Derivatives - Proof Of Second Derivative Test

Slide 1: Derivatives - Proof of second derivative test

The second derivative test helps to determine the nature of critical points on a function.
It helps in classifying the critical points as maximum, minimum, or neither.
The second derivative test is based on the concavity of the function at the critical point.
Let’s understand the proof of the second derivative test.

Slide 2: Second Derivative Test - Definition

The second derivative test states that if f’(c) = 0 and f’’(c) > 0, then f(c) is a local minimum.
If f’(c) = 0 and f’’(c) < 0, then f(c) is a local maximum.
If f’’(c) = 0, the test is inconclusive and may be a point of inflection.

Slide 3: Second Derivative Test - Proof for Local Maximum

Let’s consider a function f(x) that is twice differentiable near c.
If f’(c) = 0 and f’’(c) < 0, we will prove that f(c) is a local maximum.
We will use Taylor series expansion to prove this.

Taylor’s Theorem

Taylor’s theorem states that for a function f(x) that is n times differentiable in an interval containing c, the function can be approximated by its Taylor series expansion.
Taylor’s series expansion of f(x) about c is given by: f(x) = f(c) + f’(c)(x - c) + (1/2)f’’(c)(x - c)^2 + … + (1/n!)f^n(c)(x - c)^n + Rn(x)

Slide 4: Second Derivative Test - Proof for Local Maximum (contd.)

Considering the Taylor expansion of f(x) about the point c: f(x) = f(c) + f’(c)(x - c) + (1/2)f’’(c)(x - c)^2 + … + (1/n!)f^n(c)(x - c)^n + Rn(x)
Since f’(c) = 0, the first derivative term disappears, and we are left with: f(x) = f(c) + (1/2)f’’(c)(x - c)^2 + … + (1/n!)f^n(c)(x - c)^n + Rn(x)
This equation represents an expansion around the point c. We can see that only the second derivative term contributes near the critical point.

Slide 5: Second Derivative Test - Proof for Local Maximum (contd.)

We want to prove that if f’(c) = 0 and f’’(c) < 0, then f(c) is a local maximum.
Using the Taylor expansion, we have: f(x) = f(c) + (1/2)f’’(c)(x - c)^2 + … + (1/n!)f^n(c)(x - c)^n + Rn(x)
For x greater than c, (x - c)^2 is always positive. Thus, the terms with positive powers of (x - c) contribute positively to f(x).
In contrast, for x less than c, (x - c)^2 is also positive, but all the terms after (x - c)^2 contribute negatively.
As f’’(c) < 0, the term (1/2)f’’(c)(x - c)^2 will be negative for values of x less than c.

Slide 6: Second Derivative Test - Proof for Local Maximum (contd.)

We can now write the inequality for f(x): f(x) > f(c) + (1/2)f’’(c)(x - c)^2, for x > c, f(x) < f(c) + (1/2)f’’(c)(x - c)^2, for x < c
Since f’’(c) < 0, the inequality changes to: f(x) > f(c) - (1/2)|f’’(c)|(x - c)^2, for x > c, f(x) < f(c) - (1/2)|f’’(c)|(x - c)^2, for x < c
From the inequality, we can see that for x > c, f(x) is always greater than f(c) and for x < c, f(x) is always less than f(c).
Therefore, f(c) is a local maximum.

Slide 7: Second Derivative Test - Proof for Local Minimum

Let’s consider a function f(x) that is twice differentiable near c.
If f’(c) = 0 and f’’(c) > 0, we will prove that f(c) is a local minimum.
We will use a similar approach as in the previous case.

Taylor’s Theorem (Recap)

Taylor’s theorem states that for a function f(x) that is n times differentiable in an interval containing c, the function can be approximated by its Taylor series expansion.
Taylor’s series expansion of f(x) about c is given by: f(x) = f(c) + f’(c)(x - c) + (1/2)f’’(c)(x - c)^2 + … + (1/n!)f^n(c)(x - c)^n + Rn(x)

Slide 8: Second Derivative Test - Proof for Local Minimum (contd.)

Considering the Taylor expansion of f(x) about the point c: f(x) = f(c) + f’(c)(x - c) + (1/2)f’’(c)(x - c)^2 + … + (1/n!)f^n(c)(x - c)^n + Rn(x)
Since f’(c) = 0, the first derivative term disappears, and we are left with: f(x) = f(c) + (1/2)f’’(c)(x - c)^2 + … + (1/n!)f^n(c)(x - c)^n + Rn(x)
This equation represents an expansion around the point c. We can see that only the second derivative term contributes near the critical point.

