Find where the derivative's sign can change
Where the slope changes sign
The upper graph shows a function \(f\). The lower graph shows its derivative \(f'\), which gives the slope of \(f\) at each input.
Drag the orange point. The curve is green where it rises, because its slope is positive, and red where it falls, because its slope is negative.
The color changes only at \(x = 1\) and \(x = 2\), where the tangent is horizontal, so \(f'\) is 0.
The graph of \(f'\) has no break, so \(f'\) can change sign only where it crosses the axis, at a zero.
Cut points divide the line
The numbers are gone: \(f\) is now any function, and \(f'\) is its derivative where it exists.
A cut point is an input where the sign of \(f'\) might change. The button marks each cut point on the number line, then writes the sign of \(f'\) between the marks.
An interval is the part of the number line between two neighboring cut points. Drag the orange point along one: the sign of \(f'\) and the color of \(f\) do not change inside it.
Notation
The first type of cut point is an input \(c\) where \(f'(c) = 0\), so the tangent at \(c\) is horizontal.
\[\cy{f'(c)} = 0\]The second type is an input \(c\) where \(f(c)\) has a value but \(f'(c)\) does not. The graph has a corner, a cusp or a vertical tangent at \(c\), so it has no single slope there.
\[\begin{gathered} f(c)\text{ exists} \\ \cy{f'(c)}\text{ does not exist} \end{gathered}\]The third type is an input \(c\) where \(f(c)\) has no value, for example because a formula divides by 0. With no point at \(c\), the graph has no slope there.
\[f(c)\text{ is undefined}\]Between two neighboring cut points, \(f'\) exists, is continuous and is never 0, so it keeps one sign on each interval. A filled dot marks a zero of \(f'\), and an open ring marks an input where \(f'\) or \(f\) has no value. The signs show + in green and − in red.
The sign can change at a cut point, but it does not have to.
When it fails
The button first rounds the bottom of the curve. Then \(f'\) crosses the axis at the bottom, so the sign changes at a zero.
When the bottom sharpens into a corner again, \(f'\) jumps from \(-1.5\) to 1.5 there, with no zero. Its sign still changes, so an input where \(f'\) does not exist is a cut point too.
Example 1: A derivative with zeros
Find the inputs where the sign of \(f'\) can change for \(f(x) = x^4 - 8x^2\).
Both \(f\) and \(f'\) have a value at every input, because they are polynomials, so the only cut points are the zeros of \(f'\).
\[f'(x) = 4x^3 - 16x\]Factor \(f'\). A product is 0 only when one of its factors is 0.
\[f'(x) = 4x(x - 2)(x + 2)\]The zeros are \(-2\), 0 and 2, so these are the cut points.
\[x = -2, \qquad x = 0, \qquad x = 2\]The three cut points divide the number line into four intervals, and \(f'\) has one sign on each.
Example 2: A function with a limited domain
Find the inputs where the sign of \(f'\) can change for \(f(x) = x^2 - 8\ln x\).
The logarithm \(\ln x\) has a value only for \(x > 0\), so the graph of \(f\) lies only to the right of the vertical axis.
\[\text{domain: } x > 0\]Write \(f'\) as one fraction to see where its numerator is 0.
\[\begin{aligned} f'(x) &= 2x - \dfrac{8}{x} = \dfrac{2x^2 - 8}{x} \\ &= \dfrac{2(x - 2)(x + 2)}{x} \end{aligned}\]A tempting wrong move: A fraction is 0 when its numerator is 0. The numerator is 0 at \(x = 2\) and at \(x = -2\), so both are cut points.
\[x = 2, \qquad x = -2\]The graph has no curve at \(x = -2\), because \(f\) is not defined for \(x \le 0\), so there is no slope there.
Keep only the zero inside the domain \(x > 0\): the only cut point is \(x = 2\). The open ring at 0 marks where the domain begins.
\[x = 2\]The cut point divides the domain into two intervals, and \(f'\) has one sign on each.
Practice
Find the inputs where the sign of \(f'\) can change for \(f(x) = 3x^{2/3} - 2x\).
The graph of \(f\) has a sharp point and a top with a horizontal tangent. Each one is a cut point.
Solution
A cube root has a value for every number, so \(f\) has a value at every input.
\[f'(x) = \dfrac{2}{\sqrt[3]{x}} - 2\]The derivative has no value at \(x = 0\), because its denominator \(\sqrt[3]{x}\) is 0 there, and the tangent is vertical. But \(f(0) = 0\), so \(x = 0\) is a cut point of the second type.
\[f(0) = 0, \qquad f'(0)\text{ does not exist}\]To find the zeros of \(f'\), set it equal to 0. Then \(2/\sqrt[3]{x} = 2\), so \(\sqrt[3]{x} = 1\) and \(x = 1\).
\[\dfrac{2}{\sqrt[3]{x}} - 2 = 0 \;\Rightarrow\; x = 1\]The cut points are \(x = 0\) and \(x = 1\).
Questions
The graph shows \(p(x) = |x^2 - 4|\) and its derivative \(p'\). The graph of \(p\) has sharp points at \(x = -2\) and \(x = 2\).
At which input is \(p'(x) = 0\)?
- 0
- 2
- −4
Answer
0
Why can the sign of \(p'\) change at \(x = 2\)?
- \(p'(2)\) does not exist, because the graph has a sharp point at \(x = 2\)
- \(p'(2) = 0\), because the tangent at \(x = 2\) is horizontal
- \(p\) has no value at \(x = 2\)
Answer
\(p'(2)\) does not exist, because the graph has a sharp point at \(x = 2\)
Which inputs are cut points for \(f(x) = \ln(x^2 - 9)\)?
- \(-3\) and 3
- 0
- \(-3\), 0 and 3
Answer
\(-3\) and 3
Summary
You list the inputs where \(f'\) is 0, the inputs where \(f'\) does not exist, and the breaks in the domain of \(f\).
The sign of \(f'\) can change only at these cut points, because between two of them \(f'\) exists, is continuous and is never 0.
Divide the domain at the cut points before you decide where f rises and falls.
This is one of four skills in topic 5.3 of Which & Why: AP Calculus.