jessie's 5th post

This week in calculus we learned about absolute MAXS and MINS.
1. first derivative test

2. plug the critical values into origional function to get y-values.

3. plug endpoints in to origional function to get y-values.

4. highest y-value is absolute max.

5. lowest y-value is absolute min.

-absolute maxs or mins or written as a point or simply as the y-value.

For example:find the absolute max or min of f(x)=3x^4-4x^ on [-1,2].
f1(x)=12x^3-12x^2=0
12x^2(x-1)=0
x=1,0
(-1, 0)U(0,1)U(1,2)
f1(-.5)=-ve f1(.5)=-ve f1(1.5)=+ve
min @x=1
3(1)^4-4(1)^3=-1 (1,-1)
3(-1)^4-4(-1)^3=-7 (-1,-7)
3(2)^4-4(2)^3 (2,16)
-32=16
abs min:(1,-1)or -1
-1 at x=1

i actually understand the max and min thing but doing it with the trig functions like that long problem we did in class confuses me. all the steps are throwing me way off so if someone could kind of help me with that it would be nice.

Post #5

So the fifth week of calculus was alot of work with justifications. The good thing is..justifications weren't the hard part, you just have to know what you did in order to get your answer and put the vocabulary to the test! =)
Well I've always known that any math is accumulative, but for Calculus..it's that much more. Everything we learned since last year was involved this week. I was overwhelmed, and to be honest I still am, but I plan on gettin alot of help this weekend with maybe a whole lot of explanations for how to start the problems.
My biggest problem is to know WHERE TO BEGIN.

We learned how to find an absolute MAX or MIN , also know as a global MAX/MIN:
1. First Derivative Test
2. Plug Critical Values into Original Function to get Y-values.
3. Plug Endpoints into Original Function to get Y-values.
4. Highest Y-value is Absolute MAX. Lowest Y-value is Absolue MIN.
Remember that absoute MAXs and MINs are written as a point or simply as the y-value.
Remember not to give x=_____ ...put the y value instead.

An Example would be:
Find the absolute MIN or MAX of f(x) = 2(x)^4 - 4(x)^3 on [0,5]::

so first, we take the first derivative which is:
8(x)^3 - 8(x)^2
we can take out an 8x^2 leaving us with:
8x^2[(x)^2-1]
now we set equal to zero giving us:
x = 0,-1,1...
using these numbers and the point we have in the problem we'll plug them all in for x and which ever is the smallest is the MIN and whichever is the largest is the MAX.
Our MIN would be [1,-2] and our MAX would be [5,750].

We also learned how to look at a graph of the original and then know where the MAX/MIN, or points of infliction are.

What I don't understand is anything with Differentability. Also, for just about everything we have to take a derivative and solve and stuff..i get that..but most of the time..[besides the vocabulary we had for the first thing we learned of original, first derivative, and second derivative]..i don't understand what the question is asking me, which makes it hard for me to begin anything.

Soo...ALL IN ALL..the fifth week of calculus was..well it was what everyone thinks calculus is..PRETTY TOUGH, wowzers, this week was hard for me.
I'd love some help with this, because after working on everything for almost 6 hours thursday i'm still not understanding it. so tomorrow [saturday] at 10:00 i'm goin to PJ's with Mal, Trina, and Ashley for sure..if anyone else is able to come to help us it would be appreicated! [by the way...we stayed for 4 hours .. struggling, but i think it sort of helped a bit] THEN..tonight [sunday] ricky had a study group at his house .. mher called and asked for my help so everyone came to stephanie's house until i got back home to try and help them...EVEN WITH the FULL WEEKEND of calculus..i'm sooo worried about the test tomorrow =(


ElliE

Post #5

Calculus Week #5
So to cut it rather short, this week we learned and reviewed a few things and then near the end, we got a study guide and multiple choice and free response take home test.
I don't know if anyone else has looked at this stuff but it's pretty intense if you ask me.
One thing I want to cover in this blog is a few relationships between the graphs of the original, first derivative, and second derivative that I don't think everyone grasped exactly.

If I were to give you the graph of f(x) and ask "What are the zeros of f'(x)," what would you do?
Well what you should do is first recognize what we have been doing. For the longest time now we have been focusing on two tests: the First Derivative Test and the Second Derivative Test. In the first derivative test (what this question is asking about) we take the derivative and set it equal to 0. The 0 is important because that is the key to how we solve this question. It asks for the zeros of the first derivative, right? So when you do the first derivative test, you set it equal to 0 to solve for critical values or possible max's and min's, correct? So for the problem I introduced, you would take a look at the max's and min's of the original and those x-values would be where your zeroes of the derivative function would be. (By the way, the zeroes, the x-intercepts, and roots are all the same: it's where the function crosses the x-axis.)

