Distinguish local and absolute extrema
A peak and the whole interval
The curve is the graph of \(f(x) = 2x^3 - 9x^2 + 12x\) on the interval from 0 to 3. The orange point is at the peak \(x = 1\), where \(f(1) = 5\).
The orange dashed line marks the height 5, and the window is the stretch of the \(x\)-axis between the two faint vertical lines. Inside the window, the region is green where the curve is higher than 5 and red where it is lower.
With a narrow window, the region is all red, so no nearby value is higher than 5, which makes 5 a local maximum.
Widen the window to cover the whole interval. Green appears near \(x = 3\), where the curve reaches 9, so 5 is not the absolute maximum, the highest value on the whole interval.
Near the point, or on the whole interval
Move the point to a top. Then change the width of the window.
If some narrow window around a top shows no green, no nearby value is higher, so the top is a local maximum. If a window over the whole interval shows no green, no value on the interval is higher, so the top is the absolute maximum. For a minimum, the same test uses red.
Notation
The value \(f(c)\) is a local maximum if \(f(c) \ge f(x)\) for every \(x\) in some open interval around \(c\).
\[\cy{f(c)} \ge f(x) \quad \text{for } x \text{ near } c\]An open interval around c has points on both sides of c, so a local maximum is never at an end point.
The value \(f(c)\) is the absolute maximum on the interval if \(f(c) \ge f(x)\) for every \(x\) in the interval.
\[\begin{aligned} &\cy{f(c)} \ge f(x) \\ &\text{for every } x \text{ in the interval} \end{aligned}\]For a minimum, reverse the inequality. A maximum or a minimum is also called an extremum.
\[f(c) \le f(x)\]An absolute extremum inside the interval is also local, because the interval contains an open interval around c. At an end point it is not local.
When it fails
The button ends the interval at 3 and widens the window at \(x = 1.1\) over all of it. The curve near 1.1 does not change, so its value is still a local maximum.
The window shows no green, because the higher top at 3.9 is outside the new interval. The same value is now also the absolute maximum.
Example 1: a double well
Classify the extreme values of \(g(x) = x^4 - 8x^2\) on \([-1, 3]\).
The derivative is 0 at \(x = -2\), 0 and 2, and \(-2\) is outside the interval.
\[g'(x) = 4x(x^2 - 4)\]Compare the values at the critical points, where the derivative is 0, and at the end points.
\[\begin{aligned} g(-1) &= -7 \\ g(0) &= 0 \\ g(2) &= -16 \\ g(3) &= 9 \end{aligned}\]The absolute maximum is 9, at the end point \(x = 3\). It is not local, because the interval has no points to the right of 3.
The absolute minimum is \(-16\) at \(x = 2\), and it is also local because 2 is inside the interval. The value 0 at \(x = 0\) is a local maximum only, because 9 is higher.
Example 2: an end point
Find the absolute maximum of \(r(x) = x^3 - 9x\) on \([-2, 4]\). Is it a local maximum?
The derivative \(3x^2 - 9\) is 0 at \(x = \pm\sqrt3\), and both are in the interval. Compare the values.
\[\begin{aligned} r(-2) &= 10 \\ r(-\sqrt3) &= 6\sqrt3 \approx 10.4 \\ r(\sqrt3) &= -6\sqrt3 \\ r(4) &= 28 \end{aligned}\]The largest value is 28, at \(x = 4\), so this is the absolute maximum.
A tempting wrong move: The point \(x = 4\) is also a local maximum, because no nearby value is higher.
A local maximum needs points on both sides of \(x = 4\). The interval stops at 4, so the window can reach only to the left.
The value \(r(4) = 28\) is the absolute maximum but not a local maximum.
Practice
Classify the values of \(s(x) = x + \dfrac{4}{x}\) at its critical point and at the end points of \([1, 5]\).
Move the point to 2, to 1 and to 5. Each time, compare a narrow window with a window that covers the whole interval.
Solution
The derivative is 0 at \(x = \pm 2\), and \(-2\) is outside the interval.
\[s'(x) = 1 - \dfrac{4}{x^2}\]Compare the values at the critical point and at the end points.
\[\begin{aligned} s(1) &= 5 \\ s(2) &= 4 \\ s(5) &= \dfrac{29}{5} \end{aligned}\]The value 4 at \(x = 2\) is the absolute minimum, and also local. The value \(\tfrac{29}{5}\) at \(x = 5\) is the absolute maximum, but not local because 5 is an end point. The value 5 at \(x = 1\) is neither, because the curve is lower at 2 and higher at 5.
Questions
The graph shows \(f\) on the interval from \(x = 0\) to \(x = 8\). It has peaks at \(x = 2\) and \(x = 6\) and a valley at \(x = 4\).
Which value is a local maximum but not the absolute maximum?
- \(f(6) = 5\)
- \(f(2) = 7\)
- \(f(4) = 3\)
- \(f(8) = 1\)
Answer
\(f(6) = 5\)
What is the absolute minimum value of \(f\) on \([0, 8]\)?
- 1
- 3
- There is none, because 8 is an end point.
Answer
1
For \(h(x) = x^2 - 4x + 1\) on \([0, 5]\), which statement is correct?
- \(h(5) = 6\) is the absolute maximum but not a local maximum.
- \(h(5) = 6\) is both the absolute maximum and a local maximum.
- \(h(0) = 1\) is a local maximum.
- \(h(2) = -3\) is a local minimum but not the absolute minimum.
Answer
\(h(5) = 6\) is the absolute maximum but not a local maximum.
Summary
A value is a local maximum when no nearby value is higher, and the absolute maximum when no value on the whole interval is higher.
This works because a narrow window shows what is nearby, and a window over the whole interval shows every value.
You use this when a problem asks for a local or an absolute extremum on a named interval.
This is one of five skills in topic 5.2 of Which & Why: AP Calculus.