Sketching f' from f
Sketching $f'$ from $f$
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.2: “The Derivative as a Function” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First: Predict the Slope Graph
Imagine the graph of $f(x) = x^2$ (a parabola opening upward, vertex at the origin).
Before any calculation:
- Where on the parabola is the tangent horizontal? (Where is $f' = 0$?)
- To the left of the vertex, is $f$ increasing or decreasing? What sign should $f'$ have?
- To the right, does the slope increase or decrease as you move right?
Sketch what you think the graph of $f'(x)$ looks like based on these observations. Then compute $f'(x) = 2x$ and compare.
Quantity-First Framing
The graph of $f'$ is the graph of slopes. Each $x$-value on the $f'$ graph tells you the slope of $f$ at that $x$: how steeply the original curve is rising or falling there.
Reading $f'$ from the graph of $f$ is a translation exercise: convert each feature of the curve (rising, falling, flat, steep, gentle) into the corresponding value of $f'$ (positive, negative, zero, large, small).
Prerequisite Check
Quick Reference
Reading $f'$ from the graph of $f$:
| Feature of $f$ | Corresponding feature of $f'$ |
|---|---|
| $f$ is increasing (rising) | $f' > 0$ (above the axis) |
| $f$ is decreasing (falling) | $f' < 0$ (below the axis) |
| $f$ has a local max or min (horizontal tangent) | $f' = 0$ (crosses the axis) |
| $f$ is steep (rising or falling fast) | $|f'|$ is large |
| $f$ is gentle | $|f'|$ is small |
| $f$ has an inflection point (concavity changes) | $f'$ has a local max or min |
| $f$ is a straight line with slope $m$ | $f'$ is the constant $m$ |
Key Concepts
1. The Translation from $f$ to $f'$
Example 1. Describe the graph of $f'$ for $f(x) = x^3 - 3x$.
Before computing:
$f$ has a local maximum near $x = -1$ (curve rises then falls) and a local minimum near $x = 1$ (falls then rises). So $f' = 0$ at $x = -1$ and $x = 1$.
For $x < -1$: $f$ is rising, so $f' > 0$.
For $-1 < x < 1$: $f$ is falling, so $f' < 0$.
For $x > 1$: $f$ is rising again, so $f' > 0$.
The graph of $f'$ is above the $x$-axis for $x < -1$, below for $-1 < x < 1$, above for $x > 1$, with zeros at $x = \pm 1$.
After computing: $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$. This is a parabola opening upward, zero at $x = \pm 1$, positive outside those values, negative between them. Matches the sketch.
2. Two Representations and Translation
For a general function:
Graph of $f$: shows the height of the function. The slope at each point is what you read off visually.
Graph of $f'$: shows the slope of $f$ as a height. Where $f$ is steepest (rising most steeply), $f'$ is highest. Where $f$ is steepest falling, $f'$ is most negative.
Translation prompt. Given a graph of $f'$, sketch $f$:
- Where $f' > 0$ (above axis): $f$ is increasing.
- Where $f' < 0$ (below axis): $f$ is decreasing.
- Where $f' = 0$ (crossing axis): $f$ has a critical point (flat).
- Where $f'$ is at a local max: $f$ has an inflection point (transitions from increasing slowly to increasing faster, or from concave-down to concave-up increase).
This reverse translation is the basis of curve sketching.
3. Features of $f$ Encoded in $f'$
Example 2. A function $f$ has the following features:
- $f(-2) = 0$ (a zero), $f(0) = 2$ (local max), $f(3) = -1$ (local min)
- $f$ is increasing on $(-\infty, 0)$ and $(3, \infty)$
- $f$ is decreasing on $(0, 3)$
Sketch the graph of $f'$.
$f'(x) > 0$ on $(-\infty, 0)$: $f'$ is above the axis there.
$f'(0) = 0$ (local max of $f$): $f'$ crosses zero from positive to negative.
$f'(x) < 0$ on $(0, 3)$: $f'$ is below the axis.
$f'(3) = 0$ (local min of $f$): $f'$ crosses zero from negative to positive.
$f'(x) > 0$ on $(3, \infty)$: $f'$ is above the axis.
