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October 29th, 2012, 03:39 PM   #1
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Graphs of derivatives, determining max/mins with critical #s

Hello everyone. I can not post the graph of this problem, but I am hoping someone can remind me what to do to solve this problem correctly.

Here's the problem:

Suppose that c = -1 is a critical number for a function f. Determine if f(c) is a local maximum, local minimum or neither if the graph of f ' (x) is shown below.


Ok so let's just imagine a graph here, what exactly is it that I am looking for to determine if f(c) is a local max, local min, or neither? On the graph at c=-1, (or x=-1), y=0.

It is a parabola opening upwards with x=-2 as it's vertex. Passing through the points -3 and -1.

I have several problems similar to this, with either the graph of the derivative, the 2nd derivative, or the original function, that ask for a number of different things, and can not recall what it is I need to look for in each type of graph, to determine the value they are asking for.

Help is appreciated.
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October 29th, 2012, 03:45 PM   #2
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Re: Graphs of derivatives, determining max/mins with critica

If f'(c) = 0 and if:

a) f'(c-h) < 0 and f'(c + h) > 0 then f(c) is a local minimum. This means f''(c) > 0.

b) f'(c-h) > 0 and f'(c + h) < 0 then f(c) is a local maximum. This means f''(c) < 0.
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October 29th, 2012, 04:03 PM   #3
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Re: Graphs of derivatives, determining max/mins with critica

Actually, what would really help me out tremendously, is if someone could just take a look at the file I attached to this message. It is an "EMCF" or "daily quiz" we have to turn in before each class, this one was from last Friday. I can not seem to remember how to find the answers to the problems with each of these graphs, or I keep getting them mixed up. If someone doesn't mind, you'd be of the biggest help to take a look at each of the problems (I believe it's 10 or so), and remind me how to go about finding the values that they are asking for (for example, when given the graph of a 2nd Derivative, it will ask to give the number of critical numbers for the 1st derivative). We did this in class on Friday and I can not stop getting the procedures mixed up for how to solve for each type of graph. Just to be clear I am not asking anyone to just "give me the answers", just take a look at each problem (there are 2 graphs, and 5 questions with each graph), and remind me exactly what it is on the graph that I am looking for (for example, "look for the places where concavity changes to find.......")

I am pretty sure how to solve these types of problems mathematically, but for some reason when given graphs and asked to pinpoint values I can never seem to remember how.

Help is appreciated.
Attached Files
File Type: pdf 1431EMCF 25.pdf (41.6 KB, 9 views)
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October 29th, 2012, 05:58 PM   #4
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Re: Graphs of derivatives, determining max/mins with critica

1.) Count the number of roots that f''(x) has. A function has critical numbers at the roots of its derivative. f''(x) is the derivative of f'(x).

2.) Look at the places where f''(x) changes sign. These are the places at which f(x) has points of inflection.

3.) Look for the places where f''(x) goes from negative to positive.

4.) Look for the intervals where f''(x) is positive.

5.) Look for the intervals where f''(x) is negative.

6.) Count the number of places where the tangent line would be horizontal.

7.) Count the number of intervals of alternating concavity. The number of inflection points is one less than this.

8.) I think the problem should be asking for the number of places where f has a local minimum. Use the given hint.

9.) This problem has the same typo as 8.) Use the given hint.

10.) Count the number of changes in concavity.

11.) Count the number of intervals where concavity is up.
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October 29th, 2012, 06:10 PM   #5
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Re: Graphs of derivatives, determining max/mins with critica

Thanks Mark, that helps tremendously.

I am stuck on another problem here, here is the problem:

Find the critical numbers of f and classify the extreme values given x ? [0, 7] and f(x) = (x-7)(x-2)

so f(x) = x^2-9x+14

fprime(x) = 2x-9

2x-9=0
2x=9
x= 9/2 <<< Critical Number (right?)

So my answer choices are as follows:

a) Critical no. 9?2; absolute min f(7) = 0; local min f(9?2) = - 25?4; absolute max f(0) = 14.

b) Critical no. 9?2; local and absolute min f(9?2) = - 25?4; absolute max f(0) = 14.

^^^ there are other answer choices but these are the only ones that list 9/2 as the critical #

When putting this on a number line between 0 and 7 I picked the test points 1 and 5

f(1) = 6 (++++)
f(5) = -6 (----)

So maybe I am just not quite understanding the difference between local and absolute max/mins, I believe I am getting it confused with the increasing/decreasing parts of the function (in which you plug the test points into the first derivative, which I believe would result graphically as a parabola opening downward, with it's maximum at x=9/2, but this is different and I am having a difficult time visualizing it).

If anyone can clarify this for me I would greatly appreciate it. Thanks..
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October 29th, 2012, 06:16 PM   #6
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Re: Graphs of derivatives, determining max/mins with critica

You want to test the sign of the derivative, not the function itself on the two intervals.
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October 30th, 2012, 01:14 PM   #7
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Re: Graphs of derivatives, determining max/mins with critica

3log5(x-7)=log5^125
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