Classify a positive or negative result
Peaks and valleys bend
The curve is \(j(x) = 2x^3 - 9x^2 + 12x\). Its derivative is \(j'(x) = 6(x - 1)(x - 2)\), so the tangent is horizontal at \(x = 1\) and at \(x = 2\).
The faint vertical lines mark a window around the orange point, and the dashed orange line is the height of the point. The window is green where the curve is higher than that line and red where it is lower. Below, the orange bar on the graph of \(j''\) is its value at the point.
Move the point to \(x = 1\). The curve lies below its tangent, so the window is red. The curve bends down there, and \(j''(1) = -6\).
Move the point to \(x = 2\). The curve lies above its tangent, so the window is green. The curve bends up there, and \(j''(2) = 6\).
A horizontal tangent where the curve bends down marks a peak, and where it bends up it marks a valley.
The sign of the bend
The curve has a flat point at \(c\), so \(f'(c) = 0\). The slider sets the sign and the size of \(f''(c)\).
Move the slider above 0. The curve bends up and lies above its tangent, so the window is green. The point is lower than the curve on both sides, so it is a local minimum.
Move the slider below 0. The curve bends down and lies below its tangent, so the window is red. The point is higher than the curve on both sides, so it is a local maximum.
At 0, \(f''(c)\) has no sign and the curve does not bend at \(c\), so the test gives no answer.
Notation
Start at a stationary point \(c\), where \(f'(c) = 0\). The second derivative \(f''(c)\) must exist.
\[f'(c) = 0\]If \(f''(c)\) is positive, the graph is concave up at \(c\), so it lies above the horizontal tangent and \(f(c)\) is a local minimum.
\[\begin{gathered} \cg{f''(c) > 0} \\ \Rightarrow\ f(c)\ \text{is a local minimum} \end{gathered}\]If \(f''(c)\) is negative, the graph is concave down at \(c\), so it lies below the horizontal tangent and \(f(c)\) is a local maximum.
\[\begin{gathered} \cr{f''(c) < 0} \\ \Rightarrow\ f(c)\ \text{is a local maximum} \end{gathered}\]A positive second derivative at a stationary point gives a local minimum, and a negative one gives a local maximum.
When it fails
Press each button. All three curves have \(f'(c) = 0\) and \(f''(c) = 0\) at the orange point.
The first has a local minimum, the second has none, and the third has a local maximum. A second derivative of 0 does not decide between them.
Example 1: one valley
Classify the stationary point of \(q(x) = x^4 - 4x\).
Set the derivative equal to 0. The equation \(x^3 = 1\) has one real solution, \(x = 1\).
\[\begin{aligned} q'(x) = 4x^3 - 4 &= 0 \\ x^3 &= 1 \\ x &= 1 \end{aligned}\]Find the second derivative and substitute \(x = 1\).
\[q''(x) = 12x^2 \qquad q''(1) = 12\]The value 12 is positive, so the curve bends up and the window is green. The point is a local minimum.
The height of the valley is \(q(1) = -3\).
\[(1,\,-3)\]
Example 2: a peak and a valley
Classify the stationary points of \(k(x) = 3x - x^3\).
Set the derivative equal to 0 and solve.
\[\begin{aligned} k'(x) &= 3 - 3x^2 \\ &= 3(1 - x)(1 + x) \\ x &= -1 \ \text{ or }\ x = 1 \end{aligned}\]Differentiate again and substitute both points.
\[\begin{gathered} k''(x) = -6x \\ k''(1) = -6 \qquad k''(-1) = 6 \end{gathered}\]A tempting wrong move: At \(x = 1\) the second derivative \(-6\) is negative. A negative number is below zero, so the point is a local minimum.
\[k''(1) < 0 \ \Rightarrow\ \text{local minimum}\]At \(x = 1\) the window is red: the nearby heights are lower than \(k(1) = 2\), so the point is a peak.
A negative second derivative means that the curve bends down, so \(x = 1\) is a local maximum with height \(k(1) = 2\).
\[(1,\,2)\]At \(x = -1\) the second derivative is positive, so the point is a local minimum with height \(k(-1) = -2\).
\[(-1,\,-2)\]
Practice
Classify the stationary points of \(m(x) = x^3 - 6x^2 + 5\). Give the height of each.
At each flat point, decide whether the curve bends up or down. The sign of \(m''\) gives the bend.
Solution
Set the derivative equal to 0.
\[\begin{aligned} m'(x) &= 3x^2 - 12x = 3x(x - 4) \\ x &= 0 \ \text{ or }\ x = 4 \end{aligned}\]Find the second derivative and substitute both points.
\[\begin{gathered} m''(x) = 6x - 12 \\ m''(0) = -12 \qquad m''(4) = 12 \end{gathered}\]At \(x = 0\) the second derivative is negative, so the point is a local maximum with height \(m(0) = 5\).
\[(0,\,5)\]At \(x = 4\) the second derivative is positive, so the point is a local minimum with height \(m(4) = -27\).
\[(4,\,-27)\]
Questions
The graph shows \(b(x) = x^4 - 2x^2\). The tangent is horizontal at three points. When a window appears around the orange point, it is green where the curve is higher than the point and red where it is lower.
The tangent is horizontal at the orange point, \(x = 0\). Which statement about this point is correct?
- \(b''(0) < 0\), so \(b\) has a local maximum
- \(b''(0) > 0\), so \(b\) has a local maximum
- \(b''(0) < 0\), so \(b\) has a local minimum
- \(b''(0) = 0\)
Answer
\(b''(0) < 0\), so \(b\) has a local maximum
At which flat points is the second derivative positive?
- \(x = -1\) and \(x = 1\)
- \(x = 0\) only
- \(x = -1\) only
- All three
Answer
\(x = -1\) and \(x = 1\)
The only stationary point of \(g(x) = x + \dfrac{25}{x}\) for \(x > 0\) is at \(x = 5\). What is it?
- A local minimum with height \(g(5) = 10\)
- A local maximum with height \(g(5) = 10\)
- A local minimum with height \(g(5) = 5\)
- A local minimum with height \(\dfrac{2}{5}\)
Answer
A local minimum with height \(g(5) = 10\)
Summary
At a stationary point, a positive \(f''(c)\) gives a local minimum and a negative \(f''(c)\) gives a local maximum.
This works because the sign gives the bend: a curve that bends up lies above its tangent, and one that bends down lies below it.
You use the test when \(f''(c)\) is not 0, because a value of 0 decides nothing.
This is one of four skills in topic 5.7 of Which & Why: AP Calculus.