B. Derivatives and Their Applications · Curve Sketching: Connecting f, f′ and f″

Intervals of Increase and Decrease, and Critical Numbers

B1.3 B1.5About 70 minutes

Learning goals

I can find critical numbers where f′(x) = 0 or f′(x) does not exist.
I can build a sign chart for f′ from its factors.
I can state the intervals where a polynomial or simple rational function is increasing or decreasing.

Words to know

critical number
A number c in the domain of f where f′(c) = 0 or f′(c) does not exist.

critical point
The point (c, f(c)) for a critical number c.

sign chart
A number line or table showing the sign of each factor of f′, and of f′ itself, on each interval between critical numbers.

test value
Any number chosen inside an interval to find the sign of f′ on that whole interval.

From pictures to algebra

In the last lesson you read the sign of f′ from a graph. But usually you have an equation and no graph. Can you tell exactly where f rises and falls without plotting a single point? Yes: find f′, find where it can change sign, and test each piece.

This is like planning a hike on a trail map. The places where the trail goes flat are the only places where it can switch from climbing to descending. Find those, and you know the shape of the whole trail.

The theorem behind it

For a function f that is differentiable on an interval:

If f′(x) > 0 for every x in the interval, then f is increasing on the interval.
If f′(x) < 0 for every x in the interval, then f is decreasing on the interval.

(The proof uses a result called the Mean Value Theorem, which you will see in first-year university calculus.)

Where can f′ change sign? Only at x-values where f′(x) = 0 or where f′(x) does not exist. Between such values, a continuous f′ keeps one sign, so one test value tells you the sign of the whole interval.

Critical numbers

A critical number of f is a number c in the domain of f where either f′(c) = 0 or f′(c) does not exist. The point (c, f(c)) is a critical point.

Critical numbers are the only places where a function can turn around. They are the borders in your sign chart. Also put in the chart any x-value not in the domain (such as a vertical asymptote), because the function can also change behaviour there.

Sign chart method for intervals of increase and decrease

Find f′(x) and factor it completely.
Find the critical numbers (f′ = 0 or f′ undefined) and any x-values not in the domain.
Mark them on a number line, in order. They split it into intervals.
Find the sign of each factor in each interval (or substitute one test value into f′).
f′ > 0 → f increasing; f′ < 0 → f decreasing.
Write the intervals with open brackets, for example (−1, 2). (Some textbooks include endpoints; at a single point a function is neither increasing nor decreasing, so open intervals avoid the question.)

Worked example

Example 1: A cubic

Find the intervals of increase and decrease for f(x) = 2x³ − 3x² − 12x + 1.

Differentiate: f′(x) = 6x² − 6x − 12.
Factor: f′(x) = 6(x² − x − 2) = 6(x + 1)(x − 2).
f′ is a polynomial, so it exists everywhere. Critical numbers: x = −1 and x = 2.
Test x = −2: f′(−2) = 6(−1)(−4) = 24 > 0. Test x = 0: f′(0) = 6(1)(−2) = −12 < 0. Test x = 3: f′(3) = 6(4)(1) = 24 > 0.
Find the critical points: f(−1) = −2 − 3 + 12 + 1 = 8 and f(2) = 16 − 12 − 24 + 1 = −19.

f is increasing on (−∞, −1) and (2, ∞) and decreasing on (−1, 2). It turns at (−1, 8) and (2, −19).

Each factor changes sign only at its own zero. Multiply the signs in each column (the constant 6 is positive) to get the sign of f′, and read the arrows: up where f′ > 0, down where f′ < 0.

The graph confirms the chart: increasing for x < −1, decreasing on the shaded interval −1 < x < 2, and increasing for x > 2.

Worked example

Example 2: A critical number that is not a turning point

Find the intervals of increase and decrease for f(x) = x⁴ − 4x³.

f′(x) = 4x³ − 12x² = 4x²(x − 3).
Critical numbers: 4x² = 0 gives x = 0; x − 3 = 0 gives x = 3.
The factor 4x² is never negative. It is 0 at x = 0 and positive everywhere else.
For x < 0: 4x² > 0 and x − 3 < 0, so f′ < 0. For 0 < x < 3: still 4x² > 0 and x − 3 < 0, so f′ < 0. For x > 3: both positive, so f′ > 0.
The sign does not change at x = 0: the graph flattens for an instant and keeps falling.
f(3) = 81 − 108 = −27.

f is decreasing on (−∞, 3) (you may also write (−∞, 0) and (0, 3)) and increasing on (3, ∞). The lowest point is (3, −27); at (0, 0) the tangent is horizontal but the function keeps decreasing.

Squared factors

A factor raised to an even power, like x² or (x − 1)⁴, never changes sign. A factor raised to an odd power, like (x − 3) or (x + 2)³, changes sign at its zero. Spotting this saves you test values.

Worked example

Example 3: A rational function

Find the intervals of increase and decrease for f(x) = x + 4/x.

