Connect zeros with candidate features

Zeros of f, f′ and f″

The graph of \(f\) meets the x-axis at \(x = 0\) and \(x = 3\), the zeros of \(f\).

Move the point to \(x = 1\). The tangent is flat, so \(f'(1) = 0\), and the graph of \(f'\) crosses the axis there.

Move the point to \(x = 2\). The graph of \(f''\) crosses the axis, and the bend of \(f\) changes from down to up.

At \(x = 3\), both \(f\) and \(f'\) are 0: the curve touches the axis, and the tangent is flat.

The same zeros on any curve

Press each button in turn. At a zero of \(f\) the graph meets the x-axis. At a zero of \(f'\) the tangent is flat. At a zero of \(f''\) the tangent may cross the curve.

A zero of \(f'\) or \(f''\) is only a candidate, a point where \(f\) might have a maximum, a minimum or a change of bend.

Notation

1. A zero of \(f\) is a number \(c\) with \(f(c) = 0\). The graph meets the x-axis there.

   \[f(c) = 0\]

2. A critical point is a number \(c\) where \(f'(c) = 0\) or \(f'(c)\) does not exist. It is a candidate for a local maximum or minimum.

   \[\cy{f'(c) = 0}\]

3. Suppose \(f\) is continuous at \(c\). If \(f''(c) = 0\) or \(f''(c)\) does not exist, \(c\) is a candidate for an inflection point, where the bend changes.

   \[\cy{f''(c) = 0}\]

4. To name a candidate, test the sign on both sides of it. At a local maximum or minimum, \(f'\) changes sign, and at an inflection point, \(f''\) changes sign.

When it fails

Slide the two zeros of \(f'\) together. The local maximum and the local minimum move toward each other.

When the zeros meet, \(f'\) is still 0 there but positive on both sides. The curve is green everywhere because \(f\) rises everywhere, so there is no maximum or minimum. The graph of \(f''\) still crosses the axis there, so the point is an inflection point.

Example 1: sort the zeros of one function

For \(f(x) = (x - 2)e^x\), find the zero of \(f\), the zero of \(f'\) and the zero of \(f''\). Say what each one means.

1. Because \(e^x\) is never 0, \(f(x) = 0\) only when \(x = 2\), where the graph meets the x-axis.

2. The product rule gives \(f'\), which has the sign of \(x - 1\) because \(e^x > 0\). It is negative before 1 and positive after, so \(f\) has a local minimum at \(x = 1\), where \(f(1) = -e\).

   \[\begin{aligned} f'(x) &= e^x + (x - 2)e^x \\ &= (x - 1)e^x \end{aligned}\]

3. The product rule gives \(f''\), which has the sign of \(x\). It is negative before 0 and positive after, so the bend changes and \(f\) has an inflection point at \((0, -2)\).

   \[f''(x) = e^x + (x - 1)e^x = x\,e^x\]

Example 2: a zero of f″ that is not an inflection point

Find the inflection points of \(f(x) = x^6 - 3x^4\).

1. The candidates are the zeros of \(f''\), which are \(x = 0\) and \(x = \pm\sqrt{\tfrac65}\), about \(\pm 1.1\).

   \[\begin{aligned} f'(x) &= 6x^5 - 12x^3 \\ f''(x) &= 30x^4 - 36x^2 \\ &= 6x^2(5x^2 - 6) \\ x &= 0 \text{ or } x = \pm\sqrt{6/5} \end{aligned}\]

2. A tempting wrong move: \(f''(0) = 0\), so \((0, 0)\) is an inflection point.

   \[f''(0) = 0 \Rightarrow \text{inflection at } 0\]

   The curve stays below its tangent, the x-axis, on both sides of 0, so \(f\) bends down on both sides. The graph of \(f''\) touches the axis at 0 and does not cross it, so the bend does not change.

3. The factor \(6x^2\) is positive except at 0, so \(f''\) has the sign of \(5x^2 - 6\), which is negative on both sides of 0. Because \(f''\) does not change sign there, 0 is not an inflection point.

4. At \(x = \pm\sqrt{\tfrac65}\), the factor \(5x^2 - 6\) changes sign, so \(f''\) changes sign. These two points are the inflection points of \(f\).

Practice

Let \(f'(x) = (x - 1)^2(x - 4)\). Which zero of \(f'\) gives a local maximum or minimum of \(f\)?

Move the point to both sides of each zero, close to it. Read the sign of \(f'\) there.

Questions

The graphs show \(f\), \(f'\) and \(f''\) for one function.

1. What is at \(x = 1\)?

   • A local minimum
   • A local maximum
   • An inflection point
   • A zero of \(f\)
   Answer

   A local minimum

2. At \(x = 0\), both \(f'(0) = 0\) and \(f''(0) = 0\). What does \(f\) have there?

   • An inflection point with a flat tangent
   • A local maximum
   • A local minimum
   • No feature, because both derivatives are 0
   Answer

   An inflection point with a flat tangent

3. Let \(f''(x) = (x - 1)^2(x + 2)\). Where does \(f\) have inflection points?

   • At \(x = -2\) only
   • At \(x = 1\) only
   • At \(x = -2\) and \(x = 1\)
   • At no point
   Answer

   At \(x = -2\) only

Summary

The zeros of \(f\), \(f'\) and \(f''\) mark different points: the graph meets the axis, the tangent is flat, or the bend may change.

A zero of \(f'\) or \(f''\) is only a candidate, because a maximum, a minimum or an inflection point needs a sign change there.

You use this when a problem asks where \(f\) has a maximum, a minimum or an inflection point.

This is one of five skills in topic 5.9 of Which & Why: AP Calculus.