Applications of Derivatives — Optimization and Curve Sketching

⚡ Key Takeaways

Optimization and curve sketching use the first and second derivatives of a function to find its largest/smallest values and to draw its graph accurately. Key result: at a smooth maximum or minimum, the derivative equals zero ( $f'(x)=0$ ); the second derivative's sign reveals which it is and how the curve bends. Applies to: maximizing profit, minimizing material or distance, modelling motion, and sketching any differentiable curve. Who/when: rooted in the work of Fermat (1630s) and Newton/Leibniz (1660s–1680s).

💡 The Core Insight

A differentiable function can only turn around — switch from increasing to decreasing or vice versa — at a point where its tangent line is horizontal. Therefore the interesting features of a graph (its hills, valleys, and bends) are encoded in the places where $f'(x)=0$ , where $f'(x)$ is undefined, and where $f''(x)$ changes sign. Reading these three pieces of information lets you reconstruct the entire shape of a curve and pinpoint its optimal values without plotting hundreds of points.

DefinitionCritical Point

A critical point of a function $f$ is a number $c$ in the domain of $f$ at which either

1

\quad f'(c)=0 \qquad\text{or}\qquad f'(c)\ \text{does not exist}.

Derivation hint: A critical point is by definition where the candidate-extremum condition can be met; the two cases correspond to smooth turning points and to corners/cusps respectively. Here $f'(c)$ is the derivative of $f$ evaluated at $x=c$ , geometrically the slope of the tangent line to the graph at $(c, f(c))$ .

DefinitionLocal and Absolute Extrema

$f$ has a local (relative) maximum at $c$ if $f(c)\ge f(x)$ for all $x$ near $c$ ; a local minimum if $f(c)\le f(x)$ for all $x$ near $c$ . It has an absolute (global) maximum at $c$ if $f(c)\ge f(x)$ for every $x$ in the domain, and similarly for absolute minimum. The symbol "near $c$ " means within some open interval containing $c$ .

TheoremFermat's Theorem (Interior Extremum)

If $f$ has a local maximum or minimum at an interior point $c$ of its domain, and $f'(c)$ exists, then

2

\quad f'(c)=0.

Derivation hint: Compare the one-sided difference quotients on each side of a peak; one is $\le 0$ , the other $\ge 0$ , and if the two-sided limit exists it must equal both, forcing $0$ .

TheoremFirst Derivative Test

Let $c$ be a critical point of a continuous function $f$ .

If $f'$ changes from positive to negative at $c$ , then $f$ has a local maximum at $c$ .
If $f'$ changes from negative to positive at $c$ , then $f$ has a local minimum at $c$ .
If $f'$ does not change sign, $c$ is neither.

Symbolically, increasing means $f'(x)>0$ and decreasing means $f'(x)<0$ on the relevant interval.

TheoremSecond Derivative Test

If $f'(c)=0$ and $f''(c)$ exists:

3

\quad f''(c)>0 \Rightarrow \text{local minimum},\qquad f''(c)<0 \Rightarrow \text{local maximum}.

Derivation hint: $f''(c)>0$ means the slope is increasing through $c$ , so $f'$ goes from negative to positive — a valley; the inequality is reversed for a peak. If $f''(c)=0$ the test is inconclusive.

DefinitionConcavity and Inflection Points

$f$ is concave up on an interval where $f''(x)>0$ (the curve lies above its tangents, shaped like $\cup$ ) and concave down where $f''(x)<0$ (shaped like $\cap$ ). An inflection point is a point where the concavity changes, requiring

4

\quad f''(x)=0 \ \text{or}\ f''(x)\ \text{undefined, AND}\ f''\ \text{changes sign there}.

Derivation hint: Inflection is the second-order analogue of an extremum: it is where $f''$ "turns around," found by setting $f''=0$ and checking for an actual sign change.

