🚀 Level 3 - Topic 3: Concavity, Inflection Points, & 2nd Derivative Test 🧮

1) Introduction: Refining Curve Analysis with the Second Derivative

In the previous topic, we used the first derivative to understand where functions increase or decrease and to find local extrema. Now, we'll explore the power of the second derivative to further analyze the shape of curves, specifically focusing on concavity and inflection points, and introducing the Second Derivative Test for finding local extrema.

This topic will delve into:

Concavity Revisited: A deeper look at concave up and concave down, and how the second derivative determines concavity.
Inflection Points: Identifying inflection points as locations where concavity changes.
The Second Derivative Test: Using the second derivative to classify critical points as local maxima or minima (or neither).
Comparison of First and Second Derivative Tests: Understanding when to use each test.

These tools provide a more complete picture of the behavior and shape of function graphs.

2) Concavity and the Second Derivative

We've briefly touched upon concavity. Let's solidify our understanding and formally link it to the second derivative.

Definition 7.1: Concavity (Formal Definition)

A function \( f \) is concave up on an interval \( I \) if \( f' \) is an increasing function on \( I \).
A function \( f \) is concave down on an interval \( I \) if \( f' \) is a decreasing function on \( I \).

This definition relates concavity to the behavior of the first derivative \( f' \), which is the slope function. If the slope is increasing, the curve is turning upwards (concave up). If the slope is decreasing, the curve is turning downwards (concave down).

Theorem 7.1: Concavity Test using the Second Derivative

Let \( f \) be twice differentiable on an interval \( I \).

If \( f''(x) > 0 \) for all \( x \) in \( I \), then the graph of \( f \) is concave up on \( I \).
If \( f''(x) < 0 \) for all \( x \) in \( I \), then the graph of \( f \) is concave down on \( I \).
If \( f''(x) = 0 \) on \( I \), then \( f \) is linear on \( I \) (not strictly concave up or down, but considered to have zero concavity).

Intuition: The second derivative \( f''(x) = (f')'(x) \) is the derivative of the first derivative. If \( f''(x) > 0 \), it means \( f'(x) \) is increasing, so concavity is up. If \( f''(x) < 0 \), it means \( f'(x) \) is decreasing, so concavity is down.

Examples: Determining Concavity Intervals

Example 1: \( f(x) = x^4 - 6x^2 + 8x + 5 \)

Find the second derivative \( f''(x) \).
First derivative: \( f'(x) = 4x^3 - 12x + 8 \).

Second derivative: \( f''(x) = 12x^2 - 12 \).
Find where \( f''(x) = 0 \) or is undefined.
\( f''(x) = 12x^2 - 12 = 0 \Rightarrow 12x^2 = 12 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 \). \( f''(x) \) is defined for all \( x \).

Create a sign chart for \( f''(x) \) using \( x = -1 \) and \( x = 1 \) to divide intervals.

Intervals: \( (-\infty, -1), (-1, 1), (1, \infty) \).

Interval	Test Value	\( f''(x) = 12(x^2 - 1) \) sign	Concavity
\( (-\infty, -1) \)	\( x = -2 \)	\( f''(-2) = 12((-2)^2 - 1) = 12(3) = + \)	Concave Up (CU)
\( (-1, 1) \)	\( x = 0 \)	\( f''(0) = 12(0^2 - 1) = 12(-1) = - \)	Concave Down (CD)
\( (1, \infty) \)	\( x = 2 \)	\( f''(2) = 12((2)^2 - 1) = 12(3) = + \)	Concave Up (CU)

State the intervals of concavity.
\( f(x) \) is concave up on \( (-\infty, -1) \) and \( (1, \infty) \). \( f(x) \) is concave down on \( (-1, 1) \).

3) Inflection Points: Points of Concavity Change

As we saw, concavity can change. Points where this change occurs are called inflection points.

Definition 7.2: Inflection Point

An inflection point of a function \( f \) is a point \( (c, f(c)) \) on its graph where the concavity changes (from concave up to concave down, or from concave down to concave up).

Finding Inflection Points - Procedure

Find the second derivative \( f''(x) \).
Find potential inflection points by solving \( f''(x) = 0 \) or finding where \( f''(x) \) is undefined. These values of \( x \) are candidates for where concavity might change.
Check for concavity change at each candidate \( x = c \). Use a sign chart for \( f''(x) \) around \( x = c \). If the sign of \( f''(x) \) changes at \( x = c \), then \( (c, f(c)) \) is an inflection point. If the sign does not change, there is no inflection point at \( x = c \).

