Evaluate the second derivative
How fast the slope changes
The upper curve is \(k(x) = x^3 - 12x + 3\), and the lower curve is its slope, \(k'(x) = 3x^2 - 12\).
Move the point to \(x = 2\). The slope of \(k\) is 0, and the lower curve rises through the x-axis.
Move the point to \(x = 2.1\). The slope of \(k\) is now 1.23, so it grew by 1.23 over a step of 0.1, about 12 per unit.
The second derivative \(k''(2) = 12\) is this rate at the point itself, the slope of the orange tangent to the lower curve at \(x = 2\).
At \(x = -2\) the lower curve falls through the axis, so the slope of \(k\) falls, at 12 per unit, and \(k''(-2) = -12\).
The slope of the slope
The curve has one valley and one peak. Move the point to the valley. Its input \(c\) is a stationary point, because the slope of \(f\) is 0 there, so the lower curve crosses the x-axis at \(c\).
The lower curve is green where it rises and red where it falls. At \(c\) it rises as it crosses, so its slope, the second derivative \(f''(c)\), is positive.
Move the point to the peak, and let \(c\) be its input instead. The lower curve falls as it crosses, so \(f''(c)\) is negative.
The second derivative at a point is the slope of the graph of \(f'\) there, so a steeper crossing gives a value farther from 0.
Notation
The second derivative is the derivative of the derivative, so you differentiate \(f\) to find \(f'\), and then differentiate \(f'\) to find \(f''\).
\[f''(x) = \frac{d}{dx}\big(f'(x)\big)\]Substitute the stationary point \(c\) last, after you find \(f''\). The value \(f''(c)\) is the slope of the orange tangent to the graph of \(f'\) at \(c\).
\[\cy{f''(c)}\]If you substitute first, \(f'(c)\) is a number, and the derivative of a number is 0.
The second derivative test reads the sign of \(f''(c)\) to classify \(c\), so the value must exist.
When it fails
Press the button. The graph of \(f'\) now crosses the axis at \(c\) with a vertical tangent, and a vertical line has no slope.
This means that \(f''(c)\) does not exist, even though \(f\) still has a local minimum at \(c\). For example, \(f(x) = x^{4/3}\) has \(f'(x) = \tfrac43 x^{1/3}\), which has a vertical tangent at 0.
Example 1: a cubic with two stationary points
Evaluate \(u''\) at each stationary point of \(u(x) = x^3 - 3x^2 - 9x + 5\).
The stationary points are the zeros of \(u'\), so differentiate and factor.
\[\begin{aligned} u'(x) &= 3x^2 - 6x - 9 \\ &= 3(x - 3)(x + 1) \end{aligned}\]The factors are 0 at \(x = 3\) and at \(x = -1\), so both are stationary points.
\[x = 3 \qquad x = -1\]Differentiate \(u'\) to find \(u''\), using the power rule on each term.
\[u''(x) = 6x - 6\]Substitute each stationary point into \(u''\). The signs match the picture: the graph of \(u'\) rises through the axis at \(x = 3\) and falls through it at \(x = -1\).
\[u''(3) = 12 \qquad u''(-1) = -12\]
Example 2: a product of two functions
Evaluate \(w''\) at each stationary point of \(w(x) = x^2 e^{-x}\).
Use the product rule on \(w\) to find \(w'\), the lower curve.
\[\begin{aligned} w'(x) &= 2x e^{-x} - x^2 e^{-x} \\ &= (2x - x^2)e^{-x} \end{aligned}\]The factor \(e^{-x}\) is never 0, so \(w'\) is 0 only where the other factor \(x(2 - x)\) is 0.
\[x = 0 \qquad x = 2\]A tempting wrong move: At \(x = 2\) the slope \(w'(2)\) is 0, and the derivative of 0 is 0, so \(w''(2) = 0\).
\[w''(2) = \frac{d}{dx}(0) = 0\]The number 0 is the height of the lower curve at 2, not its slope. The orange tangent there slopes down, so \(w''(2)\) is negative, not 0.
Differentiate \(w'\) as a function of \(x\) first, using the product rule again.
\[\begin{aligned} w''(x) &= (2 - 2x)e^{-x} \\ &\quad - (2x - x^2)e^{-x} \\ &= (x^2 - 4x + 2)e^{-x} \end{aligned}\]Substitute the two stationary points. As a check, \(w''(2) \approx -0.27\), the slope shown beside the orange tangent at \(x = 2\).
\[w''(2) = -2e^{-2} \qquad w''(0) = 2\]
Practice
Evaluate \(n''\) at each stationary point of \(n(x) = 2x^3 + 3x^2 - 12x\).
Find the inputs where the graph of \(n'\) crosses the x-axis. Then ask whether \(n'\) rises or falls at each one.
Solution
Differentiate and factor, because the stationary points are the zeros of \(n'\).
\[\begin{aligned} n'(x) &= 6x^2 + 6x - 12 \\ &= 6(x + 2)(x - 1) \end{aligned}\]The zeros are \(x = -2\) and \(x = 1\).
Differentiate \(n'\) to find the second derivative.
\[n''(x) = 12x + 6\]Substitute each stationary point. The signs match the picture: the graph of \(n'\) falls at \(x = -2\) and rises at \(x = 1\).
\[n''(-2) = -18 \qquad n''(1) = 18\]
Questions
The upper graph shows \(z(x) = x^3 - 3x\), and the lower graph shows its slope \(z'\). The orange point starts at \(x = 1\), and an orange dot on the lower graph marks the same \(x\).
The orange point is at \(x = 1\). What is the sign of \(z''(1)\)?
- Positive, because the lower graph rises through the axis
- Zero, because \(z'(1) = 0\)
- Negative, because the lower graph is below the axis to the left of 1
Answer
Positive, because the lower graph rises through the axis
What is \(z''(0)\)?
- 0
- \(-3\)
- 6
Answer
0
The only stationary point of \(t(x) = x^4 - 32x\) is at \(x = 2\). What is \(t''(2)\)?
- 48
- 0
- 24
Answer
48
Summary
You differentiate \(f\) twice and substitute the stationary point \(c\) into \(f''\).
The value \(f''(c)\) is the slope of the graph of \(f'\) at \(c\), so it tells you how fast the slope of \(f\) changes there.
The second derivative test reads the sign of this value, so you find it at every stationary point you want to classify.
This is one of four skills in topic 5.7 of Which & Why: AP Calculus.