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A short class with Sarah and her students. The other ways are under More.
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Let's find where f(x) = 2x³ − 9x² + 12x + 1 rises and falls. What is our first step after noting the domain?
Differentiate. f′(x) = 6x² − 18x + 12, which factorises to 6(x − 1)(x − 2).
Good. So the critical points are 1 and 2. How do we decide the sign of f′ between them?
Take a test value like 1.5. Then (x − 1) is positive and (x − 2) is negative, so f′ is negative.
Right, so f is decreasing on (1, 2). And outside that interval?
Both factors have the same sign, so f′ is positive. It is increasing on (−∞, 1) and (2, ∞).
Exactly. Always finish with a clear sentence giving both sets of intervals.
For f(x) = x³ − 3x² + 3x − 100, the derivative is 3(x − 1)². What do you notice?
It is never negative. It is zero only at x = 1.
So is f increasing on the whole real line, or does it stop at x = 1?
It keeps increasing. A single point where f′ is zero does not make the graph go down.
Correct. Now look at f(x) = 1/x. Its derivative is −1/x², always negative. Is f decreasing on its whole domain?
No, because f(−1) = −1 is smaller than f(1) = 1. It decreases on each side of zero separately.
Well spotted. Whenever the domain has a gap, state the intervals separately.
For f(x) = sin x + cos x on [0, 2π], where is f′(x) = cos x − sin x equal to zero?
When tan x = 1, so at π/4 and 5π/4.
Now test a value in each interval. Try 0, π/2 and 3π/2.
f′(0) = 1, f′(π/2) = −1 and f′(3π/2) = 1. So it rises, then falls between π/4 and 5π/4, then rises again.
Good. For x e^(−x), the derivative is e^(−x)(1 − x). Why can we ignore the exponential factor when finding the sign?
Because e^(−x) is always positive, so only (1 − x) decides the sign.
f′(x) > 0 on an interval means f is increasing there; f′(x) < 0 means f is decreasing.
points where f′(x) = 0 or does not exist; they divide the domain into intervals for testing.
one test value in each interval, or the signs of the factors, gives the sign of f′ on that interval.
if f′(x) ≥ 0 and is zero only at isolated points, f is still increasing, as with x³.
test only values inside the domain, and state intervals separately where the domain has a gap.
f(x) = 2x³ − 9x² + 12x + 1 has f′(x) = 6(x − 1)(x − 2), so f is increasing on (−∞, 1) and (2, ∞) and decreasing on (1, 2).
f(x) = x³ − 3x² + 3x − 100 has f′(x) = 3(x − 1)², which is zero only at x = 1, so f is increasing on the whole real line.
f(x) = x e^(−x) has f′(x) = e^(−x)(1 − x), so f is increasing on (−∞, 1) and decreasing on (1, ∞).
f(x) = 1/x has f′(x) = −1/x² < 0, so f is decreasing on (−∞, 0) and on (0, ∞) separately, but not on its whole domain.
Find where f-dash is nought, split the line as taught; test each part for plus or minus, and the rise and fall are caught.
A trekker writes down the slope of the path at a few milestones. Wherever the slope is positive she is climbing, and wherever it is negative she is going down. She only needs to check once between milestones, because the path never changes direction without levelling off first.
Plus means up and minus means down: read the sign of f′ like a road sign on a hill.
D.C.T.C. - Domain, Critical points, Test values, Conclusion.
Term
If f′(x) = 6(x − 1)(x − 2), where is f decreasing?
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Is f(x) = x³ increasing on R even though f′(0) = 0?
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What should you write down before differentiating?
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Why is one test value enough for each interval?
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Where is x e^(−x) increasing?
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A point in the domain where f′(x) = 0 or where f′(x) does not exist.
A number line marked with the critical points, showing whether f′(x) is positive or negative in each interval.
A convenient number chosen inside an interval to find the sign of f′(x) on that interval.
A function for which x₁ < x₂ in an interval always gives f(x₁) < f(x₂).
A function for which x₁ < x₂ in an interval always gives f(x₁) > f(x₂).
The set of all input values for which a function is defined.
A function that is either increasing throughout an interval or decreasing throughout it.
Worked examples showing, step by step, how to find the intervals on which a function is increasing or decreasing by studying the sign of its derivative.
This lesson applies the derivative test to a range of functions. It sets out a reliable method (find the domain, differentiate, locate the critical points, test the sign of f′(x) in each interval and conclude) and then uses it on a cubic, on a function whose derivative touches zero without changing sign, on trigonometric and exponential functions, and on functions where the domain needs care, such as x − log x and 1/x. Each example is worked in full so that you can follow the reasoning and copy the layout in your own answers.
To find where a function f is increasing or decreasing:
A continuous f′ cannot change sign without passing through zero, so one test value is enough for each interval.
Find the intervals in which f(x) = 2x³ − 9x² + 12x + 1 is increasing or decreasing.
So f is increasing on (−∞, 1) and (2, ∞), and decreasing on (1, 2). As a check, f(1) = 6 and f(2) = 5: the graph climbs to 6, dips to 5 and then climbs again.
Show that f(x) = x³ − 3x² + 3x − 100 is increasing on R.
A derivative that is zero only at isolated points, and positive everywhere else, does not stop the function from increasing. Indeed f(x) = (x − 1)³ − 99, which clearly rises as x rises. So f is increasing on R.
Find the intervals in which f(x) = sin x + cos x, 0 ≤ x ≤ 2π, is increasing or decreasing.
So f is increasing on [0, π/4) and (5π/4, 2π], and decreasing on (π/4, 5π/4).
For f(x) = x e^(−x), the product rule gives f′(x) = e^(−x) − x e^(−x) = e^(−x)(1 − x). Since e^(−x) > 0 for every x, the sign of f′(x) is the sign of (1 − x). So f is increasing on (−∞, 1) and decreasing on (1, ∞).
For f(x) = x − log x (logarithm to base e), the domain is x > 0. Then f′(x) = 1 − 1/x = (x − 1)/x. For 0 < x < 1 the numerator is negative and the denominator positive, so f′(x) < 0 and f is decreasing on (0, 1). For x > 1, f′(x) > 0 and f is increasing on (1, ∞). Values x ≤ 0 are never tested, because log x is not defined there.
A warning example is f(x) = 1/x. Here f′(x) = −1/x² < 0 for every x ≠ 0, so f is decreasing on (−∞, 0) and on (0, ∞) separately. It is not decreasing on its whole domain, because −1 < 1 but f(−1) = −1 is less than f(1) = 1. When the domain has a gap, always state the intervals separately.