Increasing and Decreasing Functions

Behavior of a Function and Its Derivative

As we read a graph from left to right, its curve may rise, fall, or briefly become flat. This behavior is called the function's monotonicity, and its first derivative tells us how to classify it.

The sign of the slope tells us the curve's local direction:

• A positive slope means the function is increasing.

• A negative slope means the function is decreasing.

• A zero slope marks a stationary point, although that point need not be a maximum or minimum.

Geometrically, $f'(x)$ is the slope of the tangent line. Its sign therefore describes the curve's local direction.

Properties of Monotonicity

The relationship between the first derivative and the behavior of a function can be summarized by the following properties:

Suppose the function $y = f(x)$ is continuous and differentiable over an interval.

• If $f'(x) > 0$ for all $x$ in that interval, then $f(x)$ is an increasing function.

• If $f'(x) < 0$ for all $x$ in that interval, then $f(x)$ is a decreasing function.

• If $f'(x) = 0$ at a specific point, then $f(x)$ has a stationary point there.

Stationary points divide the domain into intervals that we can test. They are candidates for a change in monotonicity, but the derivative must change sign for the function to switch from increasing to decreasing or vice versa.

Analyzing Function Intervals

The following example shows the complete sign-analysis process.

Determine the intervals for which the function $f(x) = x^3 - 3x$ is increasing and decreasing.

Solution:

Step 1: Find the first derivative

First, we differentiate the function $f(x)$.

Step 2: Find the stationary points

Stationary points occur when $f'(x) = 0$.

From this, we get the stationary points at $x = 1$ and $x = -1$.

Step 3: Create a number line and test intervals

We place the stationary points on a number line. These points divide the line into three intervals. We take a test point from each interval to find the sign of $f'(x)$ (positive or negative).

• Interval $x < -1$:

  Take $x=-2$. $f'(-2) = 3(-2)^2 - 3 = 9$ (Positive, function is increasing).

• Interval $-1 < x < 1$:

  Take $x=0$. $f'(0) = 3(0)^2 - 3 = -3$ (Negative, function is decreasing).

• Interval $x > 1$:

  Take $x=2$. $f'(2) = 3(2)^2 - 3 = 9$ (Positive, function is increasing).

Step 4: Conclude the intervals

Based on the tests, we can conclude:

• The function is increasing on the intervals $x < -1$ or $x > 1$.

• The function is decreasing on the interval $-1 < x < 1$.

Exercises

Each problem asks where a function increases or decreases. Find the stationary points first, because they separate the intervals.

1. Determine the intervals where the function is increasing and decreasing for the curve $f(x) = x^3 - 3x^2 - 15$.

2. If $f(x) = ax^3 + x^2 + 5$ is increasing throughout $0 < x < 2$, determine the possible values of $a$.

3. Determine the intervals where $f(x) = \sin 2x \cos 2x$ is increasing and decreasing.

Answer Key

1. Solution:

   The first derivative of $f(x) = x^3 - 3x^2 - 15$ is $f'(x) = 3x^2 - 6x$.

   Stationary points are found when $f'(x) = 0$.

   $$3x^2 - 6x = 0 \implies 3x(x - 2) = 0$$

   The stationary points are at $x=0$ and $x=2$.

   By testing the intervals on a number line:

   • For $x<0$, $f'(x)$ is positive (increasing).

   • For $0<x<2$, $f'(x)$ is negative (decreasing).

   • For $x>2$, $f'(x)$ is positive (increasing).

   So, the function is increasing on $x < 0$ or $x > 2$, and decreasing on the interval $0 < x < 2$.

2. Solution:

   For this function to increase throughout the interval, it is enough to require $f'(x) \ge 0$ at every point in the interval.

   $$f'(x) = 3ax^2 + 2x = x(3ax + 2)$$

   In the interval $0 < x < 2$, the factor $x$ is always positive. Therefore, for $f'(x) \ge 0$, the second factor, $(3ax + 2)$, must also be non-negative.

   $$3ax + 2 \ge 0$$

   This inequality must hold for every $x$ in $0 < x < 2$. Since $h(x) = 3ax + 2$ is linear, its smallest limiting value occurs toward one end of the interval.

   • If $a \ge 0$, then $3ax \ge 0$, so $3ax+2$ is positive throughout the interval.

   • If $a < 0$, then $h(x)$ decreases. Because $x=2$ is excluded, $h$ has no minimum there. Its infimum is the boundary value approached as $x \to 2^-$. Requiring that boundary value to be non-negative gives the exact condition.

   Evaluate the limiting boundary condition at $x=2$:

   $$3a(2) + 2 \ge 0$$

   $$6a \ge -2$$

   $$a \ge -1/3$$

   Combining both cases, the function increases throughout the given interval precisely when $a \ge -1/3$.

3. Solution:

   Use the double angle trigonometric identity: $\sin(2A) = 2\sin(A)\cos(A)$.

   So, $f(x) = \sin 2x \cos 2x = \frac{1}{2} \sin(4x)$.

   Its first derivative is:

   $$f'(x) = \frac{1}{2} \cdot \cos(4x) \cdot 4 = 2\cos(4x)$$

   • The function is increasing when $f'(x) > 0$, which is $2\cos(4x) > 0$ or $\cos(4x) > 0$. This occurs in quadrants $\mathrm{I}$ and $\mathrm{IV}$.

     $$-\frac{\pi}{2} + 2k\pi < 4x < \frac{\pi}{2} + 2k\pi$$

     $$-\frac{\pi}{8} + \frac{k\pi}{2} < x < \frac{\pi}{8} + \frac{k\pi}{2}$$

     This interval is valid for any integer $k$.

   • The function is decreasing when $f'(x) < 0$, which is $\cos(4x) < 0$. This occurs in quadrants $\mathrm{II}$ and $\mathrm{III}$.

     $$\frac{\pi}{2} + 2k\pi < 4x < \frac{3\pi}{2} + 2k\pi$$

     $$\frac{\pi}{8} + \frac{k\pi}{2} < x < \frac{3\pi}{8} + \frac{k\pi}{2}$$

     This interval is valid for any integer $k$.