Check that the point is stationary
A flat tangent
The curve is \(f(x) = x^3 - 6x^2 + 9x + 1\). It is green where it rises and red where it falls.
The orange line is the tangent at the orange point. It has the direction of the curve there, so its slope is \(f'(x)\).
The lower graph plots this slope, and its orange bar has the height \(f'(x)\).
Drag the point along the curve. At \(x = 2\) the tangent slopes down, and the slope is \(-3\).
At \(x = 1\) the tangent is horizontal, so the slope is 0, and at \(x = 3\) it is horizontal again.
These are the only two inputs with a horizontal tangent, and at each one the curve changes between rising and falling.
A point where the slope is zero
The curve has one peak and one valley. Drag the point along it.
Where the curve rises, the tangent slopes up, so the slope is positive. Where the curve falls, the tangent slopes down, so the slope is negative.
At the peak and at the valley, the tangent is horizontal and the slope is 0, so the bar has no height. The label \(f'(c) = 0\) appears there, where \(c\) is the input of the point.
A point with a horizontal tangent is a stationary point. The second derivative test starts at such a point.
Notation
A critical point of \(f\) is an input \(c\) in the domain where \(f'(c) = 0\) or where \(f'(c)\) does not exist.
\[\begin{gathered} f'(c) = 0 \\ \text{or } f'(c)\ \text{does not exist} \end{gathered}\]The second derivative test needs the first case, \(f'(c) = 0\). Such an input \(c\) is a stationary point, because the tangent there is horizontal.
\[\cy{f'(c)} = 0\]To find the stationary points, differentiate \(f\), set \(f'(x) = 0\) and solve for \(x\).
\[f'(x) = 0\]Keep only the solutions in the domain of \(f\), because the test applies at these inputs and at no others.
When it fails
Drag the point away from the valley. The tangent tilts, so \(f'(c)\) is no longer 0, and the test does not apply at an input that is not stationary.
Press the button. A sharp corner has one slope on the left and another on the right, so \(f'(c)\) does not exist and the test does not apply. At a corner, you compare the signs of \(f'\) on each side.
Example 1: the two stationary points of a cubic
Find the stationary points of \(g(x) = 2x^3 - 3x^2 - 12x + 4\).
The slope of the tangent is \(g'(x)\), so differentiate \(g\) with the power rule.
\[g'(x) = 6x^2 - 6x - 12\]A stationary point needs \(g'(x) = 0\), so factor the derivative to solve this.
\[6x^2 - 6x - 12 = 6(x - 2)(x + 1)\]A product is 0 only when one factor is 0, which gives two inputs where the tangent is horizontal.
\[x = -1 \qquad x = 2\]Both inputs are in the domain, because \(g\) is defined for every \(x\). The heights are \(g(-1) = 11\) and \(g(2) = -16\).
\[(-1,\,11) \qquad (2,\,-16)\]
Example 2: a zero of the function is not a stationary point
Is \(x = 3\) a stationary point of \(p(x) = x^3 - 9x\)? Find the stationary points of \(p\).
The curve crosses the x-axis at \(x = 3\), because the height there is 0.
\[p(3) = 27 - 27 = 0\]A tempting wrong move: The height is 0, so \(x = 3\) is a stationary point.
\[p(3) = 0 \ \Rightarrow\ p'(3) = 0\]The orange tangent at \(x = 3\) slopes up steeply across the x-axis. A stationary point needs a slope of 0, which is a horizontal tangent, and a height of 0 is a different fact.
A stationary point needs \(p'(3) = 0\). Differentiate and substitute 3.
\[p'(x) = 3x^2 - 9 \qquad p'(3) = 18\]The slope is 18, not 0, so the point is not stationary and the test does not apply there.
To find the stationary points, solve \(3x^2 - 9 = 0\). The tangent is horizontal at both solutions.
\[\begin{gathered} x^2 = 3 \\ x = -\sqrt{3} \ \text{ or }\ x = \sqrt{3} \end{gathered}\]
Practice
Find the stationary point of \(h(x) = x + \dfrac{9}{x}\) for \(x > 0\).
A stationary point has a horizontal tangent. Move the point to the input where the bar on the lower graph has no height.
Solution
Write \(h(x) = x + 9x^{-1}\) and differentiate each term.
\[h'(x) = 1 - \frac{9}{x^2}\]Set the slope equal to 0 and solve.
\[1 - \frac{9}{x^2} = 0 \quad\Rightarrow\quad x^2 = 9\]Both \(x = 3\) and \(x = -3\) solve this equation, but the domain is \(x > 0\), so only \(x = 3\) is allowed.
The stationary point is at \(x = 3\), and its height is \(h(3) = 6\).
\[(3,\,6)\]
Questions
The graphs show \(r(x) = x^4 - 8x^2\) and its slope \(r'\). The curve is green where it rises and red where it falls.
Is \(x = 1\) a stationary point of \(r\)?
- No, because the tangent is not horizontal
- Yes, because the point is close to a valley
- Yes, because \(r(1)\) is negative
Answer
No, because the tangent is not horizontal
How many stationary points does the graph of \(r\) have?
- 3
- 2
- 1
- 4
Answer
3
Which input is a stationary point of \(s(x) = x^3 + 3x^2 - 9x\)?
- \(x = 1\)
- \(x = 3\)
- \(x = 0\)
Answer
\(x = 1\)
Summary
You differentiate \(f\), set \(f'(x) = 0\) and keep the solutions that lie in the domain.
Each solution is a stationary point, because the tangent there is horizontal and has slope 0.
You do this before the second derivative test, which applies only at a stationary point.
This is one of four skills in topic 5.7 of Which & Why: AP Calculus.