Increasing, Decreasing, and Stationary Functions

Behavior of a Function and Its Derivative

Have you ever noticed how the graph of a function can move up, down, or even flatten out for a moment? This behavior, called the monotonicity of a function, is closely related to its first derivative.

Imagine you are walking along the curve of a graph from left to right.

• When you are climbing, it means the function is increasing.

• When you are descending into a valley, it means the function is decreasing.

• When you are at the top of a hill or the bottom of a valley, you are at a stationary point.

Geometrically, the first derivative, $f'(x)$, is the gradient of the tangent line to the curve at that point. So, we can determine the function's behavior by looking at the sign of its gradient.

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.

These stationary points are the key to finding where a function changes from increasing to decreasing, or vice versa.

Analyzing Function Intervals

Let's break down a case to see how to determine the intervals where a function is increasing or decreasing.

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

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

2. If the function $f(x) = ax^3 + x^2 + 5$ is always increasing on the interval $0 < x < 2$, determine the value of the coefficient $a$!

3. Determine the intervals where the function is increasing and decreasing if the curve is given by $f(x) = \sin 2x \cos 2x$!

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 a function to be always increasing on an interval, its first derivative must be non-negative ($f'(x) \ge 0$) for every point within that 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 all values of $x$ in the interval $0 < x < 2$. Since $h(x) = 3ax + 2$ is a linear function, its behavior is monotonic. We only need to ensure it is non-negative at the most "critical" endpoint of the interval.

   • If $a \ge 0$, then $3ax$ is also non-negative, so $3ax+2$ will definitely be positive. This condition is met.

   • If $a < 0$, then $h(x)$ is a decreasing function. Its smallest value will be at the right end of the interval ($x=2$). For $h(x)$ to always be non-negative, we just need to ensure its minimum value is greater than or equal to zero.

   We test at the critical boundary $x=2$:

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

   $$6a \ge -2$$

   $$a \ge -1/3$$

   Combining both cases, the condition for the function to be always increasing on the given interval is $a \ge -1/3$.

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 the first and fourth quadrants.

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

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

     This interval is valid for any integer $k$.

   • The function is decreasing when $f'(x) < 0$, which is $\cos(4x) < 0$. This occurs in the second and third quadrants.

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

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

     This interval is valid for any integer $k$.