Identify candidates for local extrema
Where can the curve turn?
The upper graph shows \(f(x) = x - 2\sin x\) on the interval from 0 to \(2\pi\). The lower graph shows its slope \(f'(x) = 1 - 2\cos x\), and the orange bar shows \(f'(x)\) at the point.
A local maximum is higher than every point near it, on both sides, and a local minimum is lower. The curve turns at such a point.
Drag the orange point. The tangent is horizontal at \(x = \pi/3\) and at \(x = 5\pi/3\), where \(f'\) is 0, so the curve can turn there.
The interval ends at \(x = 0\) and at \(x = 2\pi\). A point at an end has no points on one side, so it cannot be a local maximum or minimum.
Three places to look
The curve has a smooth top, a sharp bottom, a flat spot with a horizontal tangent, and an open ring where it has no value. The buttons label each place.
The curve can turn only at an input \(c\) that passes three tests. First, \(f\) has a value at \(c\). Second, \(c\) is inside the interval, so the curve has points on both sides of \(c\). Third, \(f'(c)\) is 0, or \(f'(c)\) does not exist.
An input that passes all three tests is a critical point, a candidate for a local maximum or minimum. The flat spot passes them too, so a candidate is not always a turn.
Notation
A critical point of \(f\) is an input \(c\) inside the domain where \(f(c)\) exists and \(f'(c)\) is 0 or does not exist.
\[\begin{gathered} f(c)\text{ exists} \\ \cy{f'(c)} = 0 \\ \text{or } \cy{f'(c)}\text{ does not exist} \end{gathered}\]If \(f\) has a local maximum or minimum at \(c\), then \(c\) is a critical point.
\[\begin{gathered} \text{local extremum at } c \\ \Rightarrow\; \cy{f'(c)} = 0 \\ \text{or } \cy{f'(c)} \text{ does not exist} \end{gathered}\]The reverse is false: a critical point is only a candidate, and the next skills decide whether it is a top, a bottom, or neither.
The ends of an interval are never candidates, because a local maximum or minimum needs points on both sides. An input where \(f\) has no value has no point to be a top or a bottom.
When it fails
At the right end, \(x = 2\pi\), every nearby point to the left is higher than the orange dashed line at the height \(f(2\pi)\). The end looks like a local minimum.
But the graph has no points to the right of \(2\pi\). A local minimum needs points on both sides, so the end is not a candidate.
Example 1: a break in the domain
Find the candidates for a local maximum or minimum of \(f(x) = x^2 + \dfrac{16}{x}\).
The term \(\frac{16}{x}\) has no value at \(x = 0\), so the domain is every \(x\) except 0.
\[x \neq 0\]Find the derivative and write it as one fraction.
\[f'(x) = 2x - \dfrac{16}{x^2} = \dfrac{2x^3 - 16}{x^2}\]The derivative has no value at \(x = 0\), but neither does \(f\). An input outside the domain is not a candidate.
The derivative is 0 where the top is 0. Then \(2x^3 = 16\), so \(x^3 = 8\) and \(x = 2\).
\[2x^3 - 16 = 0 \;\Rightarrow\; x = \cy{2}\]The only candidate is \(x = 2\).
Example 2: an interval that ends
Find the candidates for a local maximum or minimum of \(f(x) = x\sqrt{4 - x}\).
The square root needs \(4 - x \ge 0\). The domain is \(x \le 4\), so the graph ends at \(x = 4\).
\[x \le 4\]Use the product rule and the chain rule. Write the result as one fraction.
\[\begin{aligned} f'(x) &= \sqrt{4 - x} - \dfrac{x}{2\sqrt{4 - x}} \\ &= \dfrac{8 - 3x}{2\sqrt{4 - x}} \end{aligned}\]A tempting wrong move: The bottom is 0 at \(x = 4\), so \(f'\) does not exist there. This makes \(x = 4\) a candidate.
\[\begin{aligned} &f'(4)\text{ does not exist} \\ &\Rightarrow\; x = 4 \text{ is a candidate} \end{aligned}\]The graph stops at \(x = 4\), so it has no points to the right of 4. A local maximum or minimum needs points on both sides.
Set the top equal to 0. The input \(\frac83\) is inside the domain, because \(\frac83 < 4\).
\[8 - 3x = 0 \;\Rightarrow\; x = \cy{\dfrac83}\]The only candidate is \(x = \frac83\).
Practice
Find the candidates for a local maximum or minimum of \(f(x) = (x^2 - 1)^{2/3}\).
Find where \(f'\) is 0. Then find where \(f'\) does not exist although \(f\) has a value, as at a sharp point.
Solution
A cube root has a value for every number, so \(f\) has a value at every input. Use the chain rule.
\[\begin{aligned} f'(x) &= \tfrac23 (x^2 - 1)^{-1/3} \cdot 2x \\ &= \dfrac{4x}{3\sqrt[3]{x^2 - 1}} \end{aligned}\]The derivative is 0 where the top is 0.
\[4x = 0 \;\Rightarrow\; x = \cy{0}\]The derivative has no value where the bottom is 0. Then \(x^2 = 1\), so \(x = -1\) or \(x = 1\). The function \(f\) has the value 0 at both inputs, so both are candidates.
\[\sqrt[3]{x^2 - 1} = 0 \;\Rightarrow\; x = \cy{-1},\ x = \cy{1}\]The candidates are \(-1\), 0 and 1.
Questions
The graph shows a function \(g\) on the interval from \(-4\) to 6. It has a sharp point at \(x = 1\) and an open ring at \(x = 5\), where \(g\) has no value.
Which list gives all the candidates for a local maximum or minimum of \(g\)?
- \(-2\), 1 and 3
- \(-4\), \(-2\), 1, 3 and 6
- \(-2\), 1, 3 and 5
- \(-2\) and 3
Answer
\(-2\), 1 and 3
Why is \(x = 5\) not a candidate?
- \(g\) has no value at \(x = 5\)
- \(g'(5)\) is not 0
- \(x = 5\) is near the end of the interval
Answer
\(g\) has no value at \(x = 5\)
What is the only candidate for a local maximum or minimum of \(f(x) = x\ln x\), where \(x > 0\)?
- \(\dfrac1e\)
- \(e\)
- 0 and \(\dfrac1e\)
Answer
\(\dfrac1e\)
Summary
The candidates are the inputs where \(f'\) is 0 or does not exist, apart from the ends and any input where \(f\) has no value.
A local maximum or minimum needs points on both sides and a slope that is 0 or undefined, so no other input can be one.
You make this list first whenever a question asks for a local maximum or minimum, because the next skills test each candidate.
This is one of four skills in topic 5.4 of Which & Why: AP Calculus.