1. Critical points and extrema
At a local maximum, the function rises, levels off, and then falls. At a local minimum, it does the reverse: falls, levels off, rises. The "levels off" moment is geometric — the tangent line is horizontal — and that translates immediately into algebra: $f'(x) = 0$.
A number $c$ in the domain of $f$ where either $f'(c) = 0$ or $f'(c)$ does not exist. Every local maximum or minimum of a differentiable function occurs at a critical point — but not every critical point is an extremum.
The plan, then, is to collect candidates by solving $f'(x) = 0$, then classify each one. The cheapest classifier is the first derivative test.
The first derivative test
Walk across the critical point $c$ and watch the sign of $f'$:
| Behavior of $f'$ near $c$ | Meaning at $c$ |
|---|---|
| $f' > 0$ on the left, $f' < 0$ on the right | Local maximum (rising then falling) |
| $f' < 0$ on the left, $f' > 0$ on the right | Local minimum (falling then rising) |
| Same sign on both sides | Not an extremum — the curve flattens but keeps going |
Concretely: pick a test point just left of $c$ and just right of $c$, evaluate $f'$ at each, look at the signs.
The picture above traces a typical cubic shape. At each of the two marked points the tangent (the dashed green segment) is horizontal: the slope $f'$ has just dropped to zero. To the left of the left point the curve is rising ($f' > 0$); to the right it falls ($f' < 0$). That's a local maximum. The right-hand point is the reverse — a local minimum.
2. The second derivative
The second derivative is the derivative of the derivative: $f''(x) = \tfrac{d}{dx}\big(f'(x)\big)$. If $f'$ tells you how $f$ changes, $f''$ tells you how $f'$ changes — and that bends $f$.
A function is concave up on an interval where $f'' > 0$ — the curve opens upward, like a cup. It is concave down where $f'' < 0$ — the curve opens downward, like a frown. A point where concavity switches is an inflection point.
Two interpretations are worth holding side by side:
- Geometric. $f'' > 0$ means the slope is increasing — even if the function is going down, it's "decelerating downward" and curving back up. $f'' < 0$ means the slope is decreasing.
- Physical. If $f(t)$ is position, $f'(t)$ is velocity and $f''(t)$ is acceleration. Positive acceleration means speeding up in the positive direction (or slowing down in the negative direction).
The second derivative test
This gives a fast classifier for critical points when it applies. Suppose $f'(c) = 0$. Then:
$$ \begin{aligned} f''(c) > 0 &\;\Longrightarrow\; \text{local minimum at } c \\ f''(c) < 0 &\;\Longrightarrow\; \text{local maximum at } c \\ f''(c) &= 0 &\;\Longrightarrow\; \text{inconclusive — fall back to the first derivative test} \end{aligned} $$The intuition is direct. A horizontal tangent inside a concave-up bowl must be the bottom of the bowl. A horizontal tangent on top of a concave-down arch must be the peak. When $f''(c) = 0$, the test reveals nothing — you might have a max, a min, or an inflection point with a horizontal tangent (like $f(x) = x^3$ at $x = 0$).
If $f''$ is easy to compute and clearly nonzero at the critical point, the second derivative test is one line of algebra. If $f''$ is messy, or if it's zero at the critical point, the first derivative test (signs of $f'$ on either side) is more reliable.
3. Optimization
Most real problems aren't "find the extrema of this specific function" — they're "given some constraints in the world, maximize or minimize this quantity." The translation step is the hard part. The calculus, once you have a single-variable function, is the easy part.
A recipe that works on almost every textbook problem:
- Name the quantity to optimize. Call it $Q$. Write it as a function of whatever variables describe the situation.
- Use the constraints to eliminate variables until $Q$ depends on just one.
- Identify the domain. Real geometric quantities are typically nonnegative; some have an upper bound from the constraint.
- Find critical points by solving $Q'(x) = 0$ in the interior of the domain.
- Check the endpoints too. On a closed interval, the global max or min may sit at a boundary point even when no critical point does.
- Classify the winner with the first or second derivative test.
