Introduction

This topic explores how to use the first derivative to determine where a function is increasing or decreasing. The essential knowledge tells us that the first derivative of a function can provide information about the function and its graph, including intervals where the function is increasing or decreasing. This connection between the sign of the derivative and function behavior is fundamental to analyzing functions.

The First Derivative Test for Increasing/Decreasing

The Fundamental Connection

For a function $f$ that is differentiable on an interval:

If $f^{'} (x) > 0$ on an interval, then $f$ is increasing on that interval.

If $f^{'} (x) < 0$ on an interval, then $f$ is decreasing on that interval.

If $f^{'} (x) = 0$ on an interval, then $f$ is constant on that interval.

Interpreting the Signs

Positive derivative ( $f^{'} (x)$ > 0):

The slope of tangent lines is positive
The function values are getting larger as $x$ increases
The graph rises from left to right

Negative derivative ( $f^{'} (x)$ < 0):

The slope of tangent lines is negative
The function values are getting smaller as $x$ increases
The graph falls from left to right

Example 1:

For $f (x) = x^{2}$ on the interval $(0, \infty)$ :

$f^{'} (x) = 2 x$

For $x > 0$ : $f^{'} (x) = 2 x > 0$

Therefore, $f$ is increasing on $(0, \infty)$ .

Finding Intervals of Increase and Decrease

Step-by-Step Process

Find $f^{'} (x)$
Find critical points (where $f^{'} (x) = 0$ or $f^{'} (x)$ is undefined)
Create a sign chart using the critical points
Test the sign of $f^{'} (x)$ in each interval
Conclude where $f$ is increasing or decreasing

Using Sign Charts

A sign chart organizes intervals and the sign of the derivative in each interval.

Example 2:

Find where $f (x) = x^{3} - 3 x^{2} - 9 x + 5$ is increasing and decreasing.

Find $f^{'} (x)$ : $f^{'} (x) = 3 x^{2} - 6 x - 9 = 3 (x^{2} - 2 x - 3) = 3 (x - 3) (x + 1)$

Critical points: $x = - 1$ and $x = 3$

Test intervals:

$x = - 2$ : $f^{'} (- 2) = 3 (- 5) (- 1) = 15 > 0$
$x = 0$ : $f^{'} (0) = 3 (- 3) (1) = - 9 < 0$
$x = 4$ : $f^{'} (4) = 3 (1) (5) = 15 > 0$

Conclusion:

Increasing on $(- \infty, - 1)$ and $(3, \infty)$
Decreasing on $(- 1, 3)$

Analyzing Different Function Types

Polynomial Functions

Example 3:

Determine where $g (x) = - x^{3} + 6 x^{2} - 9 x + 1$ is increasing and decreasing.

$g^{'} (x) = - 3 x^{2} + 12 x - 9 = - 3 (x^{2} - 4 x + 3) = - 3 (x - 1) (x - 3)$

Critical points: $x = 1$ and $x = 3$

Test intervals:

$x = 0$ : $g^{'} (0) = - 3 (- 1) (- 3) = - 9 < 0$
$x = 2$ : $g^{'} (2) = - 3 (- 1) (1) = 3 > 0$
$x = 4$ : $g^{'} (4) = - 3 (3) (1) = - 9 < 0$

Conclusion:

Increasing on $(1, 3)$
Decreasing on $(- \infty, 1)$ and $(3, \infty)$

Rational Functions

Example 4:

Find where $h (x) = \frac{x ^{2}}{x - 2}$ is increasing and decreasing.

$h^{'} (x) = \frac{( x - 2 ) ( 2 x ) - x ^{2} ( 1 )}{( x - 2 ) ^{2}} = \frac{2 x ^{2} - 4 x - x ^{2}}{( x - 2 ) ^{2}} = \frac{x ^{2} - 4 x}{( x - 2 ) ^{2}} = \frac{x ( x - 4 )}{( x - 2 ) ^{2}}$

Critical points: $x = 0$ and $x = 4$ Undefined: $x = 2$ (not in domain)

Test intervals (excluding $x = 2$ ):

$x = - 1$ : $h^{'} (- 1) = \frac{( - 1 ) ( - 5 )}{9} > 0$
$x = 1$ : $h^{'} (1) = \frac{( 1 ) ( - 3 )}{1} < 0$
$x = 3$ : $h^{'} (3) = \frac{( 3 ) ( - 1 )}{1} < 0$
$x = 5$ : $h^{'} (5) = \frac{( 5 ) ( 1 )}{9} > 0$

Conclusion:

Increasing on $(- \infty, 0)$ and $(4, \infty)$
Decreasing on $(0, 2)$ and $(2, 4)$

Trigonometric Functions

Example 5:

Determine where $f (x) = sin (x) + cos (x)$ is increasing on $[0, 2 π]$ .

$f^{'} (x) = cos (x) - sin (x)$

Set $f^{'} (x) = 0$ : $cos (x) - sin (x) = 0$

$cos (x) = sin (x)$

$x = \frac{π}{4}$ or $x = \frac{5 π}{4}$

Test intervals:

$x = 0$ : $f^{'} (0) = 1 - 0 = 1 > 0$
$x = π$ : $f^{'} (π) = - 1 - 0 = - 1 < 0$
$x = 2 π$ : $f^{'} (2 π) = 1 - 0 = 1 > 0$

Conclusion:

Increasing on $[0, \frac{π}{4})$ and $(\frac{5 π}{4}, 2 π]$
Decreasing on $(\frac{π}{4}, \frac{5 π}{4})$

Justifying Conclusions

Writing Complete Justifications

A complete justification should include:

The derivative calculation
Critical points identified
Sign analysis of the derivative
Clear conclusion about increasing/decreasing intervals

Example 6:

Justify that $f (x) = x^{2} - 4 x + 3$ is decreasing on $(- \infty, 2)$ .

$f^{'} (x) = 2 x - 4 = 2 (x - 2)$

For $x < 2$ : $x - 2 < 0$ , so $f^{'} (x) = 2 (x - 2) < 0$

Since $f^{'} (x) < 0$ for all $x$ in $(- \infty, 2)$ , the function is decreasing on this interval.

Connection to Function Behavior

Relating Derivative Sign to Graph Features

Understanding where a function increases or decreases helps identify:

Local extrema: Where the function changes from increasing to decreasing (or vice versa)
Overall shape: The general behavior of the graph
Monotonicity: Whether the function is always increasing or always decreasing

Example 7:

For $f (x) = e^{x}$ :

$f^{'} (x) = e^{x} > 0$ for all $x$

Since $f^{'} (x) > 0$ everywhere, $f$ is increasing on $(- \infty, \infty)$ .

This means $e^{x}$ is a monotonically increasing function with no local extrema.

FiveHive

FiveHive

5.3 - Determining Intervals on Which a Function Is Increasing or Decreasing

Introduction

The First Derivative Test for Increasing/Decreasing

The Fundamental Connection

Interpreting the Signs

Finding Intervals of Increase and Decrease

Step-by-Step Process

Using Sign Charts

Analyzing Different Function Types

Polynomial Functions

Rational Functions

Trigonometric Functions

Justifying Conclusions

Writing Complete Justifications

Connection to Function Behavior

Relating Derivative Sign to Graph Features

Practice Section