The graph of $f'$ starts positive, crosses zero at $x = 0$ (going downward), stays negative until $x = 3$, then crosses zero going upward.
4. Ask Why: Why Does a Local Maximum of $f$ Make $f' = 0$?
At a local maximum of $f$: the function changes from increasing (positive slope) to decreasing (negative slope). At the exact peak, the slope is neither positive nor negative -- it is zero. The tangent at a local max is horizontal.
This is the key insight for optimization: where is a function biggest? It is at a point where $f' = 0$ (or where $f'$ doesn’t exist). Finding critical points is the foundation of the entire optimization chapter.
5. Corners and Non-Differentiable Points
Where $f$ has a corner (like $|x|$ at $x = 0$): $f'$ has a jump discontinuity. The slope jumps from one value to another -- the two one-sided derivatives exist but differ.
Where $f$ has a vertical tangent: $f'$ has a vertical asymptote at that point.
The graph of $f'$ “inherits” all the non-differentiable behavior of $f$.
Named Misconception: iconic-graph
A common error: when asked to sketch the graph of $f'$, students draw a reflection or a shifted copy of $f$, as if the derivative graph should look like the original but mirrored or moved.
This error comes from treating “graph” as a visual template to transform. The derivative graph is not a visual transformation of $f$ -- it is a completely different function (the slope at each point).
One way to see the error breaks: the graph of $f(x) = x^3$ goes from negative to positive as $x$ increases. The graph of $f'(x) = 3x^2$ is a parabola always above the $x$-axis -- it does not look like a transformation of $x^3$. The two graphs share the same $x$-axis, but their $y$-values have completely different meanings.
Common Errors
| Error | Specific example | Correction |
|---|---|---|
| Drawing $f'$ as a mirror of $f$ | Reflecting $x^3$ to get its “derivative” | $f'$ encodes slopes, not heights; $f'(x) = 3x^2$ looks nothing like a reflection of $x^3$ |
| Sign error | Writing $f' > 0$ where $f$ is decreasing | $f$ decreasing means $f' < 0$; the height of $f$ is irrelevant, only the direction |
| Thinking local max of $f$ means $f'$ has a local max | Putting a peak in $f'$ where $f$ peaks | Local max of $f$ means $f' = 0$; $f'$ crosses zero (does not peak) at a local max of $f$ |
Leveled Practice
Level 1 -- Sign Analysis from Graph
Problem 1. For $f(x) = \sin x$ on $[0, 2\pi]$: state where $f' > 0$, $f' < 0$, and $f' = 0$.
Show answer
$f' > 0$ on $(0, \pi/2)$ and $(3\pi/2, 2\pi)$ (rising).
$f' < 0$ on $(\pi/2, 3\pi/2)$ (falling).
$f' = 0$ at $x = \pi/2$ (local max) and $x = 3\pi/2$ (local min).
Problem 2. The graph of $f$ shows that $f$ is increasing on $(-1, 2)$, decreasing on $(2, 5)$, and has a local minimum at $x = 5$ with $f'(5) = 0$. Describe the graph of $f'$ on $[-1, 5]$.
Show answer
$f' > 0$ on $(-1, 2)$; $f'(2) = 0$ (local max of $f$); $f' < 0$ on $(2, 5)$; $f'(5) = 0$ (local min of $f$, so $f'$ changes from negative to zero -- but the problem says $f$ has a local min at 5, so $f'$ would be zero and then positive for $x > 5$, but $x > 5$ is outside our range).
The graph of $f'$ on $[-1, 5]$: starts positive (above axis), passes through zero at $x = 2$, stays negative (below axis), and returns to zero at $x = 5$.
Level 2 -- Sketching from a Description
Problem 3. Sketch a rough graph of $f'$ for $f(x) = e^{-x^2}$ (a bell-shaped curve peaking at $x = 0$).
Show answer
$f$ is increasing on $(-\infty, 0)$, decreasing on $(0, \infty)$, with peak at $x = 0$.
$f'(0) = 0$ (local max of $f$). $f' > 0$ for $x < 0$. $f' < 0$ for $x > 0$.
