Increasing and Decreasing Functions (AA SL)
The derivative doesn't just give you a gradient at a single point - its sign tells you the whole shape of the graph. Where \(f'(x)\) is positive the curve climbs, where it's negative the curve falls, and this page shows you how to turn that into a solved inequality. It's part of the broader Differentiation topic.
21 questions on this sub-topic.
Reading the sign of \(f'(x)\)
Covered under IB syllabus reference SL5.2: increasing and decreasing functions, with the graphical interpretation of \(f'(x)>0\), \(f'(x)=0\) and \(f'(x)<0\).
Increasing
\(f\) is increasing on any interval where \(f'(x) > 0\).
The graph slopes upward from left to right - solve the inequality after differentiating.
Decreasing
\(f\) is decreasing on any interval where \(f'(x) < 0\).
The graph slopes downward from left to right. The endpoints of these intervals are where \(f'(x)=0\).
Need the full differentiation syllabus and GDC graphing tips for checking these intervals visually? See Differentiation.
Worked examples
Find the values of \(x\) for which \(f(x) = x^2 - 6x + 2\) is increasing.
Worked solution
\(f'(x) = 2x - 6.\) M1 A1
Set \(f'(x) > 0\): \(2x - 6 > 0.\) M1
\(\Rightarrow x > 3.\) A1
Find the interval on which \(f(x) = x^3 - 3x\) is decreasing.
Worked solution
A function decreases where its gradient is negative, so we examine \(f'\).
\(f'(x)=3x^2-3.\) A1
Step 2 - Solve \(f'(x)<0\).
\(3x^2-3<0\) M1
\(x^2<1\;\Rightarrow\;-1<x<1.\) A1
\(f\) is decreasing on \(-1<x<1\). R1
For \(f(x) = xe^{-x}\), find the interval on which \(f\) is increasing.
Worked solution
\(f'(x) = e^{-x} + x(-e^{-x})\) M1
\(= e^{-x}(1-x).\) A1
Since \(e^{-x} > 0\) for all \(x\), the sign of \(f'\) depends on \((1-x)\). M1
\(f'(x) > 0 \Rightarrow 1 - x > 0\) M1
\(\Rightarrow x < 1.\) A1
For \(f(x) = x^3 - 6x^2 + 9x + 1\):
(a) Find the interval on which \(f\) is increasing.
(b) Find the interval on which \(f\) is decreasing.
Worked solution
(a) Increasing: \(x < 1\) and \(x > 3.\) A1
(b) Decreasing: \(1 < x < 3.\) M1
\(1 < x < 3.\) A1 - Correct decreasing interval \(1<x<3\)
Common mistakes
- Testing the original function instead of the derivative. "Increasing" and "decreasing" describe the sign of \(f'(x)\), not the sign of \(f(x)\) - the question is always answered from the derivative.
- Flipping the inequality sign. Increasing means \(f'(x)>0\) and decreasing means \(f'(x)<0\) - it's easy to write these the wrong way round when working quickly.
- Losing a root when solving a quadratic inequality. For something like \(3x^2-3<0\), factorising to \(3(x-1)(x+1)<0\) and checking the sign either side of both roots avoids reporting only half the interval.
Ready to practise properly?
21 increasing-and-decreasing questions, marked instantly like the real exam.
Quick answers
How do I decide where a function is increasing or decreasing?
Find \(f'(x)\), then solve \(f'(x) > 0\) for the increasing interval and \(f'(x) < 0\) for the decreasing interval. The boundary points are where \(f'(x) = 0\).
What does f'(x) = 0 tell you about a function?
It marks a point where the gradient is momentarily zero - usually a local maximum, local minimum, or a stationary point of inflexion - and it separates the increasing and decreasing intervals.