Find critical numbers in the domain
Where the tangent is flat or missing
The curve is the graph of \(f(x) = x^{2/3}(x - 5)\), and the orange line is the tangent at the orange point. Drag the point from left to right.
At \(x = 2\) the tangent is horizontal, so its slope is 0. At \(x = 0\) the curve has a sharp peak with a vertical tangent, and a vertical line has no slope, so \(f'(0)\) does not exist.
A curve can turn only where its tangent does not slope up or down, so only where the slope is 0 or missing. These inputs are the critical points of \(f\).
This curve has two, \(x = 0\) and \(x = 2\), because at every other point the tangent has a slope that is not 0.
Where the slope is 0 or missing
The numbers are gone. The lower graph shows \(f'\), the slope of the tangent at each \(x\), and the orange bar has that height.
Where the tangent is horizontal, the graph of \(f'\) meets the axis. At the corner the two red half-tangents have different slopes, so \(f'\) has no value there and its graph jumps.
Move the point to the top, to the corner and to the flat step on the rise. At each one, \(f'\) is 0 or has no value.
A critical point is an input \(c\) in the domain of \(f\) where \(f'(c) = 0\) or \(f'(c)\) does not exist.
Notation
Let \(c\) be an input in the domain of \(f\), the set of inputs where \(f\) has a value.
If \(f'(c) = 0\), then \(c\) is a critical point, also called a critical number. To find these inputs, set \(f'\) equal to 0. A fraction is 0 only where its numerator is 0, so first write \(f'\) as one fraction.
\[\cy{f'(c) = 0}\]If \(f(c)\) is defined but \(f'(c)\) does not exist, then \(c\) is also a critical point. A fraction has no value where its denominator is 0, so \(f'\) does not exist there.
\[\cy{f'(c)\ \text{does not exist}}\]\(f'(c)\) also does not exist at a corner, at a cusp like the opening's sharp peak, and at a vertical tangent.
A critical point is only a candidate, a place where a local maximum or minimum can occur but need not. At the flat step on the rise, \(f'(c) = 0\) but the curve does not turn.
When it fails
A critical point must be in the domain of \(f\). The button removes the top of the curve from the domain.
The ring marks a hole, where \(f\) has no value. A formula for \(f'\) may still give 0 there, but the input is not in the domain, so it is not a critical point.
Example 1: a cubic
Find the critical points of \(f(x) = x^3 - 3x^2 - 9x + 2\).
The function is a polynomial, so its domain is all real numbers and \(f'\) exists everywhere. Only a zero of \(f'\) can give a critical point.
Differentiate. Then factor the result.
\[\begin{aligned} f'(x) &= 3x^2 - 6x - 9 \\ &= 3(x - 3)(x + 1) \end{aligned}\]The derivative is 0 at \(x = -1\) and at \(x = 3\), so these are the critical points. The tangent is horizontal at \((-1, 7)\) and at \((3, -25)\).
\[\cy{x = -1}, \quad \cy{x = 3}\]
Example 2: a logarithm
Find the critical points of \(f(x) = x^2 - 8\ln x\).
The logarithm is defined only for \(x > 0\), so the domain of \(f\) is \(x > 0\).
Differentiate. Then write \(f'\) as one fraction.
\[f'(x) = 2x - \frac{8}{x} = \frac{2x^2 - 8}{x}\]A tempting wrong move: The numerator is 0 at \(x = \pm 2\), and \(f'\) does not exist at \(x = 0\), so the critical points are \(-2\), 0 and 2.
\[2x^2 - 8 = 0 \quad\Rightarrow\quad x = \pm 2\]The curve lies only to the right of the vertical axis. The inputs \(-2\) and 0, marked in red, are not in the domain, so they cannot be critical points.
Keep only the inputs in the domain, \(x > 0\). The only critical point is \(x = 2\).
\[\cy{x = 2}\]
Practice
Find the critical points of \(g(x) = |x^2 - 4|\).
The graph is the parabola \(x^2 - 4\), with the part below the axis flipped upward. Move the point to find a flat tangent and two corners.
Solution
Write \(g\) in two pieces. It equals \(x^2 - 4\) when \(|x| > 2\), and \(4 - x^2\) when \(|x| < 2\).
\[\begin{aligned} g'(x) &= 2x \quad (|x| > 2) \\ g'(x) &= -2x \quad (|x| < 2) \end{aligned}\]The slope \(-2x\) is 0 at \(x = 0\), which lies in \((-2, 2)\). The slope \(2x\) is never 0 when \(|x| > 2\).
At \(x = \pm 2\), the slope is \(-4\) on the left and 4 on the right. Two slopes mean no single tangent, so \(g'\) does not exist there.
All three inputs are in the domain, so the critical points are \(-2\), 0 and 2.
Questions
The graph shows \(f\) from \(x = -4\) to \(x = 6\). It has smooth peaks at \(x = -2\) and \(x = 5\), a corner at \(x = 1\), and an open ring, a hole, at \(x = 3\).
How many critical points does \(f\) have?
- 3
- 2
- 4
Answer
3
Which statement about \(x = 1\) is correct?
- It is a critical point, because \(f\) is defined at 1 and \(f'(1)\) does not exist.
- It is not a critical point, because the tangent is not horizontal.
- It is a critical point, because \(f'(1) = 0\).
Answer
It is a critical point, because \(f\) is defined at 1 and \(f'(1)\) does not exist.
How many critical points does \(f(x) = \dfrac{4x}{x^2 - 4}\) have?
- None
- Two, at \(x = \pm 2\)
- One, at \(x = 0\)
Answer
None
Summary
You list the inputs where \(f'\) is 0 or does not exist, and keep only those in the domain of \(f\).
A local maximum or minimum can occur only at these inputs, because everywhere else the curve is rising or falling.
You use this as the first step when you look for maximum or minimum values.
This is one of five skills in topic 5.2 of Which & Why: AP Calculus.