Handle zeros without a sign change

A flat spot that continues to rise

The upper graph shows \(f(x) = (x - 1)^3 + 2\), and it is green because it rises. The lower graph shows its derivative \(f'(x) = 3(x - 1)^2\), the slope of \(f\).

Drag the orange point to \(x = 1\). The tangent is horizontal, so \(f'(1) = 0\), and the lower graph touches the x-axis there.

Drag the point past \(x = 1\). The curve continues to rise, because \(f'\) is positive on both sides of 1. It is flat for an instant at \((1, 2)\) but does not turn.

A touch is not a crossing

The scene is the same without numbers. The slider moves the lower graph up or down.

Where \(f'\) crosses the axis, it changes sign. There \(f\) turns, and its color changes from green to red or from red to green.

Move the slider to 0. The two crossings come together in one touch, and the lower graph stays above the axis on both sides, so \(f\) rises without a turn.

A turn needs a change of sign of \(f'\). A zero of \(f'\) is not enough.

Notation

1. Let \(f'(c) = 0\). If \(f'\) has the same sign on both sides of \(c\), then \(f\) does not turn at \(c\).

   \[\begin{gathered} \cg{+} \to 0 \to \cg{+} \\ \text{or} \\ \cr{-} \to 0 \to \cr{-} \end{gathered}\]

2. The dashed line is at the height \(f(c)\). Left of \(c\) the curve is below this line, and right of \(c\) it is above the line, so \(f\) increases through \(c\).

   \[\begin{gathered} p < c < r \;\Rightarrow\; f(p) < f(c) < f(r) \\ (p \text{ and } r \text{ near } c) \end{gathered}\]

3. Write one interval through \(c\). It ends where the sign of \(f'\) changes, here at \(q\), not at \(c\).

   \(f\) can be increasing on an interval even if \(f'\) is 0 at single points of it.

4. If \(f'\) is a product of factors, find the power of the factor that is 0 at \(c\). An even power is never negative, so it keeps the sign of \(f'\), and an odd power changes it.

   \[\begin{gathered} (x - c)^n: \\ n \text{ even, no sign change} \\ n \text{ odd, sign change} \end{gathered}\]

When it fails

The sign of \(f'\) is positive on both sides of \(c\), as before.

But \(f\) now jumps down at \(c\), so it is not increasing on any interval that contains \(c\). The rule needs \(f\) to be continuous at \(c\), with no jump.

Example 1: One zero turns, one does not

Find the inputs where \(f(x) = 3x^4 + 4x^3\) changes direction.

1. Factor \(f'\) to find its zeros.

   \[\begin{aligned} f'(x) &= 12x^3 + 12x^2 \\ &= 12x^2(x + 1) \end{aligned}\]

2. The zeros are \(x = 0\) and \(x = -1\).

   \[x = -1, \qquad x = 0\]

3. The factor \(x^2\) is positive on both sides of 0, because a square is never negative. Near 0, \(f'\) has the sign of \(x + 1\), which is positive.

   \[\cg{+}\ \text{then}\ \cg{+}\ \text{at}\ x = 0\]

4. The factor \(x + 1\) changes sign at \(-1\), so \(f'\) goes from negative to positive and \(f\) turns there. At 0, \(f\) flattens and continues to rise, so \(f\) changes direction at \(x = -1\) only.

   \[\cr{-}\ \text{then}\ \cg{+}\ \text{at}\ x = -1\]

Example 2: A factor with an even power

The derivative of \(f\) is \(f'(x) = x(x - 3)^2\), and \(f\) has a value at every \(x\). Find the inputs where \(f\) changes direction, and state where \(f\) is increasing.

1. The zeros of \(f'\) are \(x = 0\) and \(x = 3\).

   \[x(x - 3)^2 = 0 \;\Rightarrow\; x = 0,\ x = 3\]

2. A tempting wrong move: The derivative is 0 at both zeros, so \(f\) turns at both \(x = 0\) and \(x = 3\).

   \[\begin{gathered} f'(0) = 0,\ f'(3) = 0 \\ \Rightarrow\; \text{turns at both} \end{gathered}\]

   The curve rises on both sides of \(x = 3\), because \(f'(2.5) = 0.625\) and \(f'(3.5) = 0.875\) are both positive.

3. At \(x = 0\) the factor \(x\) changes sign, and \((x - 3)^2\) is positive. The sign of \(f'\) goes from negative to positive, so \(f\) turns.

   \[\cr{-}\ \text{then}\ \cg{+}\ \text{at}\ x = 0\]

4. Near \(x = 3\) the factor \(x\) is positive, and \((x - 3)^2\) is positive on both sides, so \(f'\) stays positive and \(f\) does not turn.

   \[\cg{+}\ \text{then}\ \cg{+}\ \text{at}\ x = 3\]

5. The function \(f\) changes direction at \(x = 0\) only. It has a derivative at \(x = 3\), so it is continuous there, and the two rising intervals join: \(f\) is increasing on \((0, \infty)\).

Practice

The derivative of \(g\) is \(g'(x) = (x + 3)(x - 1)^2(x - 5)\), and \(g\) has a value at every \(x\). Find the inputs where \(g\) changes direction, and state where \(g\) is decreasing.

On the lower graph, compare the sign of \(g'\) on both sides of each zero.

Questions

The graph shows a function \(h\) and its derivative \(h'\). The lower graph touches the axis at one input and crosses it at another.

1. At which input is \(h'(x) = 0\) but \(h\) does not change direction?

   • 1
   • −2
   • 0
   Answer

   1

2. What is the largest interval on which \(h\) is increasing?

   • \((-2, \infty)\)
   • \((-2, 1)\)
   • \((1, \infty)\)
   Answer

   \((-2, \infty)\)

3. Let \(f'(x) = x^2(x - 3)^3(x + 1)^4\). At which input does \(f\) change direction?

   • 3 only
   • \(-1\), 0 and 3
   • \(-1\) and 0
   Answer

   3 only

Summary

A zero of \(f'\) gives a horizontal tangent, but \(f\) turns there only if \(f'\) changes sign.

Here \(f' = 3(x - 1)^2\) is positive on both sides of 1, so \(f\) is increasing on \((-\infty, \infty)\).

When you state intervals, end each one only where \(f'\) changes sign, not at every zero.

This is one of four skills in topic 5.3 of Which & Why: AP Calculus.