TheoremExtreme Value Theorem

If $f$ is continuous on a closed, bounded interval $[a,b]$ , then $f$ attains both an absolute maximum and an absolute minimum on $[a,b]$ . These occur either at critical points inside $(a,b)$ or at the endpoints $a$ or $b$ .

ProofFermat's Theorem

Suppose $f$ has a local maximum at an interior point $c$ and $f'(c)$ exists. 1. By definition of local maximum, $f(c+h) \le f(c)$ for all sufficiently small $h$ , so $f(c+h)-f(c)\le 0$ . 2. For $h>0$ : the difference quotient satisfies $\dfrac{f(c+h)-f(c)}{h}\le 0$ (negative over positive). Taking the limit as $h\to 0^+$ gives the right-hand derivative $f'(c)\le 0$ . 3. For $h<0$ : now $\dfrac{f(c+h)-f(c)}{h}\ge 0$ (negative over negative). Taking the limit as $h\to 0^-$ gives the left-hand derivative $f'(c)\ge 0$ . 4. Since $f'(c)$ exists, the left and right limits are equal. The only number satisfying both $f'(c)\le 0$ and $f'(c)\ge 0$ is $f'(c)=0$ . $\blacksquare$ The minimum case follows by applying this result to $-f$ .

∎

Figure

A cubic-like curve showing a local max, local min, and an inflection point with sign data for f' and f''

Building Intuition

ExampleMinimizing the Cost of a Box

Given: An open-top box with a square base must hold a volume of $32{,}000\ \text{cm}^3$ . Find the dimensions that minimize the surface area (material used).

Setup. Let the base side be $x$ and height $h$ . Volume: $x^2 h = 32000$ , so $h = 32000/x^2$ . Surface area (base + four sides): $S = x^2 + 4xh$ . Substitute $h$ :

5

\quad S(x) = x^2 + 4x\cdot\frac{32000}{x^2} = x^2 + \frac{128000}{x}.

Derivation hint: Eliminate $h$ using the volume constraint so $S$ depends on a single variable $x$ .

Differentiate and set to zero:

6

\quad S'(x) = 2x - \frac{128000}{x^2} = 0 \;\Rightarrow\; 2x^3 = 128000 \;\Rightarrow\; x^3 = 64000.

Derivation hint: Apply the power rule to $x^2$ and to $128000x^{-1}$ , then clear the denominator. So $x = \sqrt[3]{64000} = 40\ \text{cm}$ , and $h = 32000/40^2 = 20\ \text{cm}$ .

Confirm minimum: $S''(x) = 2 + 256000/x^3 > 0$ everywhere, so the critical point is indeed a minimum.

\boxed{x = 40\ \text{cm},\quad h = 20\ \text{cm},\quad S_{\min} = 4800\ \text{cm}^2.}

ExampleSketching a Curve from Derivative Data

Given: Sketch $f(x) = x^3 - 3x^2$ .

Step 1 — Critical points. $f'(x) = 3x^2 - 6x = 3x(x-2)$ . Setting $f'=0$ gives $x=0$ and $x=2$ .

Step 2 — Classify with the second derivative. $f''(x) = 6x - 6 = 6(x-1)$ .

At $x=0$ : $f''(0) = -6 < 0$ → local maximum, value $f(0)=0$ .
At $x=2$ : $f''(2) = 6 > 0$ → local minimum, value $f(2)= 8-12=-4$ .

Step 3 — Inflection point. $f''(x)=0$ at $x=1$ ; since $f''$ changes from negative to positive there, $(1, f(1)) = (1,-2)$ is an inflection point.

Step 4 — Intercepts and behaviour. $f(x)=x^2(x-3)=0$ at $x=0$ and $x=3$ . As $x\to\infty$ , $f\to\infty$ ; as $x\to-\infty$ , $f\to-\infty$ .

