Concavity and Inflection Points
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.5: “Derivatives and the Shape of a Graph” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-5-derivatives-and-the-shape-of-a-graph |
| Textbook used in class | Stewart, Calculus, Section 3.3: “What Derivatives Tell Us about the Shape of a Graph” |
Opening Scenario
A car is accelerating: its speed is increasing, so the distance function curves upward -- the graph bends away from the road in the shape of a bowl opening upward. Then the driver hits the brakes: speed is decreasing, and the distance function curves downward -- the graph bends like an arch. Concavity measures this bending direction.
Two curves can both be increasing, yet one bends upward (accelerating) and the other bends downward (decelerating). The first derivative tells you direction; the second derivative tells you the bending.
Quick Reference
| $f''(x)$ on interval | Meaning for $f'$ | Meaning for $f$ |
|---|---|---|
| $f''(x) > 0$ | $f'$ is increasing | $f$ is concave up (bowl opening up) |
| $f''(x) < 0$ | $f'$ is decreasing | $f$ is concave down (arch opening down) |
Inflection point. A point $(c, f(c))$ where $f$ is continuous and the concavity changes direction.
Candidates for inflection points: values $c$ where $f''(c) = 0$ or $f''(c)$ does not exist. But a sign change of $f''$ at $c$ is required for an actual inflection point.
Key Concepts
1. What Concavity Means Geometrically
Concave up on an interval means the graph lies above all of its tangent lines on that interval. Equivalently, $f'$ is increasing: the slopes are getting larger as you move right.
Concave down on an interval means the graph lies below all of its tangent lines on that interval. Equivalently, $f'$ is decreasing: the slopes are getting smaller (less positive, more negative) as you move right.
Picture the parabola $y = x^2$: it opens upward (concave up) everywhere. Its tangent at $x = 0$ has slope $0$, its tangent at $x = 1$ has slope $2$, its tangent at $x = 2$ has slope $4$. The slopes keep increasing. On the other side, $y = -x^2$ opens downward (concave down): slopes decrease from $0$ to $-2$ to $-4$.
2. The Concavity Test
Since $f'' = (f')'$, the sign of $f''$ tells you whether $f'$ is increasing or decreasing, which is the same as whether $f$ is concave up or down.
Concavity Test:
- If $f''(x) > 0$ on an interval, then $f$ is concave up there.
- If $f''(x) < 0$ on an interval, then $f$ is concave down there.
Example 1. Find the intervals of concavity for $f(x) = x^4 - 4x^3$.
$f'(x) = 4x^3 - 12x^2$
$f''(x) = 12x^2 - 24x = 12x(x - 2)$
Setting $f''(x) = 0$: $x = 0$ or $x = 2$.
| Interval | Sign of $f''$ | Concavity |
|---|---|---|
| $x < 0$ | $12(-)(-) = +$ | Concave up |
| $0 < x < 2$ | $12(+)(-) = -$ | Concave down |
| $x > 2$ | $12(+)(+) = +$ | Concave up |
Boxed answer: Concave up on $(-\infty, 0)$ and $(2, \infty)$; concave down on $(0, 2)$.
3. Inflection Points
An inflection point is where the graph switches from concave up to concave down (or vice versa). This requires the concavity to actually change, not just that $f''$ is zero.
From Example 1:
- At $x = 0$: $f''$ changes from $+$ to $-$. Inflection point. $f(0) = 0$. Point: $(0, 0)$.
- At $x = 2$: $f''$ changes from $-$ to $+$. Inflection point. $f(2) = 16 - 32 = -16$. Point: $(2, -16)$.
Boxed answer: Inflection points at $(0, 0)$ and $(2, -16)$.
“$f''(c) = 0$ guarantees an inflection point at $c$.” No. The function $f(x) = x^4$ has $f''(x) = 12x^2$, and $f''(0) = 0$. But $f''(x) \geq 0$ for all $x$, so the concavity never changes: $f$ is concave up everywhere. There is no inflection point at $x = 0$ even though $f''(0) = 0$.
4. An Inflection Point Where $f''$ Does Not Exist
An inflection point can occur where $f''$ does not exist, just as a local extremum can occur where $f'$ does not exist.
Example 2. Find any inflection points of $f(x) = x^{1/3}$.
$f'(x) = \dfrac{1}{3} x^{-2/3}$, $f''(x) = \dfrac{-2}{9} x^{-5/3}$.
$f''(x)$ is undefined at $x = 0$.
For $x < 0$: $x^{-5/3} = 1/x^{5/3}$. Since $x < 0$, $x^{5/3} < 0$ (odd root and odd power preserve sign), so $f''(x) > 0$ (negative divided by negative). Concave up.
For $x > 0$: $x^{-5/3} > 0$, so $f''(x) < 0$. Concave down.
$f''$ changes sign at $x = 0$, so there is an inflection point at $x = 0$.
$f(0) = 0$. Inflection point: $(0, 0)$.
