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Question 7

This version was saved 14 years, 11 months ago View current version     Page history
Saved by Skyline
on April 19, 2009 at 8:48:43 pm
 

A Calculator MAY NOT be used to help answer this question.

 

The figure shows ƒ', the derivative of a function ƒ.

The domain of the function ƒ is the interval [0, 20].

 

(a) For what values of x, 0 < x < 20, does ƒ have a relative maximum? Justify your answer.

 

(b) For what values of x is the graph of ƒ concave down? Justify your answer.

 

(c) If ƒ(0) = 10, sketch a graph of the function ƒ on axes similar to the ones provided below. List the coordinates of all critical points and inflection points.

 

 

 

Solution

 

Text in purple will be used primarily to connotate steps that are necessary to solving the problem.

Text in light blue will be used to connotate things that one should note, or that may be of some kind of importance.

Text in black is more or less just background banter.

Text in black that is emboldened will be used solely for the questions themselves.

Text in green is used to connotate the answer to a problem.

 

ROUGH COPY. (This solution is mainly to hold this question as my own significant contribution. I still need to clean up alot of it to make it presentable and fix some minor errors and whatnot. Thanks for not taking this one :]. Justus)

 

The figure shows ƒ', the derivative of a function ƒ.

The domain of the function ƒ is the interval [0, 20]. 

 

Part A.) For what values of x, 0 < x < 20, does ƒ have a relative maximum? Justify your answer.

 

The solution of part a is for the most part a straight forward application of the first derivative test.

http://en.wikipedia.org/wiki/First_derivative_test

 

The first step to solving this problem is to find the critical numbers of the graph.

 

By definition the critical numbers of a derivative are where the derivative is either zero, or undefined.

 

 

On the sample graph provided these points are located at

 

0, 5, 10, 12.5 and 15

(because at those points, the derivative of f, illustrated in the graph is either zero, or undefined.)

However, the domain supplied by the problem eliminates 0 as a possible relative maximum or minimum and thus we are left with 5, 10, 12.5 and 15 as our critical numbers. Now to figure out if these points [(5,0), (10,0), (12.5,0) and (15,0) ] are relative maximums or not, we need to look at their context,

that is to say, whats around these points.

 

 

On this graph, everything in the blue region is positive, and everything in the red region is negative.

 

We combine this knowledge with our location of the critical numbers to begin to figure out if those points are relative minimums or maximums.

Looking to the left of (10,0), the graph of f’(x) is positive, to the right of (10,0) it is negative.

 

This means, that on the parent graph, the slope at the point to the left of the graph is positive, and on the right negative. This suggests that f(10) is a maximum. Now by the same process we check the second critical number at (15,0) and find that to the left of this point, the first derivative is negative and to the right of this point, the first derivative is positive.

 

The critical numbers at (5,0) and (12.5) are different. Looking to the left of (5,0) we see that f'(x) is positive. Looking to the right also yields the same result. Thus (5,0) is neither a relative minimum or maximum. What it is exactly, will be dealt with later. The same process reveals that at (12.5, 0) f'(x) is also negative both to the left and right and therefore it neither, is a relative min or max.

 

Thus we can say, by the first derivative test, f(10) is a relative maximum, and also by the first derivative test, f(15) is a relative minimum

 

Part B.) For what values of x is the graph of ƒ concave down? Justify your answer.

 

This question is another straight forward application of one of the derivative tests, in this case, the second derivative test

http://en.wikipedia.org/wiki/Second_derivative_test

 

. Sketching out the graph of f’’(x) we find that the result is a peace wise graph, with two portions in the positive section, and one in the negative section.

 

 

Here is an excerpt from wikipedia which will explain how we will use the second derivative test to determine concavity;

 

"The second derivative test may also be used to determine the concavity of a function as well as a function's points of inflection.

First, all points at which \ f'(x)=0 are found. In each of the intervals created, \ f''(x) is then evaluated at a single point. For the intervals where the evaluated value of \ f''(x) < 0 the function \ f(x) is concave down, and for all intervals between critical points where the evaluated value of \ f''(x) > 0 the function \ f(x) is concave up. The points that separate intervals of opposing concavity are points of inflection."

 

 

As mentioned in the wikipedia excerpt, the first thing to do is find all points at which f'(x) = 0. We've already done this in Part. A.) and know these points to be

 

f'(10)=0 and f'(15)=0 (10 and 15)

 

Now that we have that, we begin to evalute over the interval before the critical points, inbetween each of them, and after them, to find the concavity of the graph at

 

 

 

In this image the gree represents the interval between the two critical points,

 

 

Part c.) on paper

 

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