Calculus I · Chapter 5

Chapter 5: Graphs & the Derivative

Increasing & decreasing · Critical numbers · Relative extrema · Concavity & inflection · Second derivative test · Curve sketching

The derivative carries an enormous amount of information about the shape of a graph. Its sign tells you where the function rises and falls; the sign of the second derivative tells you how it curves. This chapter turns those facts into a systematic method for analyzing and sketching any function, by hand, without plotting hundreds of points.

Section 5.1

Increasing & Decreasing Functions

Theorem: The Sign of $f'$ Controls Direction
On an interval:
  • If $f'(x) > 0$ throughout, then $f$ is increasing (rising) there.
  • If $f'(x) < 0$ throughout, then $f$ is decreasing (falling) there.
  • If $f'(x) = 0$ throughout, then $f$ is constant.
The function can only switch between rising and falling where $f'(x) = 0$ or $f'(x)$ is undefined.
Top: the curve $f$, shaded green where it rises and orange where it falls. Bottom: a sign chart of $f'$. Drag the dot, the tangent's tilt matches the sign of $f'$ directly below it. Where $f'$ changes sign, $f$ turns around.
Section 5.2

Relative Extrema

Critical Numbers

Definition: Critical Number
A critical number of $f$ is a value $c$ in the domain where $$f'(c) = 0 \quad\text{or}\quad f'(c)\text{ does not exist}.$$ Relative (local) maxima and minima can occur only at critical numbers, they are the complete list of candidates for a turning point.
⚠ Critical Number ≠ Guaranteed Extremum
Every relative extremum sits at a critical number, but not every critical number is an extremum. For $f(x) = x^3$, $f'(0) = 0$, yet $x = 0$ is neither a max nor a min, the curve just flattens and keeps rising. You must test each critical number.

The First Derivative Test

Theorem: First Derivative Test
At a critical number $c$, examine the sign of $f'$ just to the left and just to the right:
  • $+$ then $-$  (rises then falls): relative maximum at $c$.
  • $-$ then $+$  (falls then rises): relative minimum at $c$.
  • no sign change: neither (a shelf or inflection).
📘 Example 5.1: First Derivative Test
Find and classify the relative extrema of $f(x) = x^3 - 3x^2 - 9x + 5$.
  1. $f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1)$.
  2. Critical numbers: $x = -1$ and $x = 3$.
  3. Sign of $f'$: positive on $(-\infty,-1)$, negative on $(-1,3)$, positive on $(3,\infty)$.
  4. At $x=-1$: $+\to-$ ⇒ relative max, $f(-1) = 10$. At $x=3$: $-\to+$ ⇒ relative min, $f(3) = -22$.
Relative max $(-1, 10)$; relative min $(3, -22)$
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Section 5.3

Higher Derivatives, Concavity & the Second Derivative Test

Higher-Order Derivatives
Differentiating $f'$ again gives the second derivative $f''(x)$, the rate of change of the slope. Notations: $f''(x)$, $\dfrac{d^2y}{dx^2}$, $y''$. In motion, if $f$ is position then $f'$ is velocity and $f''$ is acceleration.

Concavity & Inflection Points

Theorem: The Sign of $f''$ Controls Curvature
  • $f''(x) > 0$: graph is concave up (holds water, $\smile$), slopes are increasing.
  • $f''(x) < 0$: graph is concave down (spills water, $\frown$), slopes are decreasing.
An inflection point is where concavity changes, typically where $f''(x) = 0$ and switches sign.
Concave up (blue, $\smile$) versus concave down (orange, $\frown$), meeting at an inflection point where the curvature flips. Click to toggle which region is highlighted; the small tangent fans show slopes increasing on the up side and decreasing on the down side.

The Second Derivative Test

Theorem: Second Derivative Test
Suppose $f'(c) = 0$. Then:
  • $f''(c) > 0$ ⇒ concave up at $c$ ⇒ relative minimum.
  • $f''(c) < 0$ ⇒ concave down at $c$ ⇒ relative maximum.
  • $f''(c) = 0$ ⇒ test fails; fall back to the first derivative test.
Often faster than the first derivative test when the second derivative is easy to compute.
📘 Example 5.2: Second Derivative Test
Classify the critical points of $f(x) = x^3 - 3x^2 - 9x + 5$ using $f''$. From Example 5.1, critical numbers are $-1$ and $3$ $f''(x) = 6x - 6$.
  • $f''(-1) = -12 < 0$ ⇒ concave down ⇒ relative max at $x=-1$.
  • $f''(3) = 12 > 0$ ⇒ concave up ⇒ relative min at $x=3$.
Same conclusion as the first derivative test, reached faster. Inflection point where $f''=0$: $x = 1$. Max at $x=-1$, min at $x=3$, inflection at $x=1$
Section 5.4

Curve Sketching

The Complete Curve-Sketching Checklist
  1. Domain and any intercepts.
  2. Asymptotes (vertical where denominator $=0$; horizontal via limits at infinity).
  3. $f'$: critical numbers → intervals of increase/decrease → relative extrema.
  4. $f''$: intervals of concavity → inflection points.
  5. Plot the key points and connect, respecting direction and curvature.
Curve sketching assembled step by step. Click Next to layer on each piece of information, intercepts, then critical points and direction, then concavity and the inflection point, then the finished curve. Each layer is read directly off $f'$ and $f''$.
Step 1: intercepts
📘 Example 5.3: Full Curve Analysis
Analyze $f(x) = x^4 - 4x^3$.
  1. Intercepts: $f(x) = x^3(x - 4) = 0$ at $x = 0$ and $x = 4$.
  2. $f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$. Critical numbers $x = 0, 3$. Decreasing on $(-\infty,3)$, increasing on $(3,\infty)$.
  3. $x = 3$ is a relative min, $f(3) = -27$. At $x = 0$, no sign change ⇒ not an extremum.
  4. $f''(x) = 12x^2 - 24x = 12x(x - 2)$. Concave up on $(-\infty,0)$ and $(2,\infty)$, down on $(0,2)$. Inflection points at $x = 0$ and $x = 2$.
Relative min $(3,-27)$; inflection points at $x=0$ and $x=2$
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Looking Ahead
The first and second derivatives now let you find every local peak and valley. Chapter 6 pushes this to absolute extrema and real optimization problems, maximizing profit, minimizing cost, fastest, cheapest, largest, plus implicit differentiation, related rates, and approximation.
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