Source: MTH 132 Chapter 3 Lecture Notes, Michigan State University
Difficulty: Intermediate | Prerequisites: Derivative rules, critical numbers (Ch. 3 Sec. 1), inequalities.
This section is where you learn to read a function's behaviour from its derivatives. The first derivative tells you where the function is climbing or falling. The second derivative tells you how the curve bends. Together, they let you sketch a reasonably accurate graph without plotting dozens of points, classify critical numbers as local maxima or minima, and locate inflection points where the curve changes its bending direction. Solving inequalities is a prerequisite skill here, since you will constantly test where derivatives are positive or negative.
Where f'(x) > 0, the function is increasing; where f'(x) < 0, it is decreasing. Where f''(x) > 0, the curve bends concave up; where f''(x) < 0, concave down. Inflection points are where the concavity switches. The First Derivative Test uses the sign change of f' at a critical number to classify it as a local max, local min, or neither.
Increasing function
f is increasing on (a, b) if f'(x) > 0 for all x in (a, b). In simple terms, the curve is going uphill as you move left to right.
Decreasing function
f is decreasing on (a, b) if f'(x) < 0 for all x in (a, b). The curve is going downhill.
Concave up
The graph of f lies above all of its tangent lines on an interval I. Equivalently, f''(x) > 0 on I. Think of it as a bowl shape, or a curve that "holds water."
Concave down
The graph of f lies below all of its tangent lines on an interval I. Equivalently, f''(x) < 0 on I. Think of it as an upside-down bowl.
Inflection point
A point on the curve where the concavity changes (from up to down or from down to up). The function must be continuous there, and the second derivative is either zero or undefined at that point.
First Derivative Test
A method for classifying a critical number c of f as a local max, local min, or neither, based on whether f' changes sign at c.
Sign chart
A number-line diagram showing the sign (+ or -) of a derivative on each interval between critical numbers. Used to determine increasing/decreasing behaviour and to apply the First Derivative Test.
You will solve inequalities constantly in this section (to determine where f' or f'' is positive or negative). The standard method:
Move everything to one side so you have an expression > 0, < 0, >= 0, or <= 0.
Factor the expression (or find where the numerator and denominator are zero for rational expressions).
Mark the zeros on a number line. These divide the line into intervals.
Test a point in each interval to determine the sign of the expression there.
Write the solution in interval notation.
Example 1: Solve x^2 - 2x - 3 > 0.
Factor: (x - 3)(x + 1) = 0 at x = 3 and x = -1.
Test x = -10: (-13)(-9) > 0. Test x = 0: (-3)(1) < 0. Test x = 4: (1)(5) > 0.
Solution: (-infinity, -1) union (3, infinity).
Example 2: Solve (x + 5)/(x - 7) < 0.
Zeros of numerator and denominator: x = -5, x = 7 (excluded from domain).
Test x = -6: (-1)/(-13) > 0. Test x = 0: (5)/(-7) < 0. Test x = 8: (13)/(1) > 0.
Solution: (-5, 7).
Example 3: Solve 1 - 3/(x - 2) > 0.
Rewrite: (x - 2 - 3)/(x - 2) > 0, so (x - 5)/(x - 2) > 0.
Zeros: x = 5, x = 2 (excluded).
Test intervals: solution is (-infinity, 2) union (5, infinity).
If f'(x) > 0 on an interval, then f is increasing on that interval.
If f'(x) < 0 on an interval, then f is decreasing on that interval.
To apply this, find the critical numbers (where f' = 0 or DNE), then test the sign of f' in each interval between critical numbers using a test point.
Example: f(x) = x^2 + 4x + 5. f'(x) = 2x + 4. Setting 2x + 4 > 0 gives x > -2. So f is increasing on (-2, infinity) and decreasing on (-infinity, -2).
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.
Concave up: the curve bends upward, the slopes are increasing. Concave down: the curve bends downward, the slopes are decreasing.
