The Timpson Lyceum · Calculus 1 (2024)

Monotonic Function, Concavity, and Optimization - Lecture 13 (Calculus 1)

Overview

This lecture connects the algebraic properties of a function's derivatives to the shape of its graph, and then puts those tools to work. The first derivative tells us where a function is increasing or decreasing and where its local extrema lie; the second derivative introduces concavity — whether the slope itself is rising or falling — and the location of inflection points. Together these give two procedures for classifying critical points: the first-derivative test and the second-derivative test. The payoff is optimization: real applied problems where we must find the exact input that maximizes or minimizes a quantity (box volume, can surface area, inscribed area, travel time). Knowing extrema exactly — not by zooming in on a calculator — is the whole point, since these "stationary points" appear everywhere from manufacturing to physics.

Key Concepts

Increasing / decreasing from $f'$

If $f'(x) > 0$ on an open interval $I$ , then $f$ is increasing on $I$ ; if $f'(x) < 0$ on $I$ , then $f$ is decreasing on $I$ . The derivative is the slope, so a positive slope means the graph rises as $x$ increases.
This follows from the Mean Value Theorem: for $x_1 < x_2$ in $I$ there is a $c$ between them with $f'(c) = \dfrac{f(x_2)-f(x_1)}{x_2-x_1}$ . Rearranging gives $f(x_2)-f(x_1)=f'(c)\,(x_2-x_1)$ . If $f'(c)>0$ and $x_2-x_1>0$ , the right side is positive, so $f(x_2)>f(x_1)$ .

Critical points and the first-derivative test

A critical point is where $f'(x)=0$ or $f'(x)$ is undefined. Every local max or min of $f$ occurs at a critical point.
A function can only switch between increasing and decreasing by passing through a critical point. To classify one, test the sign of $f'$ just to its left and right: decreasing-then-increasing ( $-,+$ ) gives a local min; increasing-then-decreasing ( $+,-$ ) gives a local max; same sign on both sides means it is neither.
A critical point need not be an extremum. For example $f(x)=x^3$ has $f'(0)=0$ but is increasing everywhere.

Concavity from $f''$

Concavity describes how the slope changes. Concave up = slope is increasing (a smiley-face shape); concave down = slope is decreasing (a frowny-face shape).
If $f''(x) > 0$ on $I$ , then $f'$ is increasing, so $f$ is concave up on $I$ . If $f''(x) < 0$ on $I$ , then $f$ is concave down on $I$ .

Inflection points

An inflection point is where concavity changes sign (up-to-down or down-to-up) and a tangent line exists there. (We require a tangent rather than a derivative so that vertical-tangent cases still count.)
Inflection points can only occur where $f''(x)=0$ or $f''(x)$ is undefined — but not every such point is an inflection point; the concavity must actually switch. At a true inflection point the tangent line crosses the curve.

Second-derivative test — for a critical point where $f'(x)=0$ :

If $f''(x) > 0$ (concave up): local minimum.
If $f''(x) < 0$ (concave down): local maximum.
If $f''(x) = 0$ : inconclusive — could be a max, a min, or neither (again $x^3$ at $0$ : both $f'(0)=0$ and $f''(0)=0$ , yet it is neither).

Optimization (applied max–min) strategy

Write the quantity to optimize as a formula, then reduce it to a function of a single variable, using any constraint to eliminate the other variables.
Identify the feasible domain (the physically meaningful range of the variable).
Differentiate, set the derivative to $0$ , and solve for the critical point(s).
Confirm it is the desired max or min (sign of $f'$ around it, the second-derivative test, or the endpoint values), then report the exact dimensions/value.

Motion application: for a position function $s(t)$ , the velocity is $s'(t)$ and the acceleration is $s''(t)$ . So "where is velocity increasing/decreasing" is the increasing/decreasing question applied to $s'$ , and "where is acceleration positive/negative" is the concavity question applied to $s$ .

Worked Example

Open-top box (maximize volume). From a $12\text{ in} \times 12\text{ in}$ square of cardboard, cut a square of side $x$ from each corner and fold up the sides. The base measures $(12-2x)$ by $(12-2x)$ and the height is $x$ , so the volume is

$V(x) = x\,(12-2x)^2, \qquad 0 < x < 6.$

Expand, differentiate, set $V'(x)=0$ , and solve on the feasible domain:

\begin{aligned} V(x) &= x\,(12-2x)^2 = 144x - 48x^2 + 4x^3 \\ V'(x) &= 144 - 96x + 12x^2 = 12\,(x^2 - 8x + 12) = 12\,(x-2)(x-6) \\ V'(x) &= 0 \;\Longrightarrow\; x = 2 \ \text{or}\ x = 6 \\ x &= 2 \quad (x=6 \text{ is excluded: it removes the whole box}) \\ V(2) &= 2\,(12-4)^2 = 2\cdot 8^2 = 128 \ \text{in}^3. \end{aligned}

Cutting $2$ -inch corners gives the maximum volume, $128$ cubic inches.

Summary

$f' > 0 \Rightarrow$ increasing, $f' < 0 \Rightarrow$ decreasing; extrema occur only at critical points (where $f'=0$ or is undefined), and a function switches direction only by crossing one.
The second derivative gives concavity: $f'' > 0 \Rightarrow$ concave up (slope rising), $f'' < 0 \Rightarrow$ concave down (slope falling).
Inflection points are where concavity switches and a tangent exists; look for them where $f''=0$ or is undefined, then verify the sign change.
Second-derivative test: at a point with $f'=0$ , positive $f''$ means a local min, negative $f''$ means a local max, and $f''=0$ is inconclusive.
Optimization: model the target quantity, use the constraint to reduce to one variable, set the derivative to zero, and verify — this finds exact maxima/minima (box volume, minimal-surface-area can with height $=2\times$ radius, largest inscribed rectangle at a $45^\circ$ point, Snell's-law least-time path).
For motion, velocity is the first derivative and acceleration is the second derivative of position, so the same increasing/decreasing and concavity analysis describes how an object speeds up or slows down.