Examples: Finding Inflection Points

Example 1 (revisited): \( f(x) = x^4 - 6x^2 + 8x + 5 \)
From concavity analysis, \( f''(x) = 12x^2 - 12 \), and \( f''(x) = 0 \) at \( x = -1 \) and \( x = 1 \). Concavity changes at both \( x = -1 \) and \( x = 1 \).
- At \( x = -1 \): Concavity changes from Up to Down. Inflection point at \( x = -1 \). \( f(-1) = (-1)^4 - 6(-1)^2 + 8(-1) + 5 = 1 - 6 - 8 + 5 = -8 \). Inflection point is \( (-1, -8) \).
- At \( x = 1 \): Concavity changes from Down to Up. Inflection point at \( x = 1 \). \( f(1) = (1)^4 - 6(1)^2 + 8(1) + 5 = 1 - 6 + 8 + 5 = 8 \). Inflection point is \( (1, 8) \).
Inflection points are \( (-1, -8) \) and \( (1, 8) \).

Example 2: \( g(x) = x^5 \)

Find \( g''(x) \). \( g'(x) = 5x^4 \), \( g''(x) = 20x^3 \).
Solve \( g''(x) = 0 \). \( 20x^3 = 0 \Rightarrow x = 0 \).

Check for concavity change at \( x = 0 \). Sign chart for \( g''(x) = 20x^3 \).

Interval	Test Value	\( g''(x) = 20x^3 \) sign	Concavity
\( (-\infty, 0) \)	\( x = -1 \)	\( g''(-1) = 20(-1)^3 = - \)	Concave Down (CD)
\( (0, \infty) \)	\( x = 1 \)	\( g''(1) = 20(1)^3 = + \)	Concave Up (CU)

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Concavity changes at \( x = 0 \). So, \( (0, g(0)) = (0, 0^5) = (0, 0) \) is an inflection point.

Example 3: \( h(x) = x^4 \)

Find \( h''(x) \). \( h'(x) = 4x^3 \), \( h''(x) = 12x^2 \).
Solve \( h''(x) = 0 \). \( 12x^2 = 0 \Rightarrow x = 0 \).

Check for concavity change at \( x = 0 \). Sign chart for \( h''(x) = 12x^2 \).

Interval	Test Value	\( g''(x) = 20x^3 \) sign	Concavity
\( (-\infty, 0) \)	\( x = -1 \)	\( g''(-1) = 20(-1)^3 = - \)	Concave Down (CD)
\( (0, \infty) \)	\( x = 1 \)	\( g''(1) = 20(1)^3 = + \)	Concave Up (CU)

Concavity does not change at \( x = 0 \) (it's concave up on both sides). Therefore, \( (0, 0) \) is not an inflection point for \( h(x) = x^4 \), even though \( h''(0) = 0 \).

4) The Second Derivative Test for Local Extrema

The Second Derivative Test provides an alternative method to classify critical points as local maxima or minima using the second derivative. It's often easier than the First Derivative Test, but it doesn't always apply.

Second Derivative Test for Local Extrema

Suppose \( f \) is twice differentiable and \( c \) is a critical number of \( f \) such that \( f'(c) = 0 \).

If \( f''(c) > 0 \), then \( f \) has a local minimum at \( c \). (Think: concave up at \( c \), so a valley bottom).
If \( f''(c) < 0 \), then \( f \) has a local maximum at \( c \). (Think: concave down at \( c \), so a peak top).
If \( f''(c) = 0 \), the Second Derivative Test is inconclusive. We cannot conclude anything about local extrema from this test alone. In this case, we must use the First Derivative Test or other methods.

Important: The Second Derivative Test only applies when \( f'(c) = 0 \). It cannot be used when \( f'(c) \) is undefined. Also, if \( f''(c) = 0 \), it gives no information.

Examples: Using the Second Derivative Test

Example 1: \( f(x) = x^3 - 3x^2 - 9x + 4 \) (revisited)
We found \( f'(x) = 3x^2 - 6x - 9 \) and critical numbers \( x = -1, 3 \). Find the second derivative: \( f''(x) = 6x - 6 \).
- For \( x = -1 \): \( f''(-1) = 6(-1) - 6 = -12 < 0 \). Since \( f''(-1) < 0 \), by the Second Derivative Test, \( f \) has a local maximum at \( x = -1 \).
- For \( x = 3 \): \( f''(3) = 6(3) - 6 = 18 - 6 = 12 > 0 \). Since \( f''(3) > 0 \), by the Second Derivative Test, \( f \) has a local minimum at \( x = 3 \).
These conclusions match what we found using the First Derivative Test earlier. Note the Second Derivative Test quickly classified these critical points.
Example 2: \( h(x) = x^4 \) (revisited)
We found \( h'(x) = 4x^3 \) and critical number \( x = 0 \). Second derivative \( h''(x) = 12x^2 \).

At \( x = 0 \): \( h''(0) = 12(0)^2 = 0 \). The Second Derivative Test is inconclusive in this case. It tells us nothing. We had to use the First Derivative Test to find that \( x = 0 \) is a local minimum (from our earlier analysis).
Example 3: \( k(x) = x^3 \) (revisited)
We found \( k'(x) = 3x^2 \) and critical number \( x = 0 \). Second derivative \( k''(x) = 6x \).