Rewrite: f(x) = x + 4x⁻¹. Then f′(x) = 1 − 4x⁻² = 1 − 4/x².
Use a common denominator: f′(x) = (x² − 4)/x² = (x − 2)(x + 2)/x².
f′ = 0 when x = ±2. f′ is undefined at x = 0, but 0 is not in the domain of f, so it is not a critical number. It still goes in the chart.
The denominator x² is positive for x ≠ 0, so f′ has the sign of x² − 4: positive for |x| > 2, negative for |x| < 2 (x ≠ 0).
f(−2) = −2 − 2 = −4 and f(2) = 2 + 2 = 4.

f is increasing on (−∞, −2) and (2, ∞) and decreasing on (−2, 0) and (0, 2). Do not write (−2, 2): the function is not defined at 0, and f(−1) = −5 is actually less than f(1) = 5.

x = 0 is not in the domain, so it is not a critical number, but it still splits the chart. The function decreases on (−2, 0) and on (0, 2), as two separate intervals.

The graph of y = x + 4/x has a vertical asymptote at x = 0. It rises to (−2, −4), falls on both sides of the asymptote, and rises again after (2, 4).

Worked example

Example 4: A context

A small maple-syrup producer models daily profit (in hundreds of dollars) by P(x) = −x³ + 6x² + 15x − 20, where x is the number of hundreds of litres produced, 0 ≤ x ≤ 8. For which production levels is profit increasing?

P′(x) = −3x² + 12x + 15 = −3(x² − 4x − 5) = −3(x − 5)(x + 1).
Critical numbers: x = 5 and x = −1. Only x = 5 is in the domain 0 ≤ x ≤ 8.
On 0 < x < 5: test x = 1: P′(1) = −3(−4)(2) = 24 > 0. On 5 < x < 8: test x = 6: P′(6) = −3(1)(7) = −21 < 0.
P(5) = −125 + 150 + 75 − 20 = 80.

Profit increases for 0 < x < 5 and decreases for 5 < x < 8. Producing 500 L gives the highest profit, $8000, in this model. (You will solve many problems like this in Applications of Derivatives.)

Just for fun

A critical number walks into a sign chart and says, "Don't worry, I'm not judging anyone." The chart replies, "You're critical, but only of the slope."

Hands-on activity

Test-value relay

You need: a pencil, paper, a calculator and a partner (a parent, sibling or friend), or a timer if you work alone.

On separate slips, write these derivatives: f′(x) = (x − 1)(x + 4), f′(x) = x²(x − 2), f′(x) = −(x + 3)(x − 3), f′(x) = (x − 1)/(x + 2)².
Your partner picks a slip and calls out a test value of x. You say only "up" or "down" for f at that x, using the signs of the factors, without multiplying out.
After five calls, draw the full sign chart for that slip and check every call.
Swap roles.

What to look for: for (x − 1)(x + 4), f goes up, down, up with borders −4 and 1. For x²(x − 2), down, down, up (no change at 0). For −(x + 3)(x − 3), down, up, down. For (x − 1)/(x + 2)², the denominator is positive, so f follows the sign of x − 1: down, down, up, with borders −2 (where f′ is undefined) and 1.

Your turn

Find the critical numbers of f(x) = x³ − 12x + 4.

Find the intervals of increase and decrease for f(x) = x³ − 12x + 4.

Find the intervals of increase and decrease for f(x) = x² − 6x + 1.

Find the intervals of increase and decrease for f(x) = −x³ + 3x².

Explain why f(x) = x³ + 2x has no critical numbers and state where it increases.

Find the intervals of increase and decrease for f(x) = (x − 1)⁴.

Find the intervals of increase and decrease for f(x) = 1/(x − 2).

More practice: warm-up

f′(x) = (x − 4)(x + 1). What are the critical numbers?

f′(5) = −2. Is f increasing or decreasing near x = 5 (assuming f′ is continuous)?

Does the factor (x + 3)² ever change sign?

f(x) = 5x − 7. Where is f increasing?

True or false: a critical number must make f′ equal to 0.

More practice: on your own

Find the intervals of increase and decrease for f(x) = x⁴ − 8x².

Find the intervals of increase and decrease for f(x) = x³ − 3x² + 3x.

Find the intervals of increase and decrease for f(x) = x/(x + 1).

Find the intervals of increase and decrease for f(x) = x + 9/x.

Find the critical numbers of f(x) = x^(2/3)(x − 5) (that is, x to the power two-thirds, times x − 5).

Challenge

Find the values of k for which f(x) = x³ + kx has (a) two critical numbers and (b) none.

Show that f(x) = x⁵ + x³ + x − 7 is increasing everywhere and explain why it has exactly one real zero.

For f(x) = x²/(x − 1), find the intervals of increase and decrease.

Check your understanding

Quiz

What are the critical numbers of f(x) = x³ − 27x?

f′(x) = (x − 2)²(x + 1). On which interval is f decreasing?

Which statement about f(x) = 1/x is true?

A critical number of f is a number c in the domain of f where:

For f(x) = x³, what happens at the critical number x = 0?

Where is f(x) = x² − 10x increasing?

For parents and teachers

Big idea (B1.3, B1.5): to find where a function rises and falls, find f′, factor it, and make a sign chart. The only places f′ can change sign are its zeros and the x-values where it does not exist.

How to help: ask your teen to explain one sign chart out loud, column by column: "This factor is negative here because...". Explaining the chart catches most sign errors.

Common difficulties:

Factoring errors in f′ (check by expanding again).
Forgetting x-values that are not in the domain, such as vertical asymptotes, and writing one interval across them.
Assuming every critical number is a turning point (x³ and x⁴ − 4x³ show otherwise).
Mixing up zeros of f (x-intercepts) with zeros of f′ (critical numbers).