\boxed{\text{Max } (0,0),\ \text{Min } (2,-4),\ \text{Inflection } (1,-2),\ \text{roots at } x=0,\,3.}

⚠️ Common mistake

"Setting $f'(x)=0$ always gives the maximum or minimum." This is incomplete on three counts: 1. Critical points include places where $f'$ is undefined (corners, cusps, vertical tangents) — e.g. $f(x)=|x|$ has a minimum at $x=0$ where $f'(0)$ does not exist. 2. On a closed interval, the absolute extremum may be at an endpoint, where $f'$ need not be zero. 3. A critical point can be neither a max nor a min (the second-derivative test failing). The vanishing of $f'$ is necessary for an interior smooth extremum but never sufficient on its own.

Counterexample

Consider $f(x) = x^3$ . Then $f'(x) = 3x^2$ , so $f'(0)=0$ — the point $x=0$ is critical. Yet $f$ has no extremum there: it is increasing on both sides ( $f'(x)>0$ for $x\ne 0$ ). The second-derivative test is inconclusive because $f''(0)=6\cdot 0 = 0$ . In fact $(0,0)$ is an inflection point. This shows that $f'(c)=0$ does not guarantee a maximum or minimum, and that the second-derivative test can give no verdict.

connection

Optimization and curve sketching sit at the heart of differential calculus and connect outward in several directions. They are the natural pay-off of The Derivative and the rules for computing it, especially The Chain Rule for composite functions. The justification of the first-derivative test relies on The Mean Value Theorem, which guarantees a point where instantaneous slope equals average slope. Closed-interval optimization depends on Continuity through the Extreme Value Theorem. The inverse problem — recovering a function from its rate of change, needed in applications like total cost from marginal cost — is handled by The Definite Integral and the Fundamental Theorem of Calculus. Related rate-of-change problems share the same machinery under the heading Related Rates.

🔬 Where this appears

⏳ History

- c. 250 BCE — Archimedes uses tangent-like and limiting arguments on areas and the spiral, foreshadowing optimization ideas without a derivative concept.

- 1629–1636 — Pierre de Fermat develops his method of adequality to find maxima and minima and tangents, effectively setting a derivative-like expression to zero (Fermat's Theorem).

- 1660s — Isaac Newton develops the method of fluxions, treating rates of change systematically for problems of motion and extrema.

- 1684 — Gottfried Wilhelm Leibniz publishes the first account of differential calculus, introducing the notation

dx

,

dy

and the rule for finding maxima/minima where the differential vanishes.

- 1696 — L'Hôpital's Analyse des Infiniment Petits, the first textbook of differential calculus, popularizes maximum/minimum techniques.

- 1696–1697 — The brachistochrone problem (Johann Bernoulli) launches the calculus of variations, optimization over whole curves rather than single points.

- 1740s — Euler systematizes the second-derivative criterion for distinguishing maxima from minima.

- 1820s — Cauchy provides rigorous limit-based foundations, making the Extreme Value and Mean Value theorems precise.

- 1860s–1870s — Weierstrass formalizes continuity and the existence of extrema on closed intervals, completing the logical underpinning of the Extreme Value Theorem.

open-problem

The techniques here apply cleanly to smooth, single-variable functions on nice domains, but the boundaries of the subject open onto deep territory. For functions of several variables, critical points are found by setting all partial derivatives to zero, and the classification requires the Hessian matrix (a multivariable second-derivative test) — yet saddle points proliferate and global optimization becomes genuinely hard. Non-differentiable and non-convex optimization (as in modern machine learning) has no guarantee that following the derivative reaches the global minimum; gradient descent can stall in local minima or saddle regions, and characterizing when it succeeds is an active research area. Even the innocent assumption of a closed bounded interval matters: drop it and a continuous function may have no maximum at all (e.g. $f(x)=x$ on the open interval $(0,1)$ ). Constrained optimization with inequality constraints (the Karush–Kuhn–Tucker conditions) generalizes Fermat's idea but lies beyond the elementary setting.

Applications of Derivatives — Optimization and Curve Sketching

Formal Statement

Why It Holds

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also