Likewise, suppose I gave you the graph of the original and asked "What are the zeros of f''(x)," what would you do? Well looking back to what we did before, we would look at the second derivative. When we take the second derivative and set equal to 0, what are we finding? We are finding possible points of inflection, right? Points of inflection are simply where the graph changes concavity, right? So if you look at the original graph and identify where the concavity changes (like an umbrella to a bowl), at that x-value is a zero of the second derivative.

Use both of these explanations in your multiple choice to be able to rule out answers rather easily.

And a question for some of you to perhaps answer:

If I show you a graph of the first derivative, what are the max's and min's of this graph in reference to the graph of the second derivative?

Clarification Please

okay, so i'm working on the take home test, and hopefully yall can still answer this, but on the first page, number two.
I know you have to take the first derivative...but i'm not sure if i took the first derivative right. I got cos(x)+sin^2(x). can i do something more with this?

help please

i know this might sound weird but i know how to solve for discontinuities but i do not know how to identify them in a problem. For example, i might take the derivative of a function and get intervals but find out later there was a discontinuity i didn't see. Can someone explain to me how you can tell when there are discontinuities in the problems we are working now.

Question

I don't understand when and how you are suppose to check for discontinuities. I know we learned how to in advanced math, but I was looking over those notes and I'm still confused on it.

please explain :)

Please Help!

Hey yall. I've did the study guide. However, I was only able to do it due to the answers given. I know I know it's cheating, but I tried to see how they got those answers. However, I couldn't and still can't figure that out.
I'm kinda unsure on what exactly it's asking me to do and how to do it. I'd appreciate some help. Thanks!
ElliE

Post #4

I had this typed up last night, and it wouldnt copy past..so yea, I gave up.

This week we also did differentiability, which was basically looking at the pictures/graphs and being able to recognize those. The one thing that I get, let's say, 98% of the time is the First and Second Derivative Tests. Now, Thursday/Friday, we added in the justifications. This didn't make it difficult necessarily, just tedius. So, let's do an example problem!!!:

this is the second derivative test.

x^3 - 6x^2 + 12x

take the first derivative:

3x^2 -12x + 12

take the second derivative:

6x - 12

Set equal to 0 and solve:

x=2

Set up intervals:

(-inf., 2) U (2, inf.)

Plug in values within each interval:

-ve and +ve

And if the question asked you about the concavity, it would be concave up because you are going from decreasing to increasing.

Now, for the justifications, I sometimes forget that you're not really supposed to write in 1st person, so:

Using the Second Derivative Test, set the 2nd derivative equal to zero to find possible points of inflection. In this case, x=2 is one. After setting up intervals, the graph is seen to be concave up
because it goes from decreasing (negative value) to increasing (positive value).

Now time for questions:

1. Does anyone want to go over the horizontal tangent thing, because I kind of forgot?
2. And, can anyone tell me when the limits do not exist (like removable, jump etc.)?

week #4 post

This week, we learned to use the Second Derivative Test to find all possible point of inflection and intervals of concavity. Remember, points of inflection only happen where there is a change of concavity.

Example: f'(x)= 6/(x^(2)+3)

First, you have to take the derivative of that, and you have to use the quotient rule, so the beginning of the problem will look like, [(x^(2)+3)(0)-6(2x)]/(x^(2)+3)^(2), which simplifies to -12x/(x^(2)+3)^2 remember, that was just the first derivative.

Second step is to take the derivative of the first derivative, that would make this step called taking the second derivative.

Once again u need to use the quotient rule, so f''(x)={(x2+3)^(2) -(12)-[(-12x)2(x^(2)+3)2x} all that over (x^(2)+3)^4 then you get a bunch of stuff, then you simplify, then you cancel, and if I typed all that up it would be ridiculous, so I'm just going to type the end answer of the second derivative. Which is, (3)(6)(x+1)(x-1) all over (x^(2)+3)^3

The possible points of inflection are found in the numerator of the finished second derivative, in this case, if you look, it would be x=1, and x=-1

so then you set up your points, (-infinity, -1) u (-1, 1) u (1, infinity)

then you plug in. f''(-2)= positive value f''(0)=negative value f''(2)=positive value

then you know that your intervals concave up @ (-infinity, -1) u (1,infinity) or x<1,>1

and it is concave down @ (-1,1) or -1
and you're points of inflection are x=-1, and x=1

okay, and now for my question, In the example mentioned above, I understand what you do to get the points of (-infinity,-1) u (-1,1) etc.... but then it says to have to take the second derivative of (-2), (0), and (2), and you do this to find out if the value negative or positive, but where do those numbers come from? The -2, 0, and 2. This is probably a dumb question, but thanks to whoever answers it.

