This note links the sign of the derivative to whether a function is going up or down. You will find the intervals where a function is increasing or decreasing and solve problems where a parameter must make a function increase everywhere. It is the first step towards locating maximum and minimum points.
Key ideas
The derivative \(f'(x)\) is the gradient of the curve. Reading the gradient tells you the direction of the curve as you move from left to right.
- If \(f'(x)\gt 0\), the gradient is positive and \(f\) is increasing at that \(x\).
- If \(f'(x)\lt 0\), the gradient is negative and \(f\) is decreasing at that \(x\).
- If \(f'(x)=0\), the curve is momentarily flat. This is a stationary point.
Increasing and decreasing on an interval
A function is increasing on an interval if \(f'(x)\ge 0\) throughout it, so that larger inputs give larger (or equal) outputs. It is decreasing on an interval if \(f'(x)\le 0\) throughout it. A single point with zero gradient inside an interval does not spoil this. For instance \(f(x)=x^3\) has \(f'(0)=0\), yet it is increasing for all \(x\).
Method for finding intervals
- Differentiate to find \(f'(x)\).
- Solve \(f'(x)=0\) to find where the gradient could change sign. Also note any \(x\)-values where \(f\) or \(f'\) is undefined.
- Make a sign table or sketch \(f'\), testing a value in each interval between those \(x\)-values.
- State the intervals clearly, using inequalities such as \(x\lt 1\) or \(1\lt x\lt 3\).
Because \(f'\) changes sign only at a zero or at a point where it is undefined, one test value in each interval is enough.
Increasing everywhere
Sometimes a question asks for the values of a constant for which a function is increasing for all \(x\). Then \(f'(x)\ge 0\) for every \(x\). If \(f'\) is a quadratic with a positive leading coefficient, this happens when its graph never goes below the \(x\)-axis, so the discriminant satisfies \(\Delta\le 0\).
Why the turning points separate the intervals
A continuous derivative cannot jump from positive to negative without passing through zero. So between two neighbouring zeros of \(f'\), the function either rises all the way or falls all the way. This is why solving \(f'(x)=0\) gives the boundaries of every interval. The same idea applies when \(f'\) is undefined at some \(x\), such as \(x=0\) in \(f(x)=\dfrac1x\). There the interval has to be split, and the function is decreasing on \(x\lt 0\) and on \(x\gt 0\) separately.
Increasing at a point
To decide whether \(f\) is increasing at \(x=a\), calculate \(f'(a)\) and look at its sign.
Reading graphs
You may be given the graph of \(f'\) instead of a formula. Where it lies above the \(x\)-axis, \(f\) is increasing. Where it lies below the axis, \(f\) is decreasing. The zeros of \(f'\) are where \(f\) has flat tangents.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(f'(x)\gt 0\) | \(f\) is increasing: the graph rises as \(x\) increases. |
| \(f'(x)\lt 0\) | \(f\) is decreasing: the graph falls as \(x\) increases. |
| \(f'(x)=0\) | Stationary point: the tangent is horizontal. |
| Sign table | A table of the sign of \(f'\) in each interval between its zeros. |
| \(\Delta=b^2-4ac\) | Discriminant of \(ax^2+bx+c\). If \(a\gt 0\) and \(\Delta\le 0\), then \(ax^2+bx+c\ge 0\) for all \(x\). |
Worked examples
Example 1 The profit in thousands of euros from selling \(x\) hundred items is modelled by \(f(x)=2x^3-3x^2-12x+5\). Find the values of \(x\) for which \(f\) is increasing and the values for which it is decreasing.
Differentiate: \(f'(x)=6x^2-6x-12\).
Factorise: \(f'(x)=6(x^2-x-2)=6(x-2)(x+1)\).
The zeros of \(f'\) are \(x=-1\) and \(x=2\).
Test \(x=-2\): \(6(-4)(-1)=24\gt 0\). Test \(x=0\): \(6(-2)(1)=-12\lt 0\). Test \(x=3\): \(6(1)(4)=24\gt 0\).
So \(f'\gt 0\) for \(x\lt -1\), \(f'\lt 0\) for \(-1\lt x\lt 2\), and \(f'\gt 0\) for \(x\gt 2\).
Answer: increasing for \(x\lt -1\) and \(x\gt 2\); decreasing for \(-1\lt x\lt 2\).
Example 2 For \(x\gt 0\), the cost per unit of a component is \(f(x)=x+\dfrac{9}{x}\). Find where \(f\) is decreasing and where it is increasing.
Write \(f(x)=x+9x^{-1}\).
Then \(f'(x)=1-9x^{-2}=1-\dfrac{9}{x^2}\).
Solve \(f'(x)=0\): \(x^2=9\), so \(x=3\) (the value \(x=-3\) is outside \(x\gt 0\)).
Test \(x=1\): \(f'(1)=1-9=-8\lt 0\). Test \(x=6\): \(f'(6)=1-\tfrac{9}{36}=0.75\gt 0\).
Answer: decreasing for \(0\lt x\lt 3\); increasing for \(x\gt 3\).
Example 3 The function \(f(x)=x^3+ax^2+3x\) is increasing for all real \(x\). Find the possible values of \(a\).
\(f'(x)=3x^2+2ax+3\).
We need \(f'(x)\ge 0\) for all \(x\). The leading coefficient 3 is positive, so we need the discriminant \(\Delta\le 0\).
\(\Delta=(2a)^2-4(3)(3)=4a^2-36\).
Solve \(4a^2-36\le 0\): \(a^2\le 9\), so \(-3\le a\le 3\).
Answer: \(-3\le a\le 3\).
Common errors
- Common error: deciding the direction from the value of \(f(x)\) instead of the sign of \(f'(x)\). A function can be negative and still increasing.
- Common error: forgetting to test a value in each interval and assuming the signs simply alternate.
- Common error: mixing up the intervals, for example giving \(x\lt -1\) or \(x\gt 2\) as the decreasing part in Example 1. Check each interval against its test value.
- Common error: ignoring the domain. In Example 2 the solution \(x=-3\) must be rejected because the question says \(x\gt 0\).
- Common error: using \(\Delta\gt 0\) when asked for \(f'(x)\ge 0\) for all \(x\). A quadratic that is never negative has \(\Delta\le 0\).
Exam tip: a sign table or a quick sketch of \(f'\) shows your reasoning clearly. Always finish with a sentence naming the intervals in terms of \(x\).
Check yourself
- Find the intervals on which \(f(x)=x^3-3x\) is decreasing.
Answer
\(f'(x)=3x^2-3=3(x-1)(x+1)\). This is negative between its zeros, so \(f\) is decreasing for \(-1\lt x\lt 1\). - Is \(g(x)=x^2-8x\) increasing or decreasing at \(x=5\)? Justify your answer.
Answer
\(g'(x)=2x-8\), so \(g'(5)=2\gt 0\). The function is increasing at \(x=5\). - Find the values of \(k\) for which \(f(x)=x^3+kx\) is increasing for all \(x\).
Answer
\(f'(x)=3x^2+k\). Since \(3x^2\ge 0\), we need \(k\ge 0\). (If \(k\lt 0\), then \(f'\) is negative near \(x=0\).) - The graph of \(f'\) is a straight line crossing the \(x\)-axis at \(x=4\), going from below the axis to above it. Describe where \(f\) is increasing and decreasing.
Answer
\(f'\lt 0\) for \(x\lt 4\) and \(f'\gt 0\) for \(x\gt 4\). So \(f\) is decreasing for \(x\lt 4\) and increasing for \(x\gt 4\).