IB Maths AI SLDifferentiationPaper 1 & 2f′(x) > 0 / < 0~6 min read
Increasing & Decreasing Functions
The sign of the derivative tells you which way a curve is heading. Where f′(x) is positive the function climbs; where it is negative it falls; where it is zero the curve is momentarily flat. This note uses that one idea to test a function at a point and to find the exact intervals where it increases or decreases.
📘 What you need to know
A function is increasing where f′(x) > 0 — the curve rises as x increases.
A function is decreasing where f′(x) < 0 — the curve falls as x increases.
A function is stationary where f′(x) = 0 — the curve is momentarily flat.
To test at a point, substitute the x-value into f′(x) and read off the sign.
To find an interval, solve the inequality f′(x) > 0 (increasing) or f′(x) < 0 (decreasing).
Most functions are a mix — increasing on some intervals, decreasing on others, stationary at the points between.
Increasing, decreasing and stationary
The sign of the derivative tells you which way a curve is heading. Where f′(x) is positive the function climbs as x increases; where it is negative the function falls; where it is zero the curve is momentarily flat.
Increasing, decreasing, stationaryf′(x) > 0 ⇒ increasing
f′(x) < 0 ⇒ decreasingf′(x) = 0 ⇒ stationarythe sign of the derivative tells you which way the curve is heading
Where f′(x) > 0 the curve climbs; where f′(x) < 0 it falls; where f′(x) = 0 it is momentarily flat — a stationary point.
A flat point — where f′(x) = 0 — is a stationary point. Most curves are a patchwork of increasing and decreasing stretches joined at stationary points.
Testing at a point
To decide whether a function is increasing or decreasing at one particular point, substitute that x-value into f′(x) and look at the sign of the answer.
A positive value means the function is increasing there; a negative value means it is decreasing. You only need the sign — the size of the number doesn’t change the conclusion.
Finding the intervals
More often you need the whole range of x-values over which a function increases or decreases. Set up an inequality from the derivative and solve it.
Finding the intervals
increasing: solve f′(x) > 0
decreasing: solve f′(x) < 0the solution is a range of x-values — an interval — not a single point
The answer is an interval — a range like x > 4, and sometimes two ranges. For a quadratic f′(x), factorise it or sketch it to read off where it is positive or negative.
🧭 Recipe — increasing and decreasing intervals
Differentiate to find f′(x).
To test at a point, substitute the x-value into f′(x): positive means increasing, negative means decreasing.
To find intervals, set up the inequality — f′(x) > 0 for increasing, f′(x) < 0 for decreasing.
Solve the inequality for x — factorise f′(x) if it helps.
State the interval(s), note any stationary points where f′(x) = 0, and respect any domain restriction.
Worked examples
WE 1
Increasing or decreasing at a point?
Determine whether f(x) = x2 − 6x + 1 is increasing or decreasing at the point where x = 1.
test at a point — substitute into f′(x)f′(x) = 2x − 6f′(1) = 2(1) − 6 = −4−4 < 0, so f is decreasing at x = 1only the sign matters — a negative derivative means the function is falling there.
WE 2
Testing at two points
For f(x) = x3 − 12x, state whether f is increasing or decreasing at (a) x = 1 and (b) x = 3.
f′(x) = 3x2 − 12(a) f′(1) = 3 − 12 = −9 < 0 ⇒ decreasing(b) f′(3) = 27 − 12 = 15 > 0 ⇒ increasingdecreasing at x = 1 · increasing at x = 3the same curve can be decreasing in one place and increasing in another.
WE 3
Finding an increasing interval
Find the values of x for which f(x) = x2 − 8x + 5 is an increasing function.
increasing means f′(x) > 0f′(x) = 2x − 82x − 8 > 0 ⇒ 2x > 8f is increasing for x > 4the answer is an interval — a whole range of x-values, not a single point.
WE 4
A decreasing interval in two parts
Find the values of x for which f(x) = 12x − x3 is a decreasing function.
decreasing means f′(x) < 0f′(x) = 12 − 3x212 − 3x2 < 0 ⇒ 3x2 > 12 ⇒ x2 > 4f is decreasing for x < −2 and for x > 2x2 > 4 splits into two intervals — the function decreases on both outer stretches.
WE 5
Increasing and decreasing intervals
Find the intervals on which f(x) = x3 − 3x2 is increasing and decreasing.
f′(x) = 3x2 − 6x = 3x(x − 2)stationary where f′(x) = 0: x = 0 and x = 2increasing (f′(x) > 0): x < 0 or x > 2decreasing (f′(x) < 0): 0 < x < 2increasing for x < 0 and x > 2 · decreasing for 0 < x < 2factorise the derivative — its roots split the x-axis into intervals; check the sign on each.
WE 6
Full question: a profit model
A firm’s monthly profit P (in thousand dollars) from producing x hundred items is modelled by P = x3 − 6x2 + 9x, for x ≥ 0. (a) Find dP/dx. (b) Find the values of x for which the profit is increasing. (c) State what happens to the profit between x = 1 and x = 3.
(a) differentiate the profit modeldP/dx = 3x2 − 12x + 9(b) increasing: dP/dx > 03x2 − 12x + 9 = 3(x − 1)(x − 3) > 0⇒ x < 1 or x > 3; with x ≥ 0: 0 ≤ x < 1 and x > 3(c) between x = 1 and x = 3: dP/dx at x = 2 is −3(a) 3x2−12x+9 · (b) 0 ≤ x < 1 and x > 3 · (c) the profit is decreasingfactorising the derivative makes the sign analysis quick — and always respect a restriction like x ≥ 0.
💡 Top tips
Increasing ⇔ f′(x) > 0, decreasing ⇔ f′(x) < 0 — it always comes back to the sign of the derivative.
To test at a point you need only the sign of f′ there, not its size.
To find intervals, solve an inequality — the answer is a range of x-values.
Factorise the derivative: its roots are the stationary points, and they split the x-axis into intervals to test.
A quick sketch of f′(x) — especially a parabola — shows at a glance where it is above or below the axis.
⚠ Common mistakes
Testing the sign of f(x) instead of f′(x) — it is the derivative that decides increasing or decreasing.
Giving a single x-value when the question asks for an interval.
Mishandling x2 > 4 — it gives x > 2 orx < −2, not just x > 2.
Calling a stationary point increasing — where f′(x) = 0 the function is stationary, not increasing.
Ignoring a domain restriction such as x ≥ 0 when stating the final interval.
Next up: Local Minimum & Maximum Points — pinning down those stationary points and deciding whether each is a peak or a trough. Everything in this topic rests on one habit: differentiate, then read the sign. Positive climbs, negative falls, zero is flat.
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