Week 4

This week we did many new things. We started off everyday with a derivative problem on the board. I think this is good, because I need practice simplifying. We learned about differentiability. We also learned how to justify something. We also learned more about first and second derivatives.

Something I understood this week was taking first derivatives.

Let's say you need to find the critical values of (x^3 - 1)/x.

You would first take the derivative of this and get (2x^3 + 1)/(x^2).

You would then the derivative equal to zero to find you critical values.

You're critical values are x = (-1/2)^1/3.


Something I did not understand was what Differentiability is. I don't get what it means to have a differentiabily.

I don't really get the second derivative test either, with all the concave up, concave down nonsense.
Ryan

post # four

During this past week in calculus, we continued with the First Derivative Test. It is not as hard as it seemed on the first couple days that we learned it, but I just have trouble remembering the main steps, like setting equal to 0 before setting up intervals.

We also learned the Second Derivative Test on Friday this week. It is the same as the First Derivative Test except you are finding the second derivative instead of the first one. The point of the first and second derivative tests are to find changes in concavity, whether it is positive/up or negative/down, and points of inflection. But, you cannot forget that points of inflection only happen when there is a change in concavity! It is not possible unless there is a change in concavity.

First Derivative Test:

1. Take the derivative of problem
2. Set it equal to zero
3. Solve
4. Set up intervals
5. Plug In
6. Solve intervals
7. Justify

Second Derivative Test:

1. Take the derivative of problem
2. Take the derivative what you got for step 1 ^^
3. Set it equal to zero
4. Solve
5. Set up intervals
6. Plug In
7. Solve intervals
8. Justify

I understood everything this week pretty much, I just need to keep practicing. All of the worksheets this week helped me, along with the practice with did with B ROB when she came back, and I'm glad she is feeling better! :)

if anyone sees something wrong with what i posted about the steps for the first & second derivative test, please tell me!

post # 4

this week in calculus we learned the first derivitive test and went over the second derivitive test. These concepts are confusing me. Both of them i get where you have to find the derivative and then after there my problem goes all down hill. I either have trouble with the algebra or forget to do a step. Something else that was messing me up in class is that i would put an asatope as a point and it shouldn't be. I dont know when it is used or not.

Even though i am having trouble with these concepts I do understand how to take a derivative and how to use queotient rule and product rule.
EX: x^2 +4/x^5

1. you rewrite bottom, derivative top minus top deravitve of the bottom all over bottom squared.

This gives you =
-5x^8+2x^5-20x^4/(x^5)^2
-which can be further factored and simplified.

Post 4

This week we worked with first derivatives, second derivatives, differentiability, and justification. We learned about the first derivative test last week:

1. take a derivative
2. set equal to zero
3. solve for x
4. set up intervals using step 3
5. plug in first derivative
6. to find absolute value, mad/min, plug values from #5 into origional function
*remember to check for points of continuity

The second derivative test is just like the first derivative test, except you take the second derivative. When we were first learning about the second derivatives on the graphs at the beginning of last week, I didn't really follow what it was. This week I realized you're just taking the derivative of a derivative.

EX:
f(x) = 2x^3 + 4x^2 + 3x
you would take the derivative (first derivative) of this and get
f '(x) = 6x^2 + 8x + 3
after this, pretend like the derivative you just took is some origional function, and take the derivative of it, leaving you with
f "(x) = 12x + 8

This would be the second derivative. For the second derivative test, you would just take the first derivative, the second derivative, and follow the steps for the first derivative test using the second derivative.

The second derivative can help for a few things. It can be used to find points of inflection ad intervals of concavity. It is also a shortcut to find maximums and minimums. We just have to remember that points of inflection only happen if there is a change in concavity.

We also learned to justify our answers. All this basically is, is after you finish a problem, you write a paragraph on every single step you did. Also remember to include which derivative you are using and which derivative you are plugging x values into.

My question this week is differentiability in general. I missed that day and I'm looking at my notes and I'm not following. Can someone please explain?

POSTING...#4

This week we went over The first derivative test and we also learned how to do the second derivative test.
The second dervative test is basically the second derivate after the first derivative when we use the second derivative test we looked for: points of inflection, intervals of concavity, and we did the shortcut for max and min.
We also learned what differentiability is; differentiability is like a corner, a cusp or should I say BUTT, a vertical tangent, and anywhere it’s not continuous.