$f$ is very flat (nearly horizontal) for large $|x|$, so $f'$ is near zero for large $|x|$.
The graph of $f'$: a curve that is positive for $x < 0$, crosses zero at $x = 0$ going downward, negative for $x > 0$, approaching zero as $x \to \pm\infty$. (In fact $f'(x) = -2xe^{-x^2}$: an odd function with zero at $x = 0$.)
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 4 (Extension).
(a) (Floor) Given that $f'(x) > 0$ on $(1, 4)$ and $f'(4) = 0$ and $f'(x) < 0$ on $(4, 7)$: where does $f$ have a local maximum?
(b) (Mid) The graph of $f'$ is the graph $y = (x-2)^2 - 1$. Sketch the behavior of $f$ from this information alone: where is $f$ increasing, decreasing, and where are its critical points?
(c) (Ceiling) If $f$ is a differentiable function and the graph of $f'$ has a local minimum at $x = 3$ (i.e., $(f')'(3) = 0$ and $f''(3) < 0$... wait, that would be a local max), let me correct: $f'$ has a local minimum at $x = 3$ (so $f''(3) = 0$ and $f'''(3) > 0$ would be needed). What does this mean for the graph of $f$ at $x = 3$?
Show answer
(a) $f$ has a local maximum at $x = 4$ (where $f'$ changes from positive to negative).
(b) $f'(x) = (x-2)^2 - 1 = 0$ at $x = 1$ and $x = 3$.
$f' > 0$ for $x < 1$ (since $(x-2)^2 > 1$): $f$ is increasing.
$f' < 0$ for $1 < x < 3$: $f$ is decreasing.
$f' > 0$ for $x > 3$: $f$ is increasing.
Local max of $f$ at $x = 1$; local min of $f$ at $x = 3$.
(c) If $f'$ has a local minimum at $x = 3$: the slope of $f$ is at its lowest value near $x = 3$. For $f$ itself, this means $x = 3$ is an inflection point -- the concavity of $f$ changes at $x = 3$ (the rate of slope change passes through zero or a sign change). At an inflection point, $f''(3) = 0$ (the second derivative is zero, or the slope of $f'$ is zero at $x = 3$).
Common Misconceptions
a point where $f$ is large means $f'$ is large there. A function at its maximum value has a horizontal tangent: $f'$ equals zero at the peak, not its largest value. The graph of $f$ is as high as it gets precisely where $f'$ is crossing zero. Students sometimes sketch $f'$ by tracing the shape of $f$, producing a graph that looks similar to $f$ but is shifted or scaled. The derivative graph should be read from slope, not height.
where $f$ is increasing, $f'$ must also be increasing. $f'$ is positive (not necessarily increasing) wherever $f$ is increasing. The graph of $f$ rises when $f' > 0$, and it rises faster when $f'$ is getting larger, but $f'$ can be positive and decreasing (a function that is rising but decelerating). These two conditions, $f' > 0$ and $f'$ increasing, correspond to different geometric features of $f$.
Mastery Checklist
Mental Model
The graph of $f'$ is a slope meter. At each $x$, the height of the $f'$ graph equals the slope of $f$. Where $f$ rises steeply, $f'$ is tall and positive. Where $f$ is flat, $f'$ is zero. Where $f$ falls, $f'$ is negative.
To sketch $f'$: scan the graph of $f$ from left to right, asking “is the slope increasing, decreasing, or flat?” Those observations become the height of $f'$.
This translation between a curve’s shape and its slope function is the geometric foundation of optimization, curve sketching, and the Fundamental Theorem of Calculus.
Connections
Within MATH161
- Optimization: Critical points (where $f' = 0$ or undefined) are the candidates for maxima and minima. The graph of $f'$ shows exactly where these occur.
- Curve sketching: The full curve sketch uses $f'$ (for increasing/decreasing) and $f''$ (for concavity). Sketching $f'$ from $f$ is the prerequisite for this analysis.
- Fundamental Theorem: The area under the $f'$ graph gives the net change in $f$. Understanding the relationship between the graphs of $f$ and $f'$ prepares you for this connection.