Wemazi

Topic 5.2

Recognize when a maximum or minimum does not exist

Builds on checking the Extreme Value Theorem conditions. 6 minutes.

A height that is approached

The line is the graph of \(f(x) = 3x - 1\) on the open interval from 1 to 4. The open rings show that the end points are not in the interval.

The orange dashed line marks the height 11, which \(f\) approaches at the right end. Drag the point to the right.

At \(x = 3.9\) the height is 10.7, and at \(x = 3.99\) it is 10.97. The gap to 11 shrinks but never becomes 0, because \(x = 4\) is not in the interval.

Every value of \(f\) is below 11, and a higher value always exists. Therefore \(f\) has no absolute maximum, and the same argument at the left end shows that it has no absolute minimum.

A bound that is not reached

The numbers are gone. The orange dashed line marks the height that the curve approaches at the open end on the right.

Between the point and the open end there is always a higher point, so no point is the highest. Include the end point in the interval. Then the curve reaches the dashed line, and the filled dot at the end is the highest point.

An absolute maximum is a height that the function takes. A height that the function only approaches is not a maximum.

Notation

  1. An absolute maximum is a value \(M = f(c)\) that the function takes at an input \(c\) in the domain.

    \[\cy{f(c) = M}, \quad c \text{ in the domain}\]

  2. No value of \(f\) is greater than \(M\). A height that no value goes above is called a bound, and a bound is the maximum only when the function takes it.

    \[f(x) \le M \ \text{ for } x \text{ in the domain}\]

  3. A maximum or a minimum can fail to exist in three ways. The domain can leave out an end point where the values approach a bound. The values can grow without limit. A jump can leave a height approached but not taken.

  4. Compare the values that the function takes with the values it approaches at each open end and at infinity. If an approached value is higher than every value taken, there is no maximum, and if one is lower, there is no minimum.

When it fails

Include the end point \(x = 4\). The interval is now \((1, 4]\), and the ring becomes a filled dot at height 11. The function takes the value 11, so 11 is the absolute maximum.

The left end is still open, so the function approaches 2 without taking it. There is still no absolute minimum.

Example 1: a reciprocal

Find the absolute extrema of \(f(x) = \dfrac1x\) on \((0, 2]\), if they exist.

  1. The derivative is negative on the whole interval, so \(f\) decreases.

    \[f'(x) = -\dfrac{1}{x^2}\]

  2. Near \(x = 0\) the values grow without limit, so no value is the highest and there is no absolute maximum.

    \[\lim_{x \to 0^+} \dfrac1x = \infty\]

  3. Because \(f\) decreases, its lowest value is at the right end, \(x = 2\), which is in the interval. The absolute minimum is \(f(2) = \tfrac12\).

Example 2: a bell

Find the absolute extrema of \(f(x) = \dfrac{5}{x^2 + 1}\) on all real numbers, if they exist.

  1. The derivative is 0 only at \(x = 0\), where \(f(0) = 5\). For every other input the denominator is larger than 1, so \(f(x) < 5\) and the absolute maximum is 5.

    \[f'(x) = \dfrac{-10x}{(x^2 + 1)^2}\]

  2. A tempting wrong move: As \(x\) goes to infinity, the values approach 0, so the absolute minimum is 0.

    \[\lim_{x \to \infty} f(x) = 0\]

    The curve is above the \(x\)-axis for every \(x\). It comes close to height 0 and never touches it, so no input has \(f(x) = 0\).

  3. The value 0 is approached and never taken, so there is no absolute minimum.

    \[\dfrac{5}{x^2 + 1} > 0\]

Practice

Find the absolute maximum and minimum of \(k(x) = x^2 - 4x\) on \((0, 5]\), if they exist.

Move the point toward the open end at 0. Compare the height that \(k\) approaches there with the values at \(x = 2\) and \(x = 5\).

Solution

  1. The derivative is 0 at \(x = 2\), which is in the interval. The value there is \(-4\).

    \[k'(x) = 2x - 4, \qquad k(2) = -4\]

  2. At the right end, \(k(5) = 5\), and \(x = 5\) is in the interval. At the open left end, \(k\) approaches 0 and never takes that value.

  3. The approached value 0 lies between the values taken, \(-4\) and 5, so the absolute minimum is \(-4\) and the absolute maximum is 5.

Questions

The graph shows \(h\) on the interval \((1, 5]\). The open ring at \(x = 1\) marks height 6, which the curve approaches. The filled dot at \(x = 5\) is on the curve.

  1. Does \(h\) have an absolute maximum on \((1, 5]\)?

    • No, because the curve approaches 6 near \(x = 1\) and never reaches it.
    • Yes, the maximum is 6.
    • Yes, the maximum is 4.
    Answer

    No, because the curve approaches 6 near \(x = 1\) and never reaches it.

  2. What is the absolute minimum value of \(h\) on \((1, 5]\)?

    • 2
    • 4
    • There is none.
    Answer

    2

  3. Which function has an absolute maximum on \([0, \infty)\)?

    • \(\dfrac{x}{x^2 + 1}\)
    • \(1 - e^{-x}\)
    • \(\sqrt{x}\)
    • \(x^2\)
    Answer

    \(\dfrac{x}{x^2 + 1}\)

Summary

An absolute extremum is a value the function takes at an input in the domain, so a height it only approaches is not one.

At an open end, or as \(x\) goes to infinity, the function can approach a height without taking it.

Before you report an extremum on an open or unbounded interval, you compare the heights approached with the values taken.

This is one of five skills in topic 5.2 of Which & Why: AP Calculus.