Justify that the search is complete

A line lowered onto the graph

The graph shows \(f(x) = x^3 - 3x\) from \(x = 0\) to \(x = 3\). It has no breaks, and both ends belong to the interval.

The Line slider moves the dashed orange line. Lower it from above. It first touches the graph at (3, 18), the right end, which is the highest point.

Raise the line from below. It first touches the graph at \((1, -2)\), the lowest point. The tangent there is flat, so \(x = 1\) is a critical point.

Why the line must touch the graph

This curve has no breaks, and the closed interval \([a, b]\) includes both ends. Lower the dashed line from above. It first touches the curve at the corner, the dot labelled absolute maximum.

A curve with no breaks and both ends on it cannot avoid the line. This is the Extreme Value Theorem: a continuous function on a closed interval has an absolute maximum and an absolute minimum.

If the first touching point is inside the interval, the curve lies below the line on both sides of it. Then the tangent is flat or there is a corner, so the point is a critical point. Otherwise it is an end.

Notation

1. First, say why the extremes exist. If \(f\) is continuous on the closed interval \([a, b]\), the Extreme Value Theorem gives an absolute maximum and an absolute minimum.

   \[\begin{gathered} f \text{ continuous on } [a, b] \\ \Longrightarrow\ \text{absolute maximum exists} \\ \Longrightarrow\ \text{absolute minimum exists} \end{gathered}\]

2. Second, say where they are. An absolute extremum inside \((a, b)\) is also a local extremum, so \(f'\) is 0 or does not exist there. Otherwise it is at an end.

   \[\begin{gathered} c \text{ in } (a, b): \\ \cy{f'}(c) = 0 \ \text{ or } \ \cy{f'}(c) \text{ does not exist} \end{gathered}\]

3. Third, every absolute extremum is at a candidate, so compare the values of \(f\) at the candidates.

   The sign of \(f'\) can also justify an answer. If \(f' > 0\) on \((a, c)\) and \(f' < 0\) on \((c, b)\), then \(f\) rises to \(c\) and falls after it, so \(f(c)\) is the absolute maximum.

When it fails

The theorem needs \(f\) to be continuous on a closed interval. Press the first button to put a jump in \(f\) at \(c_3\).

The graph rises toward the open ring, but \(f(c_3)\) is the lower filled dot. Press the second button. The values of \(f\) come close to the line through the ring and never reach it, so \(f\) has no absolute maximum.

Example 1: A full justification

Justify that \(f(x) = x^3 - 6x^2 + 9x\) has an absolute maximum and an absolute minimum on \([-1, 5]\), and find them.

1. A polynomial is continuous, and the interval is closed, so by the Extreme Value Theorem \(f\) has an absolute maximum and an absolute minimum.

2. The derivative exists for every \(x\), so the critical points are the zeros 1 and 3 of \(f'\). Both lie inside \((-1, 5)\).

   \[\begin{aligned} f'(x) &= 3x^2 - 12x + 9 \\ &= 3(x - 1)(x - 3) \end{aligned}\]

3. Evaluate \(f\) at the candidates \(-1\), 1, 3 and 5.

   \[\begin{gathered} f(-1) = -16, \quad f(1) = 4 \\ f(3) = 0, \quad f(5) = 20 \end{gathered}\]

4. The absolute maximum is 20, at \(x = 5\). The absolute minimum is \(-16\), at \(x = -1\).

Example 2: A break inside the interval

A student says that the absolute maximum of \(f(x) = \dfrac{1}{x - 1}\) on \([0, 3]\) is \(\tfrac12\). Decide whether the candidate method supports this claim.

1. The derivative is never 0. It does not exist at \(x = 1\), but \(f\) is not defined there either.

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

2. A tempting wrong move: There are no critical points, so the candidates are the two ends. The maximum is \(f(3) = \tfrac12\), and the minimum is \(f(0) = -1\).

   \[\begin{gathered} f(0) = -1 \\ f(3) = \tfrac12 \end{gathered}\]

   To the right of the asymptote at \(x = 1\), the graph is far above the dashed line at \(\tfrac12\). To the left, it is far below \(-1\).

3. The candidate method rests on the Extreme Value Theorem, which needs \(f\) continuous on \([0, 3]\). At \(x = 1\), inside the interval, \(f\) is not continuous, so the method does not apply.

4. To the right of \(x = 1\), \(f\) rises above any number. To the left, it falls below any number, so \(f\) has no absolute maximum and no absolute minimum.

   \[\begin{gathered} \lim_{x \to 1^+} \dfrac{1}{x - 1} = \infty \\ \lim_{x \to 1^-} \dfrac{1}{x - 1} = -\infty \end{gathered}\]

Practice

Justify that \(g(x) = 6\sqrt{x} - x\sqrt{x}\) has an absolute maximum on \([0, 6]\), and find it.

State why a maximum exists. Then evaluate \(g\) at every candidate, including the end \(x = 0\), where \(g'\) does not exist.

Questions

These questions use the function \(q\) on \([0, 6]\). The graph follows \(y = x\) up to an open circle at \((4, 4)\). The point \((4, 2)\) is filled, and the graph then falls to \((6, 0)\).

1. On \([0, 6]\), which hypothesis of the Extreme Value Theorem fails for \(q\)?

   • The function \(q\) is not continuous at \(x = 4\).
   • The interval \([0, 6]\) is not closed.
   • The function \(q\) is not defined at \(x = 4\).
   Answer

   The function \(q\) is not continuous at \(x = 4\).

2. Does \(q\) have an absolute maximum on \([0, 6]\)?

   • No. The heights come close to 4 as \(x\) approaches 4 from the left, but no \(x\) gives 4.
   • Yes, the value 4 at \(x = 4\).
   • Yes, the value 2 at \(x = 4\).
   Answer

   No. The heights come close to 4 as \(x\) approaches 4 from the left, but no \(x\) gives 4.

3. For which interval does the Extreme Value Theorem guarantee that \(h(x) = \dfrac{1}{x + 1}\) has an absolute maximum?

   • \([0, 3]\)
   • \([-2, 0]\)
   • \((0, 3)\)
   Answer

   \([0, 3]\)

Summary

You state that the function is continuous on a closed interval, list the candidates, and compare the values.

This works because a continuous graph on a closed interval has a highest and a lowest point, and both are at candidates.

Write this argument when a question asks you to justify an absolute extremum.

This is one of five skills in topic 5.5 of Which & Why: AP Calculus.