Compare function values
Heights at the candidates
The graph shows \(f(x) = x^3 - 9x^2 + 15x\) on \([0, 8]\). The derivative \(3(x - 1)(x - 5)\) is 0 at \(x = 1\) and \(x = 5\), so the candidates are 0, 1, 5 and 8.
Press the button to move the point to each candidate. The table records the value of \(f\) there, the height of the graph: 0, 7, \(-25\) and 56.
The Line slider moves the dashed orange line. Lower it to height 56. The whole graph lies on or below the line, so 56 is the absolute maximum value, and it occurs at the right end.
Raise the line from below to height \(-25\). The whole graph lies on or above the line, so \(-25\) is the absolute minimum value, and it occurs at \(x = 5\).
Largest and smallest of a few numbers
This curve is continuous on \([a, b]\). The dots mark the candidates, which are the ends \(a\) and \(b\) and the critical points \(c_1\), \(c_2\) and \(c_3\). The table gives \(f\), the height of the graph, at each one.
Move the dashed line down from above the curve. It first touches the curve at a dot, because the highest point of a continuous graph on a closed interval is at a candidate.
Move the line up from below. It first touches the dot with the smallest value of f. Comparing the few heights in the table therefore compares the whole graph.
Notation
Evaluate the original function \(f\) at every candidate. You need the height of the graph, so use \(f\) and not \(f'\).
\[f(a),\ f(c_1),\ f(c_2),\ \dots,\ f(b)\]The largest of these numbers is the absolute maximum value, and the smallest is the absolute minimum value.
\[\begin{aligned} \cy{M} &= \max\{f(a), f(c_1), \dots, f(b)\} \\ \cy{m} &= \min\{f(a), f(c_1), \dots, f(b)\} \end{aligned}\]At the corner \(c_2\), \(f\) has a height even though it has no slope, so \(f(c_2)\) joins the comparison.
This method needs \(f\) to be continuous on the closed interval \([a, b]\), because then the absolute extrema exist and occur at candidates.
When it fails
Every candidate must be in the table, including the ends. Press the toggle to leave out \(a\) and \(b\).
The dashed line sits at \(f(c_1)\), the largest value left in the table. Near \(b\), the graph rises above the line, so the true maximum is at the end \(b\).
Example 1: An interior maximum
Find the absolute maximum value and the absolute minimum value of \(f(x) = xe^{-x}\) on \([0, 3]\).
The factor \(e^{-x}\) is never 0, so the zero of \(f'\) comes from the factor \(1 - x\).
\[\begin{gathered} f'(x) = (1 - x)e^{-x} \\ f'(x) = 0 \quad\Rightarrow\quad x = 1 \end{gathered}\]The candidates are 0, 1 and 3. Evaluate \(f\) at each of them.
\[\begin{gathered} f(0) = 0 \\ f(1) = \tfrac{1}{e} \\ f(3) = \tfrac{3}{e^3} \end{gathered}\]Both values are positive, and \(3/e^3\) is smaller than \(1/e\) because \(e^2 > 3\). As a check, \(1/e \approx 0.368\) and \(3/e^3 \approx 0.149\).
The absolute maximum value is \(1/e\), and the absolute minimum value is 0.
Example 2: A function with corners
Find the absolute maximum value and the absolute minimum value of \(p(x) = |x^2 - 2x|\) on \([-1, 4]\).
Between 0 and 2, \(x^2 - 2x\) is negative, so the absolute value flips its sign there. On that stretch \(p'(1) = 0\), the top at \(x = 1\), and the slope does not exist at the corners \(x = 0\) and \(x = 2\).
\[\begin{gathered} p(x) = 2x - x^2 \ \text{on}\ (0, 2) \\ p'(x) = 2 - 2x \end{gathered}\]A tempting wrong move: The derivative does not exist at 0 and 2, so these points are not candidates. Compare the values at \(-1\), 1 and 4.
\[\begin{gathered} p(-1) = 3 \\ p(1) = 1 \\ p(4) = 8 \end{gathered}\]The dashed line at height 1 is not under the whole graph. At the corners \(x = 0\) and \(x = 2\) the graph touches the x-axis, so it reaches height 0, lower than 1.
A corner is a critical point, and \(p\) has a height there. Evaluate \(p\) at the corners as well.
\[p(0) = 0,\qquad p(2) = 0\]The values are 3, 0, 1, 0 and 8. The absolute maximum value is 8, and the absolute minimum value is 0, taken at both corners.
Practice
Find the absolute maximum value and the absolute minimum value of \(g(x) = 4x - x^3\) on \([-1, 3]\).
Only one solution of \(g'(x) = 0\) lies in the interval. Evaluate \(g\), not \(g'\), at each candidate. Then move the line to compare the heights.
Solution
The derivative is 0 at \(x = \pm\tfrac{2}{\sqrt3}\). Only \(\tfrac{2}{\sqrt3}\), about 1.15, lies in the interval.
\[\begin{gathered} g'(x) = 4 - 3x^2 \\ g'(x) = 0 \quad\Rightarrow\quad x = \tfrac{2}{\sqrt3} \end{gathered}\]Evaluate \(g\) at the three candidates.
\[\begin{gathered} g(-1) = -3 \\ g\!\left(\tfrac{2}{\sqrt3}\right) = \tfrac{16\sqrt3}{9} \\ g(3) = -15 \end{gathered}\]The absolute maximum value is \(\tfrac{16\sqrt3}{9}\), about 3.08. The absolute minimum value is \(-15\).
Questions
These questions use the graph of \(u(x) = \tfrac13 x^3 - x^2 - 3x + 5\) on \([-2, 4]\). The table lists the height of the graph at each candidate.
Which candidate gives the absolute maximum value?
- \(x = -1\), where the height is \(20/3\)
- \(x = -2\), the left end
- \(x = 3\), the valley
- \(x = 4\), the right end
Answer
\(x = -1\), where the height is \(20/3\)
The orange point is at the right end, \(x = 4\). Is the height there the absolute minimum value?
- No. The dot at \(x = 3\) is lower.
- Yes, because \(x = 4\) is an end of the interval.
- Yes, because the slope at \(x = 4\) is 0.
Answer
No. The dot at \(x = 3\) is lower.
What is the absolute maximum value of \(h(x) = x^2 - 4x + 5\) on \([0, 5]\)?
- 10
- 5
- 1
- 6
Answer
10
Summary
Evaluate the original function at every candidate, and pick the largest and the smallest of these values.
These values are the absolute maximum and minimum values, because the highest and lowest points of the graph are at candidates.
Make this comparison on any closed interval where the function is continuous, after you list the candidates.
This is one of five skills in topic 5.5 of Which & Why: AP Calculus.