Test for increasing and decreasing
Walking uphill, your height rises with each step; walking downhill, it falls. The slope tells which way you are going.
On an interval, f is increasing if f′(x) > 0 and decreasing if f′(x) < 0. If f′(x) = 0 at only a few points, the function can still be increasing or decreasing on the whole interval.
Formula: f′(x) > 0 ⇒ increasing; f′(x) < 0 ⇒ decreasing
In the figure: The graph of f(x) = x² − 4x + 5 falls until x = 2 and rises afterwards, where f′(x) = 2x − 4 changes sign.
Rule 1: Read the sign of f′ — Find f′ and decide where it is positive or negative.
- Easy: f(x) = 3x + 5 has f′ = 3 > 0, so it is increasing everywhere.
- Medium: f(x) = x² − 4x has f′ = 2x − 4: increasing for x > 2, decreasing for x < 2.
- Exam-type: f(x) = x³ − 3x has f′ = 3(x − 1)(x + 1): increasing for x < −1 and x > 1, decreasing between −1 and 1.
Rule 2: Always increasing or decreasing — Show that f′ keeps one sign.
- Easy: f(x) = x³ + x has f′ = 3x² + 1 > 0 always.
- Medium: f(x) = e⁻ˣ has f′ = −e⁻ˣ < 0 always, so it is decreasing.
- Exam-type: f(x) = sin x on (0, π/2) has f′ = cos x > 0, so it is increasing there.
Do not confuse f with f′: the sign of f′ tells the direction of f.
Reading the sign of f(x) instead of the sign of f′(x).
Plus slope climbs, minus slope falls.
Practice questions
- 1Is f(x) = 5x − 7 increasing or decreasing?
- 2Is f(x) = −2x + 3 increasing or decreasing?
- 3Find where f(x) = x² − 6x is increasing.
- 4Find the intervals where f(x) = x² + 2x is increasing or decreasing.
- 5Find the intervals where f(x) = x³ − 12x is increasing or decreasing.
- 6Find the intervals where f(x) = 2x³ − 9x² + 12x + 1 is increasing or decreasing.
- 7Show that f(x) = x³ + 3x is increasing on the real numbers.
- 8Is f(x) = cos x increasing or decreasing on (0, π)?
Show answers (try them yourself first)
- 1Increasing, since f′ = 5 > 0
- 2Decreasing, since f′ = −2 < 0
- 3f′ = 2x − 6 > 0 for x > 3
- 4f′ = 2x + 2: increasing for x > −1, decreasing for x < −1
- 5f′ = 3(x − 2)(x + 2): increasing for x < −2 and x > 2; decreasing for −2 < x < 2
- 6f′ = 6(x − 1)(x − 2): increasing for x < 1 and x > 2; decreasing for 1 < x < 2
- 7f′ = 3x² + 3 > 0 for every x
- 8f′ = −sin x < 0 on (0, π), so it is decreasing
Finding the intervals with a sign chart
A hiker's map shows where the path goes up and where it goes down. A sign chart for f′ is the same map for a function.
To find the intervals, solve f′(x) = 0 to get the critical points, mark them on a number line, and check the sign of f′ in each part. Then read off the intervals of increase and decrease.
Formula: Critical points: f′(x) = 0
Rule 1: Critical points and signs — Solve f′ = 0, then test one number in each interval.
- Easy: f(x) = x² − 6x + 5 has f′ = 2x − 6 = 0 at x = 3.
- Medium: f(x) = x³ − 3x² has f′ = 3x (x − 2): plus for x < 0, minus for 0 < x < 2, plus for x > 2.
- Exam-type: f(x) = x + 1/x for x > 0 has f′ = 1 − 1/x²: minus for 0 < x < 1, plus for x > 1.
Rule 2: Same sign on both sides — A zero of f′ does not always change direction.
- Easy: f(x) = (x − 1)³ has f′ = 3(x − 1)² ≥ 0, so it is increasing on all of R.
- Medium: f(x) = x e⁻ˣ has f′ = e⁻ˣ (1 − x): increasing for x < 1, decreasing for x > 1.
- Exam-type: f(x) = sin x + cos x on (0, π/2): f′ = cos x − sin x is positive up to π/4 and negative after.
A critical point splits the number line into regions in which f′ keeps one sign.
Testing the sign only at the critical points instead of between them.
Test a number between the critical points, not at them.
Practice questions
- 1Find the critical point of f(x) = x² − 8x + 3.
- 2Find the critical points of f(x) = x³ − 3x.
- 3Find the intervals of increase and decrease of f(x) = x³ − 3x².
- 4Find the intervals of increase and decrease of f(x) = x + 1/x for x > 0.
- 5Find the intervals of increase and decrease of f(x) = x e⁻ˣ.
- 6Is f(x) = (x − 1)³ increasing on the real numbers?
- 7Find where f(x) = sin x + cos x is increasing or decreasing on (0, π/2).
- 8Is f(x) = x⁴ increasing on the real numbers?
Show answers (try them yourself first)
- 1f′ = 2x − 8 = 0, so x = 4
- 2f′ = 3x² − 3 = 0, so x = 1 and x = −1
- 3f′ = 3x(x − 2): increasing for x < 0 and x > 2; decreasing for 0 < x < 2
- 4Decreasing on (0, 1), increasing on (1, ∞)
- 5Increasing for x < 1, decreasing for x > 1
- 6Yes, f′ = 3(x − 1)² ≥ 0 and it is zero only at x = 1
- 7Increasing on (0, π/4), decreasing on (π/4, π/2)
- 8No: f′ = 4x³ is negative for x < 0; it decreases on (−∞, 0) and increases on (0, ∞)