Absolute maximum and minimum values
In Year 12 Mathematical Methods (Queensland, QCAA), the absolute (or global) maximum and minimum of a continuous function on a closed interval \([a,b]\) are its greatest and least values over the whole interval. You find them by solving \(f'(x)=0\) for the stationary points inside \((a,b)\), evaluating \(f\) at each stationary point and at both endpoints \(a\) and \(b\), then comparing the \(y\)-values — the greatest is the absolute maximum and the least is the absolute minimum.
The absolute maximum of \(f\) on \([a,b]\) is the greatest value \(f(x)\) takes anywhere on the interval; the absolute minimum is the least. They are also called the global maximum and minimum.
This is different from a local maximum or minimum, which is only the highest or lowest point compared with nearby points. A local maximum need not be the absolute maximum.
Because the interval is closed and \(f\) is continuous, the absolute maximum and minimum are guaranteed to exist, and each occurs at one of just two kinds of place:
a stationary point inside the interval (where \(f'(x)=0\)), or an endpoint \(x=a\) or \(x=b\).
So the full list of candidate points is: every stationary point in \((a,b)\), together with the two endpoints. Compare \(f\) at all of them.
The stationary points come from setting the derivative to zero:
The absolute maximum and minimum on \([a,b]\) are then read off the candidate list — the stationary points \(x_1,x_2,\dots\) in \((a,b)\) together with the endpoints:
How to find the absolute maximum and minimum on \([a,b]\)
- Differentiate. Find \(f'(x)\) using the standard rules.
- Solve \(f'(x)=0\). Keep only the stationary \(x\)-values that lie inside the interval \((a,b)\); discard any outside it.
- Evaluate \(f\) at each stationary point. Substitute each into the original \(f(x)\) for its \(y\)-value.
- Evaluate \(f\) at both endpoints. Work out \(f(a)\) and \(f(b)\) — never skip this step.
- Compare the \(y\)-values. The greatest is the absolute maximum, the least is the absolute minimum. State the value, and where it occurs if asked.
| \(f'(x)\) | \(=\) | \(3x^{2}-3\) |
| \(3x^{2}-3\) | \(=\) | \(0\Rightarrow x=1\ (x=-1\ \text{outside})\) |
| \(f(1)\) | \(=\) | \(1-3+2=0\) |
| \(f(0)\) | \(=\) | \(2,\quad f(3)=20\) |
Compare \(0,2,20\): absolute maximum \(20\) at \(x=3\); absolute minimum \(0\) at \(x=1\).
| \(f'(x)\) | \(=\) | \(x^{3}-4x=x(x-2)(x+2)\) |
| \(f'(x)\) | \(=\) | \(0\Rightarrow x=0,2\ (x=-2\ \text{outside})\) |
| \(f(0)\) | \(=\) | \(3,\quad f(2)=-1\) |
| \(f(-1)\) | \(=\) | \(1.25,\quad f(3)=5.25\) |
The greatest is \(5.25\) at the endpoint \(x=3\) — not the local maximum \(3\).
\(f'(x)=e^{x}>0\), so there is no stationary point; the extremes are at the endpoints.
| \(f(0)\) | \(=\) | \(e^{0}=1\) |
| \(f(\ln 5)\) | \(=\) | \(e^{\ln 5}=5\) |
Absolute maximum \(5\) at \(x=\ln 5\); absolute minimum \(1\) at \(x=0\).
| \(f'(x)\) | \(=\) | \(\cos x-\sin x\) |
| \(\cos x-\sin x\) | \(=\) | \(0\Rightarrow \tan x=1\Rightarrow x=\dfrac{\pi}{4}\) |
| \(f\!\left(\tfrac{\pi}{4}\right)\) | \(=\) | \(\sqrt{2},\quad f(0)=1,\quad f(\pi)=-1\) |
Absolute maximum \(\sqrt{2}\) at \(x=\dfrac{\pi}{4}\); absolute minimum \(-1\) at \(x=\pi\).
Common pitfalls
Frequently asked questions
What is the difference between a local and an absolute maximum?
A local maximum is highest only compared with nearby points; an absolute (global) maximum is the highest value over the whole interval. On a closed interval it may be at a stationary point or an endpoint, so a local maximum is not automatically the absolute one.
How do you find the absolute maximum and minimum on a closed interval?
Solve \(f'(x)=0\) for the stationary points inside the interval, evaluate \(f\) at each of them and at both endpoints, then compare the \(y\)-values: greatest is the maximum, least is the minimum.
Why do you have to test the endpoints?
Because the extreme value can occur at an endpoint rather than a stationary point — especially when the function is still rising or falling at the end of the interval. Skipping the endpoints is the most common mistake.
What if there is no stationary point in the interval?
Then \(f\) is monotonic on the interval, so the absolute maximum and minimum are the two endpoint values. For instance \(e^{x}\) on \([0,\ln 5]\) has minimum \(1\) and maximum \(5\).
Do you report the \(x\)-value or the \(y\)-value?
The absolute maximum and minimum are \(y\)-values (outputs). The \(x\)-value tells you where they occur. Give the \(y\)-value when asked for the absolute maximum or minimum.
Does this work for \(e^{x}\), \(\ln x\) and trig functions?
Yes. The method is the same for any continuous function: differentiate, solve \(f'(x)=0\) in the interval, and compare the stationary values with the endpoint values. For example \(\sin x+\cos x\) on \([0,\pi]\) has maximum \(\sqrt{2}\) and minimum \(-1\).