At \( x = 0 \): \( k''(0) = 6(0) = 0 \). The Second Derivative Test is inconclusive again. We used the First Derivative Test to find that \( x = 0 \) is not a local extremum for \( k(x) \) (neither max nor min).

5) Comparison: First Derivative Test vs. Second Derivative Test

Both the First and Second Derivative Tests are used to find local extrema, but they have different strengths and weaknesses.

Comparison Table

Feature	First Derivative Test	Second Derivative Test
What it uses	Sign change of \( f'(x) \) around critical number.	Sign of \( f''(c) \) at critical number \( c \) where \( f'(c) = 0 \).
Always applicable?	Yes, always works for continuous functions at critical numbers.	No. Inconclusive if \( f''(c) = 0 \). Only applies if \( f'(c) = 0 \) and \( f''(c) \neq 0 \).
Procedure	Examine sign of \( f'(x) \) on intervals around critical number using sign chart.	Evaluate \( f''(c) \) at each critical number \( c \) where \( f'(c) = 0 \).
Conclusion - Local Max	\( f'(x) \) changes from + to - at \( c \).	\( f''(c) < 0 \).
Conclusion - Local Min	\( f'(x) \) changes from - to + at \( c \).	\( f''(c) > 0 \).
Conclusion - No Local Extremum	\( f'(x) \) does not change sign at \( c \).	\( f''(c) = 0 \) (test is inconclusive).
When to prefer	When \( f''(x) \) is difficult to calculate, or when \( f''(c) = 0 \), or when \( f'(c) \) is undefined at critical points. Always reliable.	When \( f''(x) \) is easy to compute and \( f''(c) \neq 0 \) for critical numbers \( c \) where \( f'(c) = 0 \). Often quicker when it applies.

In practice, if the second derivative is easy to find, and \( f''(c) \neq 0 \) for your critical numbers, the Second Derivative Test can be a faster way to classify local extrema. However, the First Derivative Test is more broadly applicable and always works.

6) Practice Questions - Concavity, Inflection Points, and Second Derivative Test

Practice using the second derivative to analyze concavity, find inflection points, and apply the Second Derivative Test.

Fundamental Practice Questions

Instructions: For each function, find the intervals of concavity and inflection points. Where applicable, use the Second Derivative Test to find local extrema. If the Second Derivative Test is inconclusive, state so.

Q1. \( f(x) = x^3 - 6x^2 + 9x + 1 \)
Q2. \( g(x) = x^4 - 4x^3 \)
Q3. \( h(x) = 3x^5 - 5x^4 + 2 \)
Q4. \( y = \frac{1}{x^2 + 3} \)
Q5. \( f(x) = x\sqrt{x + 3} \)
Q6. \( g(x) = \sin(x) - \cos(x) \) on \( [0, 2\pi] \)
Q7. \( h(x) = x e^{x} \)
Q8. \( y = x + \frac{1}{x} \)
Q9. \( f(x) = (x^2 - 1)^3 \)
Q10. \( g(x) = \arctan(x) \) (Second derivative of \( \arctan(x) \) is \( \frac{-2x}{(1+x^2)^2} \))

Challenging Practice Questions

Instructions: These problems may involve more complex functions, require combining concavity/extrema analysis, or conceptual understanding.

Q1. Find the intervals where \( f(x) = x + 2\sin(x) \) is concave up and concave down on \( [0, 2\pi] \), and identify inflection points.
Q2. For \( g(x) = \frac{x^2}{x - 2} \), find all local extrema using the Second Derivative Test where possible. For any critical points where the test is inconclusive, use the First Derivative Test.
Q3. Determine the concavity of \( y = \ln(x^2 + 1) \) for all \( x \).
Q4. Find the inflection points of \( f(x) = e^{-x^2} \).
Q5. Conceptual: Explain why the Second Derivative Test works in terms of concavity and the behavior of the first derivative. What does it mean graphically for \( f''(c) > 0 \) and \( f''(c) < 0 \) when \( f'(c) = 0 \)?

7) Summary & Cheat Sheet - Concavity, Inflection Points, & Second Derivative Test

Key takeaways and a quick reference for curve shape analysis.

7.1) Concavity and Second Derivative

\( f''(x) > 0 \) => Concave Up
\( f''(x) < 0 \) => Concave Down

7.2) Inflection Points

Occur where concavity changes. Find potential points by setting \( f''(x) = 0 \) or where \( f''(x) \) is undefined, and check for concavity change.

7.3) Second Derivative Test for Local Extrema (when \( f'(c) = 0 \))

If \( f''(c) > 0 \), Local Minimum at \( c \).
If \( f''(c) < 0 \), Local Maximum at \( c \).
If \( f''(c) = 0 \), Test is Inconclusive. Use First Derivative Test.

7.4) Comparison: Tests

First Derivative Test is always reliable. Second Derivative Test is quicker when applicable (and \( f''(c) \neq 0 \)), but not always conclusive.