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Calculus AB: Applications of the Derivative

Using the First Derivative to Analyze Functions

Problems for "The Mean Value Theorem"

Problems from "Using the First Derivative to Analyze Functions"

First, let's establish some definitions: f is said to be increasing on an interval I if for all x in I , f (x 1) < f (x 2) whenever x 1 < x 2 . f is said to be decreasing on an interval I if for all x in I , f (x 1) > f (x 2) whenever x 1 < x 2 . A function is monotonic on an interval I if it is only increasing or only decreasing on I .

The derivative can help us determine whether a function is increasing or decreasing on an interval. This knowledge will later allow us to sketch rough graphs of functions.

Let f be continuous on [a, b] and differentiable on (a, b) . If f'(x) > 0 for all x on (a, b) , then f is increasing on [a, b] . If f'(x) < 0 for all x on (a, b) , then f is decreasing on [a, b] .

This should make intuitive sense. In the graph below, wherever the slope of the tangent is positive, the function seems to be increasing. Likewise, wherever the slope of the tangent is negative, the function seems to be decreasing:

Figure %: Increasing and decreasing functions


Example: Find regions where f (x) = x 3 - x 2 - 6x is increasing and decreasing.

Solution:


f'(x) = x 2 - x - 6  
f'(x) = (x - 3)(x + 2)  

Now we find regions where f'(x) is positive, negative, or zero. f'(x) = 0 at x = 3 and x = - 2 . This can be marked on the number line:

From looking at the factors, it is clear that the derivative is positive on (- ∞, - 2) , and (3,∞) . The derivative is negative on (- 2, 3) .

This means that f is increasing on (- ∞, - 2) , and (3,∞) , and that it is decreasing on (- 2, 3) .

This can be indicated by arrows in the following way:

The points x = - 2 and x = 3 have horizontal tangents, which makes them critical points, but are they local extrema? It might be assumed that because f is increasing to the left of x = - 2 and decreasing to the right of x = - 2 that x = - 2 represents a local maximum. Similarly, it might be assumed that because f is decreasing to the left of x = 3 and increasing to the right of x = 3 that x = 3 represents a local minimum. This is in fact correct. This idea can be generalized in the following way:

The First Derivative Test for Classifying Critical Points

Let c be a critical number (i.e., f'(c) = 0 or f'(c) is undefined) of a continuous function f that is differentiable near x = c except possibly at x = c . Then if f'(x) is negative to the left of c and positive to the right of c , f has a local minimum at c . If f'(x) is positive to the left of c and negative to the right of c , then f has a local maximum at c . If f'(x) does not change sign at c , then (c, f (c)) is neither a local maximum nor a local minimum.

This should be clear from the figures below:


Example

Sketch a rough graph of f (x) = x 3 - x 2 - 6x

Based on previously collected data, f is increasing on (- ∞, - 2) , and (3,∞) f is decreasing on (- 2, 3) f has a local max at x = - 2 and a local min at x = 3

To sketch the graph, we might want to find the exact coordinates of the critical points: f (- 2) = 7 and f (3) = - 13 . Using this information, a rough sketch of f might look like:

Figure %: Rough sketch of f (x) = x 3 - x 2 - 6x based on first- derivative information

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