Curve Sketching with Derivatives
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 topographic map uses contour lines to give a complete picture of the terrain without a photograph. Similarly, a table of sign information for $f'$ and $f''$ gives a complete picture of a function’s graph without plotting individual points.
Reading that table fluently means this: given the sign of $f'$ and $f''$ on each piece of the domain, describe the shape of the graph, locate its peaks, valleys, and bends, and produce a sketch accurate enough to answer any calculus question about the function.
Quick Reference
| $f'$ | $f''$ | Shape of that piece |
|---|---|---|
| $+$ | $+$ | Rising, bending upward (like the right side of a valley) |
| $+$ | $-$ | Rising, bending downward (like the right side of a hill) |
| $-$ | $+$ | Falling, bending upward (like the left side of a valley) |
| $-$ | $-$ | Falling, bending downward (like the left side of a hill) |
Landmarks (transition points between pieces):
- $f'$ changes sign at $c$ $\Rightarrow$ local extremum at $c$.
- $f''$ changes sign at $c$ $\Rightarrow$ inflection point at $c$.
Key Concepts
1. Building the Combined Sign Chart
The first step is to find the sign-change points for both $f'$ and $f''$. These divide the domain into pieces. On each piece, the signs of $f'$ and $f''$ are constant, so the shape of that piece is one of the four shapes above.
Approach:
- Find all critical numbers ($f' = 0$ or $f'$ undefined) and all candidates for inflection points ($f'' = 0$ or $f''$ undefined).
- List all of these points together and order them on the number line. These are the dividing points.
- Determine the sign of $f'$ and $f''$ in each interval between dividers.
- Describe the shape of each interval and record the landmark at each dividing point.
2. A Complete Example
Example. Sketch a graph of $f(x) = x^4 - 4x^3$, identifying all local extrema and inflection points.
Step 1: Derivatives and critical points.
$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$.
$f'(x) = 0$ at $x = 0$ and $x = 3$.
$f''(x) = 12x^2 - 24x = 12x(x - 2)$.
$f''(x) = 0$ at $x = 0$ and $x = 2$.
Step 2: All division points: $x = 0, 2, 3$.
Step 3: Sign chart.
| Interval | $f'$ | $f''$ | Shape |
|---|---|---|---|
| $x < 0$ | $f'(-1) = 4(1)(-4) = -16 < 0$: $-$ | $f''(-1) = 12(1)(-3) < 0$... wait, $f''(-1) = 12(-1)(-1-2) = 12(-1)(-3) = 36 > 0$: $+$ | Falling, bending up |
| $0 < x < 2$ | $f'(1) = 4(1)(-2) = -8 < 0$: $-$ | $f''(1) = 12(1)(-1) = -12 < 0$: $-$ | Falling, bending down |
| $2 < x < 3$ | $f'(2.5) = 4(6.25)(-.5) < 0$: $-$ | $f''(2.5) = 12(2.5)(0.5) = 15 > 0$: $+$ | Falling, bending up |
| $x > 3$ | $f'(4) = 4(16)(1) > 0$: $+$ | $f''(4) = 12(4)(2) > 0$: $+$ | Rising, bending up |
Step 4: Classify landmarks.
At $x = 0$: $f'$ sign: $- \to -$ (no sign change). No local extremum. But $f''$ sign: $+ \to -$ (sign changes). Inflection point. $f(0) = 0$.
At $x = 2$: $f'$ sign: $- \to -$ (no change in $f'$). No local extremum. $f''$ sign: $- \to +$. Inflection point. $f(2) = 16 - 32 = -16$.
At $x = 3$: $f'$ sign: $- \to +$. Local minimum. $f''$ sign: $+ \to +$ (no change). No inflection. $f(3) = 81 - 108 = -27$.
Step 5: Key values for the sketch.
$f(0) = 0$, $f(2) = -16$, $f(3) = -27$.
As $x \to -\infty$: $f(x) \to +\infty$ (positive leading coefficient, even degree). As $x \to +\infty$: $f(x) \to +\infty$.
Boxed answer: Local minimum of $-27$ at $x = 3$. Inflection points at $(0, 0)$ and $(2, -16)$. No local maximum.
Recap. The function falls from upper-left, passes through the origin (an inflection point), continues falling while transitioning from bending-down to bending-up at $(2, -16)$ (second inflection point), reaches its lowest point $-27$ at $x = 3$, then rises to upper-right.
3. Reading Questions off the Sketch
Once you have the shape information, many questions can be answered by reading the sketch:
- Is $f$ increasing or decreasing at a given $x$? Check the sign of $f'$ in that interval.
- Does $f$ have a local max or min at a given critical number? Check for a sign change of $f'$.
- Is the graph concave up or down at a given $x$? Check the sign of $f''$.
- Does a given $y$-value appear on the graph? Use the Intermediate Value Theorem.
The sketch is a compressed record of all this information.
4. Multiple Representations
The information about $f$ can be presented in four equivalent ways:
- Formula: $f(x) = x^4 - 4x^3$.
- Table of signs: the chart from Step 3.
