Monotonicity and extrema: reading a function off the sign of f′

Monotonicity & extrema

How does the sign of f′(x) tell you where f rises, where it falls, and where it turns?

Drag a point along the cubic f(x) = x³ − 3ax and watch its tangent turn green, red or flat while the graph of f′ is drawn beneath. See why the extrema sit where f′ changes sign, and why f′ = 0 alone is not enough.

Drag the ringed points on the board.

High school
University
Sign of the derivative
Intervals of increase
Local maximum and minimum
Stationary points

a

no extrema

two extrema

1.0

x_0

-2.0

f'(x) = 3x^2 - 3a

$f'(x_0) = 3x_0^2 - 3a$: 9.00
$f(x_0)$: -2.00
$\text{local max } f(-\sqrt a)$: 2.00
$\text{local min } f(\sqrt a)$: -2.00

At P the function isrising (f′ > 0)

Extremaone maximum, one minimum

Guided tour

1Where f′ > 0, f rises
Take $f(x)=x^3-3x$ ( $a=1$ ) and put $P$ at $x_0=-2$ . The tangent slopes up: $f'(-2)=3\cdot4-3=9>0$ , and the blue graph of $f'$ is above the axis. On $(-\infty,-1)$ the curve climbs.
2Where f′ < 0, f falls
Move $P$ to $x_0=0$ : $f'(0)=-3<0$ , the tangent turns red and points down. Between the roots of $f'$ , on $(-1,1)$ , $f'$ is negative and the curve falls. That red stretch on the axis is the interval of decrease.
3The turn is where f′ changes sign
At $x_0=-1$ the tangent is flat: $f'(-1)=0$ . Left of it $f'>0$ , right of it $f'<0$ , so the curve stops rising and starts falling: a local maximum, $f(-1)=2$ . At $x=1$ the sign goes from $-$ to $+$ : a local minimum, $f(1)=-2$ .
4f′ = 0 is not enough
Set $a=0$ , so $f(x)=x^3$ . At $x_0=0$ the tangent is flat, $f'(0)=0$ , but $f'=3x^2$ is positive on both sides: the curve keeps rising and there is no extremum. For $a<0$ , $f'$ never reaches $0$ at all.

Explanation

1
The sign test
On an interval:
- $f'(x)>0$ everywhere $\Rightarrow$ $f$ is increasing there;
- $f'(x)<0$ everywhere $\Rightarrow$ $f$ is decreasing there.
Why: for any $x_1<x_2$ in the interval, the mean value theorem gives a point $\xi$ between them with $f(x_2)-f(x_1)=f'(\xi)(x_2-x_1)$ . If $f'>0$ throughout, the right side is positive, so $f(x_2)>f(x_1)$ . In pictures: some tangent between the two points is parallel to the chord, so the chord can only slope the way the tangents do. To find the intervals, solve $f'(x)=0$ , cut the number line at the roots, and test the sign of $f'$ on each piece.
2
Local extrema
A local maximum at $x_0$ means $f(x_0)$ beats every nearby value. If $f'$ changes from $+$ to $-$ at $x_0$ , $f(x_0)$ is a local maximum; from $-$ to $+$ , a local minimum; no change of sign, no extremum.
For $f(x)=x^3-3ax$ with $a>0$ : $f'(x)=3(x^2-a)=0$ at $x=\pm\sqrt a$ , giving the maximum $f(-\sqrt a)=2a\sqrt a$ and the minimum $f(\sqrt a)=-2a\sqrt a$ .
3
Stationary is not the same as extreme
A point with $f'(x_0)=0$ is a stationary point. Every extremum of a differentiable function is stationary, but not every stationary point is an extremum: $y=x^3$ is flat at $0$ and still increasing. $f'(x_0)=0$ is necessary; the change of sign is what makes it sufficient.
4
Extremum versus maximum
A local maximum need not be the largest value overall: on $[-1,3]$ the cubic $x^3-3x$ has a local maximum $2$ at $x=-1$ , but $f(3)=18$ is bigger. On a closed interval, compare the extrema with the values at the two endpoints; the largest of them is the maximum.

Check yourself

Q01

On which interval is $f(x)=x^3-12x$ decreasing?

Q02

If $f'(x_0)=0$ , then $x_0$ is…

Q03

What is the local minimum value of $f(x)=x^3-3x$ ?

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