Slide 9: Second Derivative Test - Proof for Local Minimum (contd.)

We want to prove that if f’(c) = 0 and f’’(c) > 0, then f(c) is a local minimum.
Using the Taylor expansion, we have: f(x) = f(c) + (1/2)f’’(c)(x - c)^2 + … + (1/n!)f^n(c)(x - c)^n + Rn(x)
For x greater than c, (x - c)^2 is always positive. Thus, the terms with positive powers of (x - c) contribute positively to f(x).
In contrast, for x less than c, (x - c)^2 is also positive, but all the terms after (x - c)^2 contribute negatively.
As f’’(c) > 0, the term (1/2)f’’(c)(x - c)^2 will be positive for values of x less than c.

Slide 10: Second Derivative Test - Proof for Local Minimum (contd.)

We can now write the inequality for f(x): f(x) > f(c) + (1/2)f’’(c)(x - c)^2, for x > c, f(x) < f(c) + (1/2)f’’(c)(x - c)^2, for x < c
Since f’’(c) > 0, the inequality changes to: f(x) > f(c) + (1/2)f’’(c)(x - c)^2, for x > c, f(x) < f(c) + (1/2)f’’(c)(x - c)^2, for x < c
From the inequality, we can see that for x > c, f(x) is always greater than f(c) and for x < c, f(x) is always less than f(c).
Therefore, f(c) is a local minimum.

Slide 11: Second Derivative Test - Example 1

Let’s consider the function f(x) = x^3 - 3x^2 - 9x + 5.
To apply the second derivative test, we need to find the critical points of the function.
First, we find the first derivative: f’(x) = 3x^2 - 6x - 9.
Setting f’(x) = 0, we get 3x^2 - 6x - 9 = 0.
Solving the quadratic equation, we find x = -1 and x = 3.
These are the critical points of the function.

Slide 12: Second Derivative Test - Example 1 (contd.)

Now, let’s find the second derivative: f’’(x) = 6x - 6.
Evaluating f’’(x) at the critical points:
- f’’(-1) = 6(-1) - 6 = -12,
- f’’(3) = 6(3) - 6 = 12.
Since f’’(-1) < 0 and f’’(3) > 0, we can conclude:
- f(-1) is a local maximum, and
- f(3) is a local minimum.

Slide 13: Second Derivative Test - Example 2

Let’s consider the function f(x) = x^4 + 4x^2 + 5.
The first step is to find the first derivative: f’(x) = 4x^3 + 8x.
Setting f’(x) = 0, we get 4x^3 + 8x = 0.
Factoring out 4x, we have 4x(x^2 + 2) = 0.
Solving for x, we find x = 0 as the only critical point.

Slide 14: Second Derivative Test - Example 2 (contd.)

Now, let’s find the second derivative: f’’(x) = 12x^2 + 8.
Evaluating f’’(x) at the critical point:
- f’’(0) = 12(0)^2 + 8 = 8.
Since f’’(0) > 0, we can conclude that f(0) is a local minimum.

Slide 15: Second Derivative Test - Example 3

Consider the function f(x) = x^2 + 3x + 2.
Find the first derivative: f’(x) = 2x + 3.
Setting f’(x) = 0, we get 2x + 3 = 0.
Solving for x, we find x = -3/2 as the only critical point.

Slide 16: Second Derivative Test - Example 3 (contd.)

Now, find the second derivative: f’’(x) = 2.
Evaluating f’’(x) at the critical point:
- f’’(-3/2) = 2.
Since f’’(-3/2) > 0, we can conclude that f(-3/2) is a local minimum.

Slide 17: Second Derivative Test - Equations

To summarize the second derivative test:
- If f’(c) = 0 and f’’(c) < 0, then f(c) is a local maximum.
- If f’(c) = 0 and f’’(c) > 0, then f(c) is a local minimum.
- If f’’(c) = 0, the test is inconclusive and may be a point of inflection.
Note: The second derivative test only applies to twice differentiable functions.

Slide 18: Second Derivative Test - Geometric Interpretation

The second derivative test can provide geometric information about the behavior of a function.
A local minimum occurs when the graph of a function changes concavity from concave up (positive second derivative) to concave down (negative second derivative).
A local maximum occurs when the graph changes concavity from concave down (negative second derivative) to concave up (positive second derivative).
The point of inflection occurs where the second derivative changes sign.

