Check the Extreme Value Theorem conditions
A line that must touch the curve
The curve is the graph of \(f(x) = x e^{-x}\) from \(x = 0\) to \(x = 4\). It has no gap, and both end points belong to it.
The orange line is horizontal. Lower it from above, and it first touches the curve at \(x = 1\), at the height \(\tfrac1e\), which is about 0.37.
Raise the line from below, and it first touches the curve at the end point \(x = 0\), at height 0.
The first touch from above finds the highest value, the absolute maximum, and the first touch from below finds the lowest value, the absolute minimum.
A connected curve on a closed interval
The curve is connected from the left end to the right end, and both ends belong to it.
Lower the line from the top with the slider. It meets the curve at a first point, which can be a top, a corner or an end. Raise it from the bottom, and it again meets the curve at a first point.
In every case, a connected curve over a closed interval has an absolute maximum and an absolute minimum. The curve does not have to be smooth.
Notation
Suppose \(f\) is continuous on the closed interval \([a, b]\). Continuous means that the graph has no gap and no jump. Closed means that both end points belong to the interval.
The interval is also bounded, which means that it has a finite length.
Then \(f\) has an absolute maximum and an absolute minimum on \([a, b]\). This is the Extreme Value Theorem.
\[\begin{aligned} &f \text{ continuous on } [a, b] \\ &\Rightarrow \text{ absolute maximum on } [a, b] \\ &\text{ and absolute minimum on } [a, b] \end{aligned}\]The theorem does not ask for a derivative. At the corner \(f'\) does not exist, but both absolute extrema still exist.
The theorem says that the absolute extrema exist, not where they are.
When it fails
Each condition matters. The first control removes the right end from the interval, and the second puts a jump in \(f\) at \(c\).
With the right end open, \(f\) comes as close as you like to the dashed line but never reaches it. Make the jump large enough that the right end is lower than the open ring at \(c\). Then the ring marks a height that \(f\) approaches but never reaches. In both cases \(f\) has no absolute maximum.
Example 1: a corner
Does the Extreme Value Theorem apply to \(g(x) = |x - 2|\) on \([-1, 6]\)? If it does, find the absolute extrema.
The interval \([-1, 6]\) is closed, because both end points are in it.
The graph has no gap, so \(g\) is continuous on \([-1, 6]\). It has a corner at \(x = 2\), so \(g'(2)\) does not exist, but the theorem does not need a derivative.
The theorem applies. The V is lowest at its corner and highest at one end, so compare the values at these three points.
\[g(-1) = 3, \ g(2) = 0, \ g(6) = 4\]The absolute maximum is 4, at \(x = 6\), and the absolute minimum is 0, at \(x = 2\).
Example 2: a tangent
Does the Extreme Value Theorem apply to \(h(x) = \tan x\) on \([0, 2]\)?
The interval \([0, 2]\) is closed and bounded. Check whether \(h\) is continuous.
A tempting wrong move: The tangent is a trigonometric function, and trigonometric functions are continuous, so the theorem applies.
\[\tan x \text{ is continuous}\]The graph climbs without bound as \(x\) approaches \(\tfrac\pi2\), which is about 1.57 and inside \([0, 2]\). The curve breaks there, and it rises above the orange line at any height, so \(h\) has no absolute maximum.
The tangent equals \(\tfrac{\sin x}{\cos x}\), and \(\cos\tfrac\pi2 = 0\). The function has no value at \(x = \tfrac\pi2\), so it is not continuous on \([0, 2]\), and the theorem does not apply.
\[\tfrac\pi2 \approx 1.57 < 2\]
Practice
Let \(f(x) = x^2\) for \(0 \le x < 2\) and \(f(x) = 1\) for \(2 \le x \le 3\). Does the theorem apply on \([0, 3]\)?
Compare the value of \(f\) at \(x = 2\) with the values of \(f\) near 2 on the left. Then lower the line from the top.
Solution
The interval is closed and bounded. Check continuity at \(x = 2\), where the formula changes.
As \(x\) approaches 2 from the left, \(x^2\) approaches 4, but the value at 2 is 1.
\[\lim_{x \to 2^-} x^2 = 4 \ne 1 = f(2)\]\(f\) has a jump at \(x = 2\), so it is not continuous on \([0, 3]\), and the theorem does not apply. In fact, \(f\) has no absolute maximum, because the values left of 2 come close to 4 but never reach it.
Questions
The graph shows \(g\) from \(x = 1\) to \(x = 6\). It has a corner at \(x = 2\) and a jump at \(x = 4\), where the filled dot is above the open ring.
Why can the Extreme Value Theorem not be used on \([1, 6]\)?
- \(g\) is not continuous at \(x = 4\).
- \(g\) is not differentiable at \(x = 2\).
- The interval \([1, 6]\) is not closed.
Answer
\(g\) is not continuous at \(x = 4\).
On which interval can the theorem be used?
- \([1, 3]\)
- \([1, 4]\)
- \([3, 6]\)
- \((1, 3)\)
Answer
\([1, 3]\)
Which function satisfies the conditions of the theorem on \([0, 5]\)?
- \(|x - 3|\)
- \(\dfrac{1}{x - 3}\)
- \(\ln x\)
- \(\sqrt{x - 1}\)
Answer
\(|x - 3|\)
Summary
You check that the function is continuous on a closed, bounded interval, and you do not ask for a derivative.
This works because a connected curve over a closed interval cannot skip its highest or its lowest height.
You use it before you search for extreme values, to know that they exist.
This is one of five skills in topic 5.2 of Which & Why: AP Calculus.