1st and 2nd Derivative tests


by bloodredorioo
Tags: calculus 1, critical points, derivatives, relative extrema
bloodredorioo
bloodredorioo is offline
#1
Nov13-11, 01:51 PM
P: 1
1. The problem statement, all variables and given/known data
This question is asking me to find the local max and min values of f using both the 1st and 2nd derivative tests. Then it asks me which method I prefer.
f(x)= x+ sqrt(1-x)


2. Relevant equations
I don't have a problem with finding the derivative of the function, I already did that. I just have problems with the concepts of the 2 tests; they were not explained very well in my class.


3. The attempt at a solution
My understanding is that the 1st derivative test is the mean value theorum: where there exists a point c between point a and point b, and the derivative f ' (c) = f(b)-f(a)/ b-a.
For this 1st derivative test I would find critical points and then plug them back into the original equation to get extrema. Then use the Increasing/decreasing sign test to figure out if those values are relative minimum or relative maximum.
Please correct me if you have a better way of explaining the 1st derivative test.

I thought the 2nd derivative test required me to get the second derivative (f '' ), then set f '' = 0 and solve for x to get critical numbers. Then plug the critical numbers back into the original equation ( f) to get the points of inflection.


My problem is, it asks me which method I prefer. Which is confusing because I though that both of the tests gave you something different? The first test gives you relative maximum/minimum and the second test gives you the points of inflection. So is this trick question? You can't really prefer one because they give you different data.
I also wouldn't mind if you explained it to me using the function I provided, or if you didn't, I just need some help here.
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Ray Vickson
Ray Vickson is offline
#2
Nov13-11, 03:09 PM
HW Helper
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Quote Quote by bloodredorioo View Post
1. The problem statement, all variables and given/known data
This question is asking me to find the local max and min values of f using both the 1st and 2nd derivative tests. Then it asks me which method I prefer.
f(x)= x+ sqrt(1-x)


2. Relevant equations
I don't have a problem with finding the derivative of the function, I already did that. I just have problems with the concepts of the 2 tests; they were not explained very well in my class.


3. The attempt at a solution
My understanding is that the 1st derivative test is the mean value theorum: where there exists a point c between point a and point b, and the derivative f ' (c) = f(b)-f(a)/ b-a.
For this 1st derivative test I would find critical points and then plug them back into the original equation to get extrema. Then use the Increasing/decreasing sign test to figure out if those values are relative minimum or relative maximum.
Please correct me if you have a better way of explaining the 1st derivative test.

I thought the 2nd derivative test required me to get the second derivative (f '' ), then set f '' = 0 and solve for x to get critical numbers. Then plug the critical numbers back into the original equation ( f) to get the points of inflection.


My problem is, it asks me which method I prefer. Which is confusing because I though that both of the tests gave you something different? The first test gives you relative maximum/minimum and the second test gives you the points of inflection. So is this trick question? You can't really prefer one because they give you different data.
I also wouldn't mind if you explained it to me using the function I provided, or if you didn't, I just need some help here.

I don't understand why the question asks you for a preferred method, since the first and second-derivative tests are for _different_ aspects of the problem. First of all: never mind the mean-value theorem (which you mis-wrote, anyway---you should have used brackets and written f'(c) = [f(b)-f(a)]/(b-a)); just remember the interpretation of the derivative f'(x) as the slope of the tangent line to the graph y = f(x). So, when f'(x0) = 0 the tangent line at x = x0 is horizontal, meaning that the graph is not sloping up and not sloping down at x = x0. That makes x0 a candidate for being a (local) max or min; however, it may also be an 'inflection point', which is neither a max nor a min. *Telling which is which is where the second-derivative tests come in*.

Suppose f'(x0) = 0 and f''(x0) > 0. In words, what is that saying? Well, f''(x0) > 0 means the derivative f'(x) is increasing at x = x0, so it goes from negative to positive as x increases through x0. In other words, the graph of y = f(x) is sloping down for x a bit below x0 and is sloping up for x a bit above x0; can you see what the graph must look like near x = x0?

Next suppose f'(x0) = 0 but f''(x0) < 0. Using an argument similar to the above, get a picture of what the graph y = f(x) looks like near x = x0.

Finally, let f'(x0) = 0, and suppose also that f''(x0) = 0. What can you say now? I'll let you see why you really can't say _anything_: x0 can be a max or a min or neither, depending on other information about the function f(x). In other words, when you look for points where the second derivative vanishes you are looking for points where you have NO information about whether or not the function is a max or a min. I don't know why the questioner would ask you about those points.

RGV


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