Increasing, decreasing and stationary at a point

Using the sign of the derivative to decide whether a curve is increasing, decreasing or stationary at a point, finding where a function increases or decreases, and what this says about x-intercepts.

The sign of the derivative

The sign of f′ along the curve

f(x) = x³ − 3x² − 9x + 5. Drag P and watch the sign of the gradient.

Try this

Find the two places where the readout says "stationary".

Example 1●

For $f(x) = x^3 - 3x^2 - 9x + 5$, is the curve increasing or decreasing at $x = 2$ and at $x = 4$?

$f'(x) = 3x^2 - 6x - 9$.

$f'(2) = 12 - 12 - 9 = -9 < 0$, so it is decreasing at $x = 2$.

$f'(4) = 48 - 24 - 9 = 15 > 0$, so it is increasing at $x = 4$.

Where is a function increasing?

Example 2●●

Find where $f(x) = x^3 - 3x^2 - 9x + 5$ is increasing and where it is decreasing.

$f'(x) = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1)$.

$f'(x) > 0$ for $x < -1$ or $x > 3$, and $f'(x) < 0$ for $-1 < x < 3$.

f′ is an upright parabola with zeroes −1 and 3.

So $f$ is increasing for $x < -1$ or $x > 3$, and decreasing for $-1 < x < 3$.

Increasing everywhere

Example 3●●

Show that $f(x) = x^3 + x$ is increasing for all $x$, and hence that it has only one $x$-intercept.

$f'(x) = 3x^2 + 1$, and $3x^2 \ge 0$, so $f'(x) \ge 1 > 0$ for every $x$.

$f$ is increasing everywhere, so it crosses the $x$-axis at most once. Since $f(0) = 0$, the only intercept is $x = 0$.

A curve that always goes up can cross any horizontal line at most once.

Flashcards

$f'(a) > 0$

Increasing at $x = a$.

$f'(a) = 0$

Stationary at $x = a$.

$f'(x) > 0$ for all $x$

At most one $x$-intercept.

Practice

For $f(x) = x^3 - 3x^2 - 9x + 5$, find $f'(2)$.
For the same $f$, find $f'(4)$.
Where is $f(x) = x^3 - 3x^2 - 9x + 5$ decreasing?
Find both $x$-values where $f(x) = x^3 - 3x^2 - 9x + 5$ is stationary.
$f'(x) = 3x^2 + 1$. What can you conclude?
Where is $y = x^2$ decreasing?
A curve is stationary at $x = a$ when:
For $f(x) = 2x^3 - 6x$, find $f'(1)$.
Where is $f(x) = 2x^3 - 6x$ decreasing?