Recognize an unattained extremum
Heights that the graph approaches
The graph shows \(f(x) = -x^2 + 2x + 4\) for \(x\) between \(-1\) and 3. The open circles at the ends mean that the end points are not on the graph.
Drag the point toward the right end. At x = 2.9 the height is 1.39, and at x = 2.99 it is 1.0399. The heights come close to the dashed orange line at height 1, but the point never reaches the open circle.
Whichever point you choose, the graph is lower nearer an end, so no point is the lowest and there is no absolute minimum.
The top of the hill, \((1, 5)\), is on the graph, so the absolute maximum value is 5.
Approached and taken
With closed ends, the lowest points of the graph are the end dots.
Press the first button to open the ends. The end points leave the graph, and every remaining point is higher than the dashed line at their height.
An absolute extremum must be a value that the function takes at some \(x\) in the interval. A value that is only approached is not an extremum.
Notation
An open interval leaves out its ends: on \((a, b)\), the numbers \(a\) and \(b\) are not candidates.
\[a < x < b\]A value \(L\) is approached if \(f(x)\) comes as close to \(L\) as you like when \(x\) is near the end. It is taken if \(f(c) = L\) for some \(c\) in the interval.
\[\begin{gathered} \lim_{x \to a^+} f(x) = \cy{L} \\ \text{but } f(x) \ne L \text{ for every } x \text{ in } (a, b) \end{gathered}\]If an end of the interval is open, find the limit of \(f\) there. Compare it with the values at the candidates, which are the critical points inside and any closed end.
If a limit is greater than every candidate value, or \(f\) grows without bound at the end, there is no absolute maximum. If it is less than every candidate value, or \(f\) falls without bound, there is no absolute minimum.
When it fails
Press the toggle to close the ends. The end points return to the graph and take the lowest height, so there is an absolute minimum.
Press it again to open the ends. Every remaining point is higher than \(L\), so no point is the lowest and there is no absolute minimum.
Example 1: One end is open
Find the absolute extrema of \(g(x) = \dfrac{1}{x}\) on the interval \((0, 2]\).
The derivative is negative on \((0, 2]\), so \(g\) decreases on the whole interval and has no critical points.
\[g'(x) = -\dfrac{1}{x^2}\]The right end, 2, belongs to the interval, and \(g\) decreases, so the absolute minimum value is \(g(2) = \tfrac12\).
The left end, 0, is not in the interval. As \(x\) approaches 0 from the right, the heights grow without bound, so there is no absolute maximum.
\[\lim_{x \to 0^+} \dfrac{1}{x} = \infty\]
Example 2: Counting an open end
Find the absolute extrema of \(p(x) = x^3 - 27x\) on the interval \((-8, 5]\).
The derivative is 0 at \(x = -3\) and \(x = 3\), both in the interval.
\[\begin{aligned} p'(x) &= 3x^2 - 27 \\ &= 3(x - 3)(x + 3) \end{aligned}\]A tempting wrong move: The ends are candidates, so evaluate \(p\) at \(-8\), \(-3\), 3 and 5. The absolute minimum value is \(-296\).
\[\begin{gathered} p(-8) = -296, \quad p(-3) = 54 \\ p(3) = -54, \quad p(5) = -10 \end{gathered}\]The graph has an open circle at \(x = -8\), so \(p\) takes no value there. The dashed orange line at \(-296\) lies below every point of the graph.
The left end is open, so \(x = -8\) is not a candidate. The candidates are \(-3\), 3 and 5, with values 54, \(-54\) and \(-10\).
The limit at the open end is \(-296\), less than every candidate value, and the graph never takes it, so there is no absolute minimum. The absolute maximum value is 54, at \(x = -3\).
\[\lim_{x \to -8^+} p(x) = -296\]
Practice
Find the absolute extrema of \(q(x) = x^2 - 4x + 7\) on the interval \([0, 5)\).
Find the critical point and the value at the closed end. Then drag the point toward the open end and decide whether the height it approaches is taken.
Solution
The derivative is 0 at \(x = 2\), inside the interval. The candidates are the closed end 0 and the critical point 2.
\[\begin{gathered} q'(x) = 2x - 4 \\ q(0) = 7, \quad q(2) = 3 \end{gathered}\]At the open end the limit is 12, greater than both candidate values, and \(q\) never takes it because the end is open.
\[\lim_{x \to 5^-} q(x) = 12\]The absolute minimum value is 3, at \(x = 2\). There is no absolute maximum.
Questions
These questions use the graph of \(r(x) = (x - 1)^2 + 1\) on \([0, 4)\). It has a filled dot at \((0, 2)\), a valley at \((1, 1)\) and an open circle at \((4, 10)\).
Does \(r\) have an absolute maximum on \([0, 4)\)?
- No. The heights come close to 10, but no \(x\) in the interval gives 10.
- Yes, 10 at \(x = 4\).
- Yes, 2 at \(x = 0\).
- Yes, 1 at \(x = 1\).
Answer
No. The heights come close to 10, but no \(x\) in the interval gives 10.
Does \(r\) have an absolute minimum on \([0, 4)\)?
- Yes, 1 at \(x = 1\).
- No, because the right end of the interval is open.
- Yes, 2 at \(x = 0\).
Answer
Yes, 1 at \(x = 1\).
What does \(t(x) = \ln x\) have on the interval \((0, e]\)?
- An absolute maximum of 1 at \(x = e\), and no absolute minimum.
- An absolute maximum of 1 at \(x = e\), and an absolute minimum of 0 at \(x = 1\).
- No absolute maximum and no absolute minimum.
Answer
An absolute maximum of 1 at \(x = e\), and no absolute minimum.
Summary
With an open end, you compare the limit there with the values at the critical points and at any closed end.
A limit above all of those values means no absolute maximum, and a limit below all of them means no absolute minimum.
This works because an absolute extremum must be a value that the function takes, and an open end is not on the graph.
This is one of five skills in topic 5.5 of Which & Why: AP Calculus.