Determining Intervals on Which a Function is Increasing or Decreasing: AP Calculus AB/BC Study Guide
Introduction
Strap in, future calculus wizards! Today, we're diving into the magical world of derivatives to determine where functions play nice and where they throw tantrums—aka where they increase and decrease. Might sound like math with a mood swing, but it's crucial for mastering AP Calculus! 📈🎢
When Does a Function Increase or Decrease?
First off, let’s understand what the derivative of a function tells us. The derivative is like the function’s personal trainer—it measures how quickly the function is changing at any given point.
 If the derivative is positive (imagine the function pumping those weights), the function is increasing.
 If the derivative is negative (think of the function slumping on the couch eating chips), the function is decreasing.
Picture this: the graph of our function is a roller coaster 🎢, and the derivative is your trusty sensor telling you whether the ride is going up or down.
Finding Intervals of Increase and Decrease
Now, how do we find where the function behaves like an upward thrill ride or a downward plunge? Easypeasy! We look for where the derivative is positive or negative respectively.
StepbyStep Guide

Find the Derivative (f’(x)): This is the heart of the operation. Determine the derivative using differentiation rules.

Identify Critical Points: These are points where the derivative is zero (f’(x) = 0) or where the derivative does not exist (undefined points). This is like spotting where our roller coaster changes direction.

Divide the Number Line: Split the number line into intervals based on these critical points. Each interval is like a section of our ride.

Test Points in Each Interval: Choose any point within each interval, plug it into the derivative, and see if it’s positive or negative. Positive means our ride is going up (increasing interval), negative means it's going down (decreasing interval).

Interpret the Results: Sum up where the function is increasing and where it's decreasing.
Example Walkthrough
Let’s use a roller coaster scenario to demonstrate the steps. Consider our function is a thrill ride with the derivative given by:
[ h’(x) = \frac{(x+7)}{x^2} ]
 Critical Points: Solve [ h’(x) = 0 ]
[ \frac{(x+7)}{x^2} = 0 ]
This gives us ( x = 7 ) as a critical point. The derivative and function can also be undefined at ( x = 0 ).

Intervals: Divide the number line at 7 and 0. So we have: [ (\infty, 7), (7, 0), (0, \infty) ]

Test Points:

For the interval ((∞, 7)), test ( x = 8 ): [ h’(8) = \frac{(8+7)}{(8)^2} = \frac{1}{64} ] Since the result is negative, ( h(x) ) is decreasing.

For the interval ((7, 0)), test ( x = 1 ): [ h’(1) = \frac{(1+7)}{(1)^2} = \frac{6}{1} = 6 ] Since the result is positive, ( h(x) ) is increasing.

For the interval ((0, ∞)), test ( x = 1 ): [ h’(1) = \frac{(1+7)}{(1)^2} = \frac{8}{1} = 8 ] Again, the result is positive, so ( h(x) ) is increasing.

In conclusion: [ h(x) ] is increasing on ((7, 0)) and ((0, ∞)), and decreasing on ((∞, 7)).
Practice Problems
Now let’s put this to the test with some practice problems. Grab your calculators and get ready for a brain workout!

Let ( f(x) = x^3  27x ). On which intervals is ( f ) decreasing?

Let ( g(x) = x^4  2x^2 ). On which intervals is ( g ) increasing?
Solutions:
Function Behavior: Question 1
 Derivative: ( f’(x) = 3x^2  27 )
 Critical Points: ( 3x^2  27 = 0 ) gives ( x = ±3 )
 Intervals: [ (∞, 3), (3, 3), (3, ∞) ]
Evaluate in each interval:
 At ( x = 4 ): ( f’(4) = 3(4)^2  27 = 21 ) (positive, increasing)
 At ( x = 0 ): ( f’(0) = 27 ) (negative, decreasing)
 At ( x = 4 ): ( f’(4) = 21 ) (positive, increasing)
Conclusion: ( f(x) ) is decreasing on ((3, 3)).
Function Behavior: Question 2
 Derivative: ( g’(x) = 4x^3  4x )
 Critical Points: ( 4x(x^2  1) = 0 ) gives ( x = 0, ±1 )
 Intervals: [ (∞, 1), (1, 0), (0, 1), (1, ∞) ]
Evaluate in each interval:
 At ( x = 2 ): ( g’(2) = 24 ) (negative, decreasing)
 At ( x = 0.5 ): ( g’(0.5) = 1.5 ) (positive, increasing)
 At ( x = 0.5 ): ( g’(0.5) = 1.5 ) (negative, decreasing)
 At ( x = 2 ): ( g’(2) = 24 ) (positive, increasing)
Conclusion: ( g(x) ) is increasing on ((1, 0)) and ((1, ∞)).
Conclusion
Congratulations! 🥳 You've now mastered the ability to determine when your functions go up or down—knowledge that’s sure to make your calculus journey smoother. It’s like having a map for your mathematical roller coaster. Now go forth, find those intervals, and show your AP Calculus exam who's boss! 🚀