you to solve complex problems related to derivatives effectively.

Recap

• The second derivative test determines the concavity and nature of critical points.
• It involves finding the first and second derivatives of a function.
• The sign changes of the second derivative at critical points indicate their nature.
• Inflection points occur where the curve changes its concavity.
• Examples covered various functions and their critical points.

Practice Problems

• Find the critical points and determine their nature using the second derivative test for the following functions:
  • f(x) = x^3 - 6x^2 + 9x - 3
  • f(x) = 4x^3 - 6x^2 - 8x + 12
  • f(x) = x^4 - 8x^3 + 16x^2

References

• Stewart, J. (2008). Calculus: Early Transcendentals. Cengage Learning.
• Larson, R., Hostetler, R., & Edwards, B. (2006). Calculus of a Single Variable. Cengage Learning.
• Khan Academy. (n.d.). Second derivative test. Retrieved from https://www.khanacademy.org/math/ap-calculus-ab/ab-derivative-applications/ab-analyzing-functions/a/second-derivative-test "