Example: f(x) = x^2 + 4x + 5. f'(x) = 2x + 4. f''(x) = 2. Since 2 > 0 for all x, the function is concave up everywhere on (-infinity, infinity).
An inflection point is where the curve changes concavity. To find inflection points:
Compute f''(x).
Find where f''(x) = 0 or f''(x) is undefined.
Check that the concavity actually changes sign at that point. If f'' does not change sign, there is no inflection point.
Example: f(x) = x^3 - x^2 - 5x + 3. f''(x) = 6x - 2. Setting 6x - 2 = 0 gives x = 1/3. Testing: f''(0) = -2 < 0 (concave down) and f''(1) = 4 > 0 (concave up). Concavity changes at x = 1/3, so there is an inflection point there.
Example with no inflection point at a zero of f'': f''(x) = (x + 1)^2(x + 5). f'' = 0 at x = -1 and x = -5. At x = -1, f'' does not change sign (it is negative on both sides), so x = -1 is not an inflection point. At x = -5, f'' changes from negative to positive, so x = -5 is an inflection point.
Suppose c is a critical number of f (so f'(c) = 0 or f'(c) DNE).
If f' changes from positive to negative at c, then f has a local maximum at c.
If f' changes from negative to positive at c, then f has a local minimum at c.
If f' does not change sign at c (positive on both sides, or negative on both sides), then f has neither a local max nor a local min at c.
Example 1: f(x) = 2x^3 + 3x^2 - 36x on (-infinity, infinity)
f'(x) = 6x^2 + 6x - 36 = 6(x + 3)(x - 2). Critical numbers: x = -3, x = 2.
Sign chart for f': test x = -4: f'(-4) = 6(1)(-6) > 0 (positive). Test x = 0: f'(0) = 6(3)(-2) < 0 (negative). Test x = 3: f'(3) = 6(6)(1) > 0 (positive).
f is increasing on (-infinity, -3) and (2, infinity). f is decreasing on (-3, 2).
Local max at x = -3 (f' changes + to -). Local min at x = 2 (f' changes - to +).
Example 2: f(x) = x/(x^2 + 1) on (-infinity, infinity)
f'(x) = (1 - x^2)/(x^2 + 1)^2. f'(x) = 0 when x^2 = 1, so x = 1 or x = -1.
Sign chart: test x = -2: f'(-2) = (1-4)/25 < 0. Test x = 0: f'(0) = 1 > 0. Test x = 2: f'(2) = (1-4)/25 < 0.
f is increasing on (-1, 1). f is decreasing on (-infinity, -1) and (1, infinity).
Local min at x = -1. Local max at x = 1.
Example 3: f(x) = x(x - 4)^3
f'(x) = (x - 4)^3 + 3x(x - 4)^2 = (x - 4)^2(4x - 4) = 4(x - 4)^2(x - 1). Critical numbers: x = 4, x = 1.
Sign chart: test x = 0: f'(0) = 4(16)(-1) < 0. Test x = 2: f'(2) = 4(4)(1) > 0. Test x = 5: f'(5) = 4(1)(4) > 0.
f is decreasing on (-infinity, 1). f is increasing on (1, 4) and (4, infinity).
Local min at x = 1. x = 4 is neither a max nor a min (f' does not change sign).
f''(x) = 2(x - 4)(6x - 12) = 12(x - 4)(x - 2). f'' = 0 at x = 4 and x = 2.
Concave up on (-infinity, 2) and (4, infinity). Concave down on (2, 4). Inflection points at x = 2 and x = 4.
Students assume that f''(c) = 0 automatically means there is an inflection point at c. You must verify the concavity actually changes sign. If f'' is zero at c but does not change sign, there is no inflection point.
Students forget to check where f' is undefined, not just where f' = 0. Both generate critical numbers.
A common error is stating "f is increasing because f' is positive at one test point" without establishing that f' is positive on the entire interval.
Students sometimes mix up concave up and concave down. Concave up = bowl shape (f'' > 0). Concave down = cap shape (f'' < 0).
Building sign charts for f' and f'' is the core skill tested in this section.