Well anyways an example of the second derivative test is:
6/x^2+3 the first derivative would be -12x/(x^2+3)^2 and to get the second derivative you would take the derivative of the first one and after deriving and simplifying the second derivative would be:
And the second derivative would be 36(x+1)(x-1)/(x^2+3)^3
But my question is can anyone help me find out the points of inflection, and the intervals and that stuff because I am kind of lost after that part. And Can some re-explain how to do the shortcut for Max and Min because I didn’t quite understand that.
But other than that I am pretty good but that which I just asked is most of the lesson so I’m really worried about the few test and quizzes we got coming up

this week in calc was kinnd of mind blowing. the first der test isnt hard to do and neither is the second der test but the problems gets huge and its kind of deceiving. these are the steps for the first and sec der tests.

First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)

Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection.

i know all the der formulas that we had in the beginning of the year and i know the rules to the two der tests. the one thing im haviing problems is i keep messing up in the algebra part of the test. some of the rulees tothe tests i skip over or forget.

Question

What is the homework everyone is talking about?

4th post

So this past week in calculus was a short one especially for me since I was only there 3 of 4 days. But the week started reviewing the first derivativetest. This has to do with the increasing, decreasing, positive slope, negative slope, concave up, concave down, horizontal tangent, slope=0, derivative above x axis, derivative below x axis, zero of derivative, maximum, minimum, and point of inflection in graphs. These things are used to get the derivative depending on which function they go with whether it being the original function, 1st derivative, or 2nd derivative. And the main thing these are used for is looking at a graph.

Then there are the steps to the first derivative test. The first thing to do is to take the derivative. Then set the derivative equal to 0. Next thing is to solve for x. After take the intervals from the last step and set them up. Then plug in the intervals into the first derivative. Finally in order to find an absolute maximum or minimum plug the values from the previous step into the original function.

I understand for the most part the first derivative test. But my problem is I went visit LSU Friday which means I missed calculus which means I missed everything we learned about second derivatives. Plus I just found out about a test on everything on Monday. But I would really appreciate if someone can explain the second derivative test to me before tomorrow so that I know it if it is on the test.

Post #4

Sooooo, I have many questions.

First, I don't remember even talking about a zero of derivative, is that used for anything? It is one of our words, but I don't know anything about it.

So what I do understand is the rules for the first derivative:
1. take the derivative
2. set it equal to 0
3. solve for x for max, mins, horizonal tangents, your critical points
4. set up intervals using step 3
5. plug in first derivative
6. (if needed) to find an absoulute max/ min plug values from #5 into original function, check endpoints

but I'm having troubles within the problem
Example: Find the critical points of f(x)=(x^2-4)^2/3
=2/3(x^2-4)^-1/3 * 2x=0
4x/3(x^2-4)^1/3 =0
4x=0 x=0, + or - 2 --> critical points Now something I don't get is what did she say about the 0 and how do you check for points of discontinuity?

I do get what's next you set up intervals and plug into your first derivative:
(-infinity,-2) f(-3)= -ve so decreasing
(-2,0) f(-1)= +ve so increasing
(0,2) f(1)= -ve so decreasing
(2,infinity) f(3)= +ve so increasing
So your mins are x=-2, x=2, but can your max be x=0?

Now I know for the second derivative test the main thing that is different is you plug into the second derivative and you deal with concave up or down and points of inflection instead of increasing or decreasing. Example: 6/x^2+3
(x^2+3)(0)-[6(2x)]/(x^2+3)^2 = -12x/(x^2+3)^2
then take the derivative of that which you should get = 36(x+1)(x-1)/(x^2+3)^3
Do I then set equal to 0 to get x= + or - 1? I do fully understand the interval part which I showed earlier in a different problem expect you would use concave up or down.

I know no one can help me on this, but I'm still having trouble simplifying derivatives. I guess it will still take more and more practice.

I'm out.

Post #4

This week was a really short week in calculus. We were off Monday, Tuesday and Wednesday Mrs. Robinson wasn’t at school, and Thursday I wasn’t there so that only leaves Friday. Since I missed Thursday, those notes are what I am most confused about. I copied the notes, but I don’t understand how to find differentiability or what we do with it when we find it if we do anything at all. Friday we learned the second derivative test and I think I understood it pretty well. It is the same steps as the first derivative test, but instead of using the first derivative, you have to find the second derivative.

