Handle an inconclusive result
Three curves, one second derivative
Press a button to show each curve. All three have a horizontal tangent at \(x = 2\), and each is green where it rises and red where it falls.
Curve A is \((x - 2)^4 + 1\), curve B is \(5 - (x - 2)^4\), and curve C is \((x - 2)^3 + 3\).
For all three, \(f'(2) = 0\) and \(f''(2) = 0\), so the second derivative cannot tell them apart.
Press the play button to read the slope at \(x = 1\) and at \(x = 3\). On curve A it is \(-4\), then 4. On curve B it is 4, then \(-4\). On curve C it is 3 on both sides.
The signs of the slope decide, as the number line under the lower graph shows. Curve A has a local minimum, curve B a local maximum, and curve C neither.
The slope on each side
When \(f''(c) = 0\), the test gives no answer, so you use the slope instead.
Choose an input close to \(c\) on each side. Read the sign of the slope at each from the orange bar on the lower graph.
If the slope is negative, then positive, the curve falls and then rises, so the point is a local minimum. Positive, then negative gives a local maximum.
If both slopes have the same sign, the curve moves in one direction through \(c\), so the point is neither.
Notation
When \(f''(c) = 0\) or \(f''(c)\) does not exist, the test is inconclusive: the point can be a local minimum, a local maximum or neither.
\[\begin{gathered} f'(c) = 0, \quad f''(c) = 0 \\ \Rightarrow\ \text{no conclusion} \end{gathered}\]The slope can change sign only at a critical point, where \(f'\) is 0 or does not exist. Choose a test input close to \(c\) on each side, with no other critical point between \(c\) and either input.
Find the sign of \(f'\) at each test input.
\[\begin{gathered} \cr{-} \to \cg{+}:\ \text{local min} \\ \cg{+} \to \cr{-}:\ \text{local max} \\ \text{same sign:}\ \text{neither} \end{gathered}\]You can also compare \(f(c)\) with the heights at the two test inputs. If \(f(c)\) is lower than both, the point is a local minimum, and if higher, a local maximum.
When it fails
Press the button to show a curve with a second flat point at \(x = 3\). The left test point stays at \(x = 1\), where the slope is negative.
Move the right test point from 2.5 past 3. At 2.5 the slope is negative, but past 3 it is positive, so the signs read negative, then positive, as for a local minimum. That reading is wrong, because the sign changes at 3, another critical point. The slope is negative on both sides of 2, so that point is neither.
Example 1: the slope has the same sign on both sides
Classify the stationary point of \(p(x) = x^3 - 6x^2 + 12x\).
Find the derivative. The point \(x = 2\) is stationary.
\[\begin{aligned} p'(x) &= 3x^2 - 12x + 12 \\ &= 3(x - 2)^2 \\ p'(2) &= 0 \end{aligned}\]The second derivative is 0 at \(x = 2\), so the test gives no answer.
\[p''(x) = 6x - 12 \qquad p''(2) = 0\]\(p'\) has no other zero, so use the test inputs 1 and 3.
\[p'(1) = 3 \qquad p'(3) = 3\]Both slopes are positive, so the curve rises through \(x = 2\) and the point is neither. The heights also rise: \(p(1) = 7\), \(p(2) = 8\), \(p(3) = 9\).
\[7 < 8 < 9\]
Example 2: a valley with a zero second derivative
Classify the stationary point \(x = 0\) of \(q(x) = \cos x + \dfrac{x^2}{2}\).
Find the derivative. The point \(x = 0\) is stationary.
\[q'(x) = x - \sin x \qquad q'(0) = 0\]The second derivative is 0 at \(x = 0\), so the test gives no answer.
\[q''(x) = 1 - \cos x \qquad q''(0) = 0\]A tempting wrong move: The second derivative is 0, so the point is neither.
\[q''(0) = 0 \ \Rightarrow\ \text{neither}\]A second derivative of 0 gives no answer, so it rules nothing out. On the graph, \(q(0) = 1\) and the heights at \(x = \pm 1\) are about 1.04, so the curve has a valley at 0.
Use the signs of \(q'\) at the test inputs \(-1\) and 1. The only solution of \(x = \sin x\) is \(x = 0\), so no other stationary point lies between them.
\[\begin{gathered} q'(-1) = -1 + \sin 1 \approx -0.16 \\ q'(1) = 1 - \sin 1 \approx 0.16 \end{gathered}\]The slope is negative, then positive, so the point \(x = 0\) is a local minimum.
Practice
Find the local extrema of \(f(x) = \dfrac{x^5}{5} - \dfrac{x^4}{2}\).
Use the second derivative where it is not 0. Where it is 0, read the sign of the slope on each side from the lower graph.
Solution
The stationary points are \(x = 0\) and \(x = 2\).
\[f'(x) = x^4 - 2x^3 = x^3(x - 2)\]At \(x = 2\) the second derivative is positive, so \(f\) has a local minimum there.
\[\begin{aligned} f''(x) &= 4x^3 - 6x^2 \\ &= 2x^2(2x - 3) \\ f''(2) &= 8 \\ f(2) &= -\tfrac85 \end{aligned}\]At \(x = 0\) the second derivative is 0. Use the test inputs \(-1\) and 1, with no other stationary point between them.
\[f'(-1) = 3 \qquad f'(1) = -1\]The slope is positive, then negative, so the point \(x = 0\) is a local maximum.
\[f(0) = 0\]
Questions
The graphs show \(c(x) = \dfrac{x^4}{4} - x^3\) and its slope \(c'\). The curve is green where it rises and red where it falls.
The second derivative is 0 at \(x = 0\). What does the graph show there?
- Neither a local maximum nor a local minimum
- A local minimum, because the tangent is horizontal
- A local maximum, because \(c''(0) = 0\)
- It cannot be decided from the graph
Answer
Neither a local maximum nor a local minimum
Where does the sign of the slope change?
- At \(x = 3\) only, from negative to positive
- At \(x = 0\) and at \(x = 3\)
- At \(x = 0\) only
- Nowhere
Answer
At \(x = 3\) only, from negative to positive
For \(g(x) = 10 - x^6\), the derivative is 0 at \(x = 0\) and \(g''(0) = 0\). What is the point?
- A local maximum
- A local minimum
- Neither, because the second derivative is 0
- It cannot be classified
Answer
A local maximum
Summary
When \(f''(c) = 0\), you compare the signs of \(f'\) at a test input on each side of \(c\).
A change from negative to positive gives a local minimum, positive to negative gives a local maximum, and no change gives neither.
The same signs decide when \(f''(c)\) does not exist, because the test gives no answer there either.
This is one of four skills in topic 5.7 of Which & Why: AP Calculus.