- Words: “falls steeply from upper left, bends upward near the origin, bends downward between $x=0$ and $x=2$, bends upward again before the minimum at $x=3$, then rises.”
- Sketch: the graph drawn from the table.
Each representation captures the same information in a different form. When a question gives you one representation, translating to another often reveals the answer quickly.
a positive function value means the graph is rising at that point.
This is the height-vs-slope error in curve-sketching context. $f(x) > 0$ says the graph is ABOVE the x-axis; $f'(x) > 0$ says the graph is RISING (slope is positive). These are independent. For $f(x) = x^2 - 1$: at $x = -2$, $f(-2) = 3 > 0$ (above the axis) and $f'(-2) = -4 < 0$ (falling). The function is positive but decreasing. When sketching, track $f$ values (for where the graph sits) and $f'$ sign (for which way it goes) as separate columns in your sign chart.
a complete sketch can be drawn just by plotting several points.
This is the iconic-graph error: trusting a visual picture without derivative information. Plotting $f$ at $x = -2, -1, 0, 1, 2$ produces dots, but connecting them in a smooth curve requires knowing where the function increases, decreases, is concave up, and is concave down. Without that information, the sketch may miss local extrema between sampled points, draw the wrong curvature, or misplace inflection points. The derivative analysis (sign charts for $f'$ and $f''$) provides the structural information; the plotted points then fill in the heights.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Confusing inflection points with extrema | Marking $x=0$ as a local min on $x^4-4x^3$ | $f'$ does not change sign at $x=0$; it is an inflection point, not an extremum |
| Omitting a division point | Forgetting $x=2$ in the example above | List all zeros of $f'$ and $f''$ before building the chart |
| Drawing the wrong curvature | Sketching concave down when $f''>0$ | $f''>0$ means concave up (bowl), not down (arch) |
| Confusing the $y$-value with the shape | Saying “the function is zero so it is flat” | $f(0)=0$ says the output value; $f'(0)=0$ says the slope is zero; they are different |
Leveled Practice
Level 1: Direct Application
Problem 1. Use the information below to match the shape on $(0, 1)$:
$f' > 0$ and $f'' < 0$ on $(0, 1)$.
Which of the following describes the graph on $(0, 1)$?
(A) Falling, bending upward. (B) Rising, bending downward. (C) Falling, bending downward. (D) Rising, bending upward.
Show answer
$f' > 0$ means rising. $f'' < 0$ means concave down (bending downward).
Boxed answer: (B) Rising, bending downward.
Problem 2. For $f(x) = x^3 - 3x$, find all critical numbers, classify them (First or Second Derivative Test), and find all inflection points.
Show answer
$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$. Critical numbers: $x = 1, -1$.
$f''(x) = 6x$.
$f''(-1) = -6 < 0$: local max. $f(-1) = -1 + 3 = 2$. $f''(1) = 6 > 0$: local min. $f(1) = 1 - 3 = -2$.
Inflection candidates: $f''(x) = 6x = 0$ at $x = 0$. Sign of $f''$ changes ($-$ for $x<0$, $+$ for $x>0$): inflection point at $(0, f(0)) = (0, 0)$.
Boxed answer: Local max of $2$ at $x=-1$; local min of $-2$ at $x=1$; inflection point at $(0, 0)$.
Level 2: Multiple Steps
Problem 3. A function $f$ is continuous and twice differentiable. A sign chart gives:
| Interval | $f'$ | $f''$ |
|---|---|---|
| $x < -2$ | $+$ | $-$ |
| $-2 < x < 0$ | $-$ | $-$ |
| $0 < x < 3$ | $-$ | $+$ |
| $x > 3$ | $+$ | $+$ |
Identify all local extrema and inflection points from the chart alone.
Show answer
Local extrema (where $f'$ changes sign):
- At $x = -2$: $f'$ goes $+ \to -$: local maximum.
- At $x = 3$: $f'$ goes $- \to +$: local minimum.
- At $x = 0$: $f'$ goes $- \to -$: no sign change, no extremum.
Inflection points (where $f''$ changes sign):
- At $x = -2$: $f''$ goes $- \to -$: no sign change, no inflection point.
- At $x = 0$: $f''$ goes $- \to +$: sign change. Inflection point at $x = 0$.
- At $x = 3$: $f''$ goes $+ \to +$: no sign change, no inflection point.
Boxed answer: Local max at $x=-2$; local min at $x=3$; inflection point at $x=0$.
Mastery Checklist
Mental Model
Think of the combined sign chart as a terrain map with two layers of information: direction (are you going uphill or downhill?) and curvature (is the slope getting steeper or gentler?). A single piece of the chart, one sign of $f'$ and one sign of $f''$, describes one basic section of terrain.
The transition points between pieces are where the terrain type changes. A sign change of $f'$ is a peak or valley (the terrain reverses slope). A sign change of $f''$ is where the slope shifts from steepening to easing (the terrain changes from valley-shape to hill-shape or back).
Reading the full chart from left to right gives the shape of the graph across the entire domain without ever computing a numerical value.
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