5. Four Curve Shapes
Combining the sign of $f'$ and $f''$ gives four basic shapes:
| $f'$ | $f''$ | Shape |
|---|---|---|
| $+$ | $+$ | Increasing and concave up (like a right half of a bowl) |
| $+$ | $-$ | Increasing and concave down (like a right side of an arch) |
| $-$ | $+$ | Decreasing and concave up (like a left side of a bowl) |
| $-$ | $-$ | Decreasing and concave down (like a left half of an arch) |
These four shapes are the building blocks for every curve sketch.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Concluding $f''(c)=0$ gives an inflection point | $f(x)=x^4$, $f''(0)=0$ | Check that $f''$ actually changes sign at $c$; here it does not |
| Confusing concavity with direction of increase | “Concave up means the function is going up” | Concavity is about bending, not direction: a decreasing function can be concave up (like the right half of $1/x$ for $x > 0$) |
| Forgetting points where $f''$ does not exist | Only solving $f''(x) = 0$ | Also check where $f''$ fails to exist as candidates for inflection points |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Find the intervals on which $f(x) = x^3 - 6x$ is concave up and concave down. Find any inflection points.
Show answer
$f'(x) = 3x^2 - 6$, $f''(x) = 6x$.
$f''(x) = 0$ at $x = 0$.
For $x < 0$: $f''(x) < 0$ (concave down). For $x > 0$: $f''(x) > 0$ (concave up).
Sign changes at $x = 0$: inflection point. $f(0) = 0$.
Boxed answer: Concave down on $(-\infty, 0)$; concave up on $(0, \infty)$; inflection point at $(0, 0)$.
Problem 2. Show that $f(x) = x^4$ has no inflection point even though $f''(0) = 0$.
Show answer
$f''(x) = 12x^2 \geq 0$ for all $x$. The second derivative is zero at $x = 0$ but non-negative everywhere, so it does not change sign. The concavity is always “up” (or flat at $x=0$ but immediately returning to up). There is no change in concavity, so no inflection point.
Boxed answer: $f''(0) = 0$ but $f''$ does not change sign, so there is no inflection point at $x = 0$.
Level 2 -- Multiple Steps
Problem 3. Find all inflection points of $f(x) = x^5 - 5x^4$.
Show answer
$f'(x) = 5x^4 - 20x^3$, $f''(x) = 20x^3 - 60x^2 = 20x^2(x - 3)$.
$f''(x) = 0$ at $x = 0$ and $x = 3$.
Sign chart for $f''$:
- $x < 0$: $20(+)(-) = -$. Concave down.
- $0 < x < 3$: $20(+)(-) = -$. Concave down.
- $x > 3$: $20(+)(+) = +$. Concave up.
At $x = 0$: $f''$ goes from $-$ to $-$. No sign change. No inflection point.
At $x = 3$: $f''$ goes from $-$ to $+$. Sign change. Inflection point. $f(3) = 243 - 405 = -162$.
Boxed answer: One inflection point at $(3, -162)$. There is no inflection point at $x = 0$ despite $f''(0) = 0$.
Level 3 -- Deeper Problems
Problem 4. A student says: “The graph of $f$ is concave down, so $f$ must be decreasing.” Find a specific function and interval that shows this is wrong.
Show answer
Let $f(x) = -x^2 + 4$ on $(0, 1)$.
$f'(x) = -2x < 0$ on $(0, 1)$: decreasing. But wait, we need a function that is concave down and increasing.
Let $f(x) = \sqrt{x}$ on $(0, \infty)$.
$f'(x) = \dfrac{1}{2\sqrt{x}} > 0$: increasing. $f''(x) = -\dfrac{1}{4x^{3/2}} < 0$: concave down.
So $\sqrt{x}$ is increasing and concave down simultaneously. The student’s claim is false.
Boxed answer: $f(x) = \sqrt{x}$ on $(0, \infty)$ is increasing (since $f' > 0$) and concave down (since $f'' < 0$). Concave down does not require decreasing.
Mastery Checklist
Mental Model
Think of concavity as the curvature of a road. Concave up is like the bottom of a valley: the road curves upward, holds water in its center, and the tangent lines are below the road surface. Concave down is like the top of a hill: the road curves downward, and the tangent lines are above the road surface.
The second derivative measures the rate of change of the slope. If slopes are getting steeper (increasing), the road is bending upward (concave up). If slopes are getting gentler (decreasing), the road is bending downward (concave down).
An inflection point is where the road switches from valley-shaped to hill-shaped or back. The sign of $f''$ must actually change; a zero of $f''$ is only a candidate, not a confirmation.
Connections
Within Chapter 3
- First Derivative Test (Section 3.3): The first and second derivatives give two different types of information. The first tells you direction; the second tells you bending. Together they describe the full shape.
- Second Derivative Test (Section 3.3): Uses concavity to classify critical numbers without a sign chart for $f'$.
- Curve sketching (Section 3.5): Inflection points are the landmarks where the sketch changes curvature. Plot them alongside the local extrema.
Back to Applications of Differentiation | Previous: First Derivative Test | Next: Second Derivative Test