Worked setup: minimum-surface can
Design a closed cylindrical can holding a fixed volume $V$, using as little metal as possible. Let $r$ be the radius and $h$ the height. Then
$$ V = \pi r^2 h \quad\text{(constraint)}, \qquad S = 2\pi r^2 + 2\pi r h \quad\text{(surface area to minimize)}. $$Solve the constraint for $h$ to eliminate it from $S$:
$$ h = \frac{V}{\pi r^2} \;\;\Longrightarrow\;\; S(r) = 2\pi r^2 + 2\pi r \cdot \frac{V}{\pi r^2} = 2\pi r^2 + \frac{2V}{r}. $$Differentiate and set $S'(r) = 0$:
$$ S'(r) = 4\pi r - \frac{2V}{r^2} = 0 \;\;\Longrightarrow\;\; 4\pi r^3 = 2V \;\;\Longrightarrow\;\; r^3 = \frac{V}{2\pi}. $$Substituting back gives $h = 2r$ — the optimal can is as tall as it is wide (height equals diameter). A second derivative check, $S''(r) = 4\pi + 4V/r^3 > 0$, confirms this is a minimum.
Real soda cans aren't proportioned this way, and the reason is informative: the optimization above ignores that the top and bottom of a can are thicker than the side wall, ignores manufacturing constraints, and ignores how the can will be held. Whenever a calculus answer disagrees with the world, the model — not the math — is what to interrogate.
5. Linear approximation
Zoom in on a smooth curve far enough and it stops looking curved — it looks like its tangent line. That's the whole content of differentiability, restated. Turn it into a formula by taking the tangent line at $x = a$ and using it to estimate $f$ at nearby inputs:
$$ f(x) \approx f(a) + f'(a)(x - a) \qquad \text{(for } x \text{ near } a\text{)}. $$The right-hand side is called the linearization of $f$ at $a$, often written $L(x)$.
The accuracy of the approximation deteriorates as $x$ wanders away from $a$. The error is governed by $f''$: the bigger the curvature, the faster the tangent line drifts from the curve. (Make that quantitative and you've discovered Taylor's theorem.)
Linearization is the gateway to a much bigger idea: if a single line is a decent local model, a parabola fit to the same point is even better, and a polynomial that matches $f, f', f'', \dots$ at $a$ is better still. That's the Taylor series, and every infinite series for $\sin x$, $e^x$, and $\ln(1+x)$ in your future descends from this one move.
6. Mean value theorem, Rolle's theorem, and Newton's method
A few classical results round out the toolkit. The first two formalize the picture of "slopes must do something specific in between two points"; the third is a numerical algorithm that uses derivatives to chase roots.
If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists at least one $c$ in $(a, b)$ with
$$ f'(c) = \frac{f(b) - f(a)}{b - a}. $$Somewhere between the endpoints, the instantaneous slope equals the average slope across the interval.
Rolle's theorem is the special case $f(a) = f(b)$: under the same continuity and differentiability hypotheses, there is some $c \in (a, b)$ with $f'(c) = 0$. Geometrically, a smooth curve that starts and ends at the same height must level off somewhere in between.
Rolle's theorem is the engine behind many uniqueness arguments — if two solutions existed, $f'$ would have to vanish between them. The MVT is the engine behind error bounds for linear approximation and the proof that "derivative zero everywhere $\Rightarrow$ function is constant."
Curve sketching, in one breath
Stitching together the first and second derivatives produces a complete qualitative picture of a function. The checklist:
- Domain and intercepts — where is $f$ defined, where does it hit the axes.
- Asymptotes — vertical (where the denominator vanishes), horizontal (limit as $x \to \pm\infty$).
- $f'$ sign chart — where the function rises and falls; critical points become local maxes and mins.
- $f''$ sign chart — where the curve is concave up vs down; sign changes mark inflection points.
- Plot the skeleton — extrema, inflection points, intercepts, asymptotes — then connect with the right concavity.
Newton's method
To solve $f(x) = 0$ when no algebraic shortcut works, start from a guess $x_0$ and iteratively replace it by the x-intercept of the tangent line at the current guess:
$$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}. $$Each step assumes the function is well-approximated by its tangent line near $x_n$ — exactly the linearization idea from §5, applied iteratively. When the method converges, it does so quadratically: the number of correct digits roughly doubles per step. When it fails — flat derivative, oscillation, divergence — the failure is loud and easy to spot.
If $f'(x_n) = 0$ the update divides by zero. If the starting guess is far from a root, iterates can leap to a different root or fail to converge at all. Sketch the function first; pick $x_0$ where the tangent points toward the root.