Expect problems asking: find all intervals of increase/decrease, classify critical points, find intervals of concavity, and identify inflection points.
The First Derivative Test is a favourite exam tool. Know all three cases (sign change + to -, sign change - to +, no sign change).
Curve sketching problems (Section 5) combine everything from this section into one multi-part question.
True or False: If f'(x) > 0 on (a, b), then f is increasing on (a, b).
Answer: True.
Fill in the blank: f is concave up on I when f''(x) ____ 0 for all x in I.
Answer: > (greater than).
True or False: If f''(c) = 0, then c is always an inflection point.
Answer: False. The concavity must actually change at c.
Fill in the blank: If f' changes from negative to positive at a critical number c, the First Derivative Test says f has a local ____ at c.
Answer: minimum.
Q: Find the intervals where f(x) = 2x^3 + 3x^2 - 36x is increasing and decreasing. Identify local extrema.
A: f'(x) = 6(x + 3)(x - 2). f' > 0 on (-infinity, -3) and (2, infinity), so f is increasing there. f' < 0 on (-3, 2), so f is decreasing. Local max at x = -3, local min at x = 2.
Q: Find where f(x) = x^3 - x^2 - 5x + 3 is concave up and concave down. Identify inflection points.
A: f''(x) = 6x - 2. f'' = 0 at x = 1/3. f'' < 0 for x < 1/3 (concave down). f'' > 0 for x > 1/3 (concave up). Inflection point at x = 1/3.
Q: Suppose f''(x) = (x + 1)^2(x + 5). Where is f concave up? Where is f concave down? Where are the inflection points?
A: f'' = 0 at x = -1 and x = -5. Test regions: f''(-6) = (25)(-1) < 0 (concave down). f''(-3) = (4)(2) > 0 (concave up). f''(0) = (1)(5) > 0 (concave up). Concave down on (-infinity, -5). Concave up on (-5, -1) and (-1, infinity). Inflection point only at x = -5 (concavity changes). x = -1 is not an inflection point.
Q: For f(x) = x/(x^2 + 1), find the intervals of increase and decrease, and classify the critical points.
A: f'(x) = (1 - x^2)/(x^2 + 1)^2. The denominator is always positive, so the sign of f' depends on the numerator: 1 - x^2 = (1 - x)(1 + x). f'(x) = 0 at x = 1 and x = -1. Sign chart: f' < 0 for x < -1, f' > 0 for -1 < x < 1, f' < 0 for x > 1. f is decreasing on (-infinity, -1), increasing on (-1, 1), decreasing on (1, infinity). Local min at x = -1, local max at x = 1.
Q: Solve the inequality x^2 - 2x - 3 > 0.
A: Factor: (x - 3)(x + 1) > 0. Zeros at x = -1 and x = 3. Test x = -2: (-5)(-1) = 5 > 0. Test x = 0: (-3)(1) = -3 < 0. Test x = 4: (1)(5) = 5 > 0. Solution: (-infinity, -1) union (3, infinity).
Q: For f(x) = x(x - 4)^3, find the intervals of concavity and all inflection points.
A: f''(x) = 12(x - 4)(x - 2). f'' = 0 at x = 4 and x = 2. Test x = 0: 12(-4)(-2) = 96 > 0 (concave up). Test x = 3: 12(-1)(1) = -12 < 0 (concave down). Test x = 5: 12(1)(3) = 36 > 0 (concave up). Concave up on (-infinity, 2) and (4, infinity). Concave down on (2, 4). Inflection points at x = 2 and x = 4.
This section connects directly to curve sketching (Section 5), where you put together domain, intercepts, asymptotes, increasing/decreasing, concavity, and inflection points to draw a complete graph.
The First Derivative Test is an alternative to the Second Derivative Test (which uses f'' to classify critical points). Both are valid; choose whichever is simpler for the problem at hand.
Concavity and inflection points reappear in optimisation (Section 7) when confirming whether a critical point is a maximum or minimum.
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