Find all candidates on the interval
The highest and lowest points of a hill-and-valley graph
The graph shows \(f(x) = x^3 - 12x\) on the interval from \(x = -3\) to \(x = 5\). The filled dots mark the two ends, which belong to the interval.
The highest point of the graph on this interval is the absolute maximum, and the lowest point is the absolute minimum. A candidate is an x-value where one of them can be.
Drag the orange point along the curve, or press the button to visit the four candidates. The orange line is the tangent, which has the same slope as the curve.
At \(x = -2\) and at \(x = 2\), the tangent is flat, so the slope is 0. A point where the slope is 0 is a critical point, and the curve turns around there.
The candidates are the two critical points and the two ends: \(x = -3, -2, 2\) and \(5\).
Where an extreme can be
This curve is continuous on the closed interval \([a, b]\). The small dots mark the ends, the flat tangents, and the corner, a sharp point with no single slope.
The two large dots mark the highest and the lowest point. Move the Tilt slider, and the large dots move from one small dot to another. They are never anywhere else.
If the highest point were inside the interval with a tilted tangent, a higher point would lie next to it. So the highest point is at an end, at a flat tangent or at a corner.
Notation
Start with a function \(f\) that is continuous on a closed interval from \(a\) to \(b\). Closed means that both ends belong to the interval.
\[f \text{ continuous on } a \le x \le b\]A critical point is an x-value \(c\) where \(f\) is defined and the slope is 0 or does not exist. The curve here has two flat tangents, at \(c_1\) and \(c_2\), and a corner at \(c_3\).
\[\begin{gathered} \cy{f'}(c) = 0 \\ \text{or } \cy{f'}(c) \text{ does not exist} \end{gathered}\]To find the critical points, solve \(f'(x) = 0\). Then find where \(f'(x)\) is undefined, for example where a denominator is 0.
\[f'(x) = 0\]Keep a critical point only if it lies between \(a\) and \(b\), because the problem is about that interval. The candidates are the two ends and the critical points you kept.
\[\text{candidates: } a,\ c_1,\ c_2,\ \dots,\ b\]If \(f\) is continuous on \([a, b]\), its absolute maximum and minimum are at candidates. At any other point, the tangent is tilted, so the curve is higher on one side and lower on the other.
When it fails
A zero of \(f'\) counts only inside the interval. Press the button to move the point to the flat tangent at \(d\), beyond the end \(b\).
The slope at \(d\) is 0, but \(d\) is not in \([a, b]\). The red row shows that \(d\) is not a candidate.
Example 1: Zeros of the derivative from a sine
List all candidates for \(f(x) = x + 2\cos x\) on the interval \([0, 2\pi]\).
The derivative exists for every \(x\), so the critical points are the zeros of the derivative.
\[f'(x) = 1 - 2\sin x\]Set the derivative equal to 0 and solve for the sine.
\[1 - 2\sin x = 0 \quad\Rightarrow\quad \sin x = \tfrac{1}{2}\]On \([0, 2\pi]\), the sine is positive only between 0 and \(\pi\), and there it equals \(\tfrac12\) at two x-values. Both lie inside the interval.
\[x = \tfrac{\pi}{6}\quad\text{and}\quad x = \tfrac{5\pi}{6}\]Add the two ends.
\[\text{candidates: } 0,\ \tfrac{\pi}{6},\ \tfrac{5\pi}{6},\ 2\pi\]
Example 2: A point where the slope does not exist
List all candidates for \(g(x) = x^{2/3}(x - 5)\) on the interval \([-1, 4]\).
Differentiate with the product rule. Then factor out \(x^{-1/3}\) to show where \(g'\) is 0 or undefined.
\[\begin{aligned} g'(x) &= \tfrac23 x^{-1/3}(x - 5) + x^{2/3} \\ &= \tfrac53\, x^{-1/3}(x - 2) \end{aligned}\]A tempting wrong move: Set the derivative equal to 0. The only solution is \(x = 2\), so the candidates are \(-1\), 2 and 4.
\[\begin{aligned} \tfrac53\, x^{-1/3}(x - 2) &= 0 \\ x &= 2 \end{aligned}\]The graph has a sharp peak at \(x = 0\), and the peak is the highest point of the graph. This list misses it.
The factor \(x^{-1/3}\) is 1 over the cube root of \(x\), so it divides by 0 at \(x = 0\) and \(g'(0)\) does not exist. The function \(g\) is defined at 0, so \(x = 0\) is a critical point.
\[g(0) = 0\]Both critical points lie between \(-1\) and 4. Add the two ends.
\[\text{candidates: } -1,\ 0,\ 2,\ 4\]
Practice
List all candidates for \(h(x) = x^4 - 8x^2 + 3\) on the interval \([-1, 3]\).
Factor the derivative to find every zero. Keep only the zeros on the solid part of the curve, which lies between \(-1\) and 3.
Solution
The derivative exists for every \(x\). Factor it to find its zeros.
\[\begin{aligned} h'(x) &= 4x^3 - 16x \\ &= 4x(x - 2)(x + 2) \end{aligned}\]The zero at \(-2\) lies outside the interval, so it is not a candidate. The zeros at 0 and 2 lie inside.
Add the two ends.
\[\text{candidates: } -1,\ 0,\ 2,\ 3\]
Questions
These questions use the graph of \(f(x) = |x^2 - 4|\) on the interval from \(-3\) to 3. The graph has a hump at \(x = 0\) and sharp corners at \(x = -2\) and \(x = 2\).
How many candidates does \(f\) have on this interval?
- Five: the two ends, the corners at \(-2\) and 2, and the hump at 0
- Three: the two ends and the hump at 0
- Four: the two ends and the two corners
Answer
Five: the two ends, the corners at \(-2\) and 2, and the hump at 0
Why is the corner at \(x = 2\) a candidate?
- The slope \(f'(2)\) does not exist.
- The slope \(f'(2)\) is 0.
- The point \(x = 2\) is an end of the interval.
Answer
The slope \(f'(2)\) does not exist.
What are the candidates for \(k(x) = x + \dfrac{9}{x}\) on the interval \([1, 6]\)?
- \(1, 3, 6\)
- \(-3, 1, 3, 6\)
- \(1, 6\)
Answer
\(1, 3, 6\)
Summary
The candidates are the two ends and every critical point inside the interval.
On a closed interval, a continuous graph has its highest and lowest points at candidates, because the tangent is tilted anywhere else.
To find an absolute maximum or minimum on a closed interval, make this list first.
This is one of five skills in topic 5.5 of Which & Why: AP Calculus.