State increasing and decreasing intervals

From signs to intervals

The curve of \(f\) is green where it rises and red where it falls. The lower graph shows \(f'\), which is 0 at \(x = 0\) and at \(x = 2\). These two inputs are cut points, where the sign of \(f'\) can change, as the sign chart at the bottom shows.

Drag the orange point from left to right. Between 0 and 2, \(f'\) is above the axis and \(f\) rises. To the left of 0 and to the right of 2, \(f'\) is below the axis and \(f\) falls.

The button labels each colored part with an interval of \(x\) whose ends are cut points. The first and the last part have only one cut point, so their other end is \(-\infty\) or \(\infty\).

An interval between cut points

This curve has a break at the dashed vertical line, called an asymptote, where \(f\) has no value. It is still green where \(f\) rises and red where it falls.

A function is increasing on an interval if a larger input always gives a larger output. It is decreasing on an interval if a larger input always gives a smaller output.

The sign chart has three cut points: the two inputs where the tangent is horizontal, and the asymptote. If \(f'\) is positive on a whole interval, \(f\) is increasing on it, and if \(f'\) is negative there, \(f\) is decreasing.

The two red parts lie on opposite sides of the asymptote, where the curve is not connected, so they are two separate intervals.

Notation

1. The tangent at the point gives the sign of \(f'\) on each interval. Where \(f' > 0\), \(f\) is increasing, and where \(f' < 0\), \(f\) is decreasing, as the arrows under the sign chart show.

   \[\begin{gathered} f' > 0 \text{ on } (p, q) \\ \Rightarrow\; f \text{ increasing on } (p, q) \end{gathered}\]

2. Write each interval from one cut point to the next, with parentheses. Intervals use inputs, never outputs. If no cut point ends an interval, write \(-\infty\) or \(\infty\) there.

3. At a break, keep the intervals apart. \(f\) is decreasing on \((a, b)\) and on \((b, c)\), but not on \((a, c)\), because \(f\) has no value at \(b\).

4. If \(f\) is continuous at an end, a closed end with a square bracket, as in \([c, \infty)\), is also correct. An end at a break keeps its parenthesis, because \(f\) has no value there.

When it fails

The derivative is negative on \((a, b)\) and on \((b, c)\), so \(f\) is decreasing on each of them.

The line joins a point on each side of the asymptote. It rises, so a larger input gives a larger output, and \(f\) is not decreasing on any interval that contains \(b\). Write the two intervals separately.

Example 1: A rational function

State the intervals on which \(f(x) = \dfrac{x}{x^2 + 1}\) is increasing and on which it is decreasing.

1. Use the quotient rule to find \(f'\). The denominator is positive for every \(x\), so the sign of \(f'\) is the sign of \(1 - x^2\).

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

2. The cut points are \(x = -1\) and \(x = 1\). Test one point in each interval: \(f'(-2)\) is negative, \(f'(0) = 1\) is positive and \(f'(2)\) is negative.

   \[\begin{gathered} \cr{-}\ (x < -1), \quad \cg{+}\ (-1 < x < 1), \\ \cr{-}\ (x > 1) \end{gathered}\]

3. Write each interval: \(f\) is increasing where \(f'\) is positive and decreasing where \(f'\) is negative.

   \[\begin{aligned} &\text{increasing on } (-1, 1) \\ &\text{decreasing on } (-\infty, -1) \\ &\text{and on } (1, \infty) \end{aligned}\]

Example 2: Sign of f or sign of f′

State the intervals on which \(f(x) = 12x - x^3\) is increasing and on which it is decreasing.

1. The derivative factors, and its zeros \(-2\) and 2 are the cut points.

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

2. A tempting wrong move: The graph of \(f\) is above the x-axis from \(x = 0\) to \(x = 2\sqrt{3}\), which is about 3.46, so \(f\) is increasing on \((0, 2\sqrt{3})\).

   \[f > 0 \;\Rightarrow\; f \text{ increasing}\]

   The curve is above the axis there, but it is red from \(x = 2\) to \(x = 2\sqrt{3}\) because it falls. The sign of \(f\) is a height, and the direction depends on the sign of \(f'\).

3. Test \(f'\) in each interval: \(f'(-3) = -15\), \(f'(0) = 12\) and \(f'(3) = -15\).

   \[\begin{gathered} \cr{-}\ (x < -2), \quad \cg{+}\ (-2 < x < 2), \\ \cr{-}\ (x > 2) \end{gathered}\]

4. The function \(f\) is increasing on \((-2, 2)\) and decreasing on \((-\infty, -2)\) and on \((2, \infty)\).

Practice

State the intervals on which \(f(x) = x - 4\sqrt{x}\) is increasing and on which it is decreasing, for \(x \ge 0\).

Find the cut point inside the domain first. Then test one point on each side, and compare with the graph.

Questions

The graph shows the function \(g(x) = \dfrac{1}{x^2 - 1}\) and its derivative. Dashed vertical lines mark \(x = -1\) and \(x = 1\).

1. On which interval is \(g\) decreasing?

   • \((0, 1)\)
   • \((0, \infty)\)
   • \((-1, 0)\)
   Answer

   \((0, 1)\)

2. The function \(g\) is increasing on \((-\infty, -1)\) and on \((-1, 0)\). Why is the answer not the single interval \((-\infty, 0)\)?

   • \(g\) has no value at \(x = -1\), so the graph breaks there
   • \(g'(-1) = 0\), so the curve turns at \(x = -1\)
   • \(g\) is decreasing at \(x = -1\)
   Answer

   \(g\) has no value at \(x = -1\), so the graph breaks there

3. On which interval is \(f(x) = x^3 - 6x^2 - 15x + 2\) decreasing?

   • \((-1, 5)\)
   • \((-\infty, -1)\) and \((5, \infty)\)
   • \(x = -1\) and \(x = 5\)
   Answer

   \((-1, 5)\)

Summary

You write each interval between cut points, and you call \(f\) increasing where \(f'\) is positive and decreasing where \(f'\) is negative.

This works because a positive slope on a whole interval means a larger input gives a larger output, and a negative slope means the opposite.

Use it when a question asks where \(f\) rises or falls, and list separate intervals where the graph breaks.

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