Reading Graphs with Derivatives

Turn rates into a graph description

The sign of $f'$ controls local increase or decrease on intervals. A positive derivative throughout an interval implies increasing behavior; a negative derivative implies decreasing behavior. Critical numbers are inputs in the domain where $f'=0$ or the derivative does not exist.

A critical number is a candidate, not a verdict. A change in $f'$ from positive to negative produces a local maximum; negative to positive produces a local minimum. No sign change can mean neither. For instance, $x^3$ increases through its stationary point at zero.

Visual guide

VISUAL GUIDEDerivative sign locates turning points

f(x) = x³ − 3x
f′(x) = 3x² − 3

For f(x) = x³ − 3x, f′ = 3x² − 3 changes sign at ±1. Positive derivative means increasing; negative means decreasing. The left turning point is a maximum and the right one a minimum.

Worked example: keep local and absolute separate

Let $f(x)=x^3-3x$ on $[-2,2]$ . Its derivative is $3(x-1)(x+1)$ , so the interior critical numbers are $-1$ and $1$ . The derivative is positive before $-1$ , negative between $-1$ and $1$ , and positive after $1$ . Thus $-1$ is a local maximum and $1$ a local minimum.

For absolute extrema, evaluate every critical candidate and both endpoints:

f(-2)=-2,\quad f(-1)=2,\quad f(1)=-2,\quad f(2)=2.

The absolute maximum is $2$ , attained at $-1$ and $2$ . The absolute minimum is $-2$ , attained at $-2$ and $1$ . A closed-interval problem can have tied answers and endpoint extrema.

Concavity and inflection

The second derivative describes how the first derivative changes. Where $f''>0$ , tangent slopes increase and the graph is concave up; where $f''<0$ , it is concave down. For this example, $f''=6x$ , so the concavity changes at zero. The point $(0,0)$ is an inflection point even though $f'(0)=-3\ne0$ .

Solving $f''=0$ only produces candidates. For $x^4$ , the second derivative is $12x^2$ and does not change sign at zero, so there is no inflection there. A candidate outside the domain is not a point on the graph.

Worked example: a nondifferentiable minimum

On $[-1,2]$ , $|x|$ has a critical number at zero because its derivative does not exist there. Its value is zero, less than its values at either endpoint, so it is the absolute minimum. A search that only solves $f'=0$ misses it entirely.

For a stationary point where $f''$ exists, the second-derivative test is convenient: positive means a local minimum and negative a local maximum. Zero is inconclusive, so return to a sign chart or another argument.

Practice

Find the increasing and decreasing intervals of $x^2-4x$ .
Does $x^3$ have an extremum at zero?
Find the maximum and minimum of $x^2$ on $[-1,3]$ .

Show worked solutions

$f'=2(x-2)$ is negative below $2$ and positive above it, so the graph decreases then increases.
No. Its derivative $3x^2$ is positive on either side, so the function keeps increasing. Its concavity does change.
Check $x=0$ and the endpoints. The minimum is $0$ at zero; the maximum is $9$ at $3$ .