Slide 19: Second Derivative Test - Graphical Representation

Let’s consider a graphical representation of the second derivative test.
The red graph represents the original function f(x).
The blue graph represents the first derivative f’(x).
The green graph represents the second derivative f’’(x).
The critical points (local maximum and local minimum) are shown where f’(x) = 0 and f’’(x) changes sign.

Slide 20: Conclusion

The second derivative test is a powerful tool to determine the nature of critical points on a function.
It helps us classify critical points as local maximum, local minimum, or points of inflection.
Understanding the proof and examples of the second derivative test will enable you to apply it to various real-world applications and problem-solving situations.

Slide 21: Example of Second Derivative Test

Let’s consider the function f(x) = x^3 - 3x^2 - 9x + 5.
First, find the critical points by setting f’(x) = 0.
Solving 3x^2 - 6x - 9 = 0, we get x = -1 and x = 3.
Next, find the second derivative f’’(x) = 6x - 6.
Evaluate f’’(x) at the critical points:
- f’’(-1) = -12
- f’’(3) = 12
Based on the second derivative test, f(-1) is a local maximum and f(3) is a local minimum.

Slide 22: Example of Second Derivative Test

Let’s consider the function f(x) = x^4 + 4x^2 + 5.
Find the critical points by setting f’(x) = 0.
Solving 4x^3 + 8x = 0, we find x = 0.
Compute the second derivative f’’(x) = 12x^2 + 8.
Evaluate f’’(x) at the critical point:
- f’’(0) = 8
According to the second derivative test, f(0) is a local minimum.

Slide 23: Example of Second Derivative Test

Consider the function f(x) = x^2 + 3x + 2.
Find the critical points by setting f’(x) = 0.
Solving 2x + 3 = 0, we find x = -3/2.
Compute the second derivative f’’(x) = 2.
Evaluate f’’(x) at the critical point:
- f’’(-3/2) = 2
Applying the second derivative test, f(-3/2) is a local minimum.

Slide 24: Second Derivative Test and Inflection Points

It is important to note that the second derivative test cannot determine if a critical point is a point of inflection.
An inflection point is a point on a function’s graph where the concavity changes.
An inflection point occurs when the second derivative changes sign at that point.
To find inflection points, look for sign changes in the second derivative.
For example, if f’’(x) changes from positive to negative or negative to positive, there is an inflection point.

Slide 25: Example of Inflection Point

Consider the function f(x) = x^3.
Find the second derivative: f’’(x) = 6x.
Solve f’’(x) = 0, we get x = 0.
The graph of f(x) changes concavity at x = 0, indicating an inflection point.
At x < 0, the graph is concave down (f’’(x) < 0), and at x > 0, the graph is concave up (f’’(x) > 0).

Slide 26: Second Derivative Test - Summary

The second derivative test helps determine the nature of critical points on a function.
If f’(c) = 0 and f’’(c) < 0, then f(c) is a local maximum.
If f’(c) = 0 and f’’(c) > 0, then f(c) is a local minimum.
If f’’(c) = 0, the test is inconclusive and may indicate a point of inflection.
The second derivative test can be used to analyze functions in calculus and optimization problems.

Slide 27: Applications of Second Derivative Test

The second derivative test has practical applications in various fields.
It is applied in optimization problems to maximize or minimize a function.
The test is used in economics for determining the price and quantity that would maximize profit.
It has applications in physics to analyze the motion of objects and determine maximum or minimum values.
Engineers utilize the second derivative test to optimize systems and make them more efficient.

Slide 28: Importance of Second Derivative Test

The second derivative test is significant in understanding the behavior of functions at critical points.
It helps us classify critical points as maximum, minimum, or points of inflection.
The test provides valuable information about the shape and behavior of a function at specific points.
By analyzing the concavity and curvature of a function, we can make informed decisions in various real-world applications.
It is essential to understand the second derivative test to solve problems involving optimization, rate of change, and curvature.

Slide 29: Recap

The second derivative test is used to classify the nature of critical points on a function.
It helps determine local maximums and local minimums.
The second derivative test can be proven using Taylor series expansion.
It has practical applications in various fields, including optimization and physics.
Understanding the second derivative test is crucial for solving calculus problems and interpreting the behavior of functions.

Slide 30: Thank You!

Thank you for attending this lecture on the second derivative test.
Understanding the second derivative