The steps include:
1. Take the first derivative
2. Take the second derivative
3. Set it equal to 0 and solve for x
4. Set up intervals
6. Plug in second derivative
7. Find points of inflection

Ex: f(x) = 6/ x^2=3

First derivative: (x^2+3) (0) – [(2x) (6)]/(x^2+3)^2 which equals -12x/(x^2+3)^2

Second derivative: (x^2+3) ^2 (-12) – [(-12x) (2 (x^2+3)) (2x)]/(x^2+3) ^4When simplified it comes out to 36 (x+1) (x-1)/(x^2+3) ^3

Set the top equal to zero: 36(x+1) (x-1) =0 and solve for x which equals x=1 and x= -1

Set up intervals: (-infinity, -1) (-1, 1) (1, infinity)

Pick a number between and plug into second derivative: f (-2) = positive = concave up, f (0) = negative = concave down, f (2) = positive = concave up.

Points of inflection is where the graph changes concavity which is at x=-1 and x=1.

The last step is to justify your answer.

It would also help if someone can tell me what is on the quiz tomorrow.

Post Number Four

Hello Calculus :)

This past week we continued to learn the First Derivative Test. I understand the steps and how to do it, yet i still have trouble taking the derivative and simplifying it. It's like I do that part wrong then the WHOLE problem is wrong from there. I am so stressed out about this it's not even funny. I work problems over and over and still cannot seem to get it right. But anyways one of my questions is how to take a derivative when it is like cos^3(x^5) or something like that. If anyone can help with that i would greatly apprectiate it.

Thursday i was absent and of course Mrs. Robinson came back on that day--just my luck. I got the notes from someone and it was on differentiability. Can anyone explain to me what that's all about like in general? What exactly are we looking for and why?

Friday we learned the second derivative test. Doing this test finds the points of inflection and intervals of concavity (concave up-bowl, concave down-umbrella). Just remember that points of inflection only happen when there is a CHANGE in concavity.

Another question is which way to put the intervals when you find whether it is concave up or concave down? Should i put it in the parentheses like the intervals or put it in the lessthan/greater than form. I'll probably put it in both if i do get to that step oh goddd.

Okay now for another question, sorry guys. When do you check for points of discontinuities and what do you do if you find them?

All in all i understand the steps and formulas for everything, putting them to play is my problem. I will keep working on simplifying derivatives and such but if anyone is seriously up for tutoring i would take it in a heartbeat. Please try to comment on my blog and help me :)

Post #4

After three days of school last week, I felt as if those days were spent on reviewing what we have already covered, First Derivative Test and Second Derivative Test. The steps should already be ingrained in people's heads but if not here are the steps:

First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)

Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)

One thing I am still having troubles grasping is the chain rule. Say you have (x+1)^(2). I see the problem as x^n, so I bring the 2 out in front then subtract the exponent by 1 leaving me with 2(x+1)^(1). It does not make sense to me that you have to multiply by the derivative of the inside also because the rule does not call for it. Can someone please explain why you have to take the derivative of the inside also?

Quiz on Monday

Since I missed Thursday, can someone PLEASE tell me what's on this quiz Monday? Until Malerie just told me about it today (Sunday, at 4 in the afternoon) I had no idea about this quiz. I would really like to not fail this quiz, although I haven't a chance anyway. I just want to have a heads up on what's on this. Thanks!! :)

Also, what's up with the wills? What happened to those? I was really looking forward to the help.

Post #4

Calculus - Weekly Reflection #4
So this week in Calculus was... fairly simple and to the point. Things we learned include taking the first derivative, using the first derivative test, taking the second derivative, using the second derivative test, finding intervals on which functions increase and decrease, finding max's and min's of a function, and justifying our answers in Calculusspeak. :-) (yes, Calculusspeak).

Anyway, this coming Monday we have a test on everything we've learned so far in Calculus. EVERYTHING. So..., I think I'm going to help some of you out with a problem that might come up. Some people are getting confused when using the Chain Rule with trigonometric functions. The Chain Rule, if you don't know it, is when you have something like (2x+4)³. For this, you would bring the 3 to the front, copy the 2x+4 and make it squared, then times by the derivative of the inside (that's the chain rule part) which is 2 to make it a whopping end function looking like 6(2x+4)². Right? That's all fine and dandy. Some people may understand that part just fine, which is good, but for some reason they don't understand it when you throw in a trigonometric function. So let's look at this:
sin³(x²). Some people are thinking to themselves, uh-oh. sin³? We don't have a derivative formula for that. Well you are wrong. :-) The way to help with this type of problem is simply this. Rewrite the previous as this:

(sin(x²))³ Now--doesn't this look REALLY similar to the problem we did before? Bring the 3 to the front...make it squared...and times by the derivative of the inside (yes, i love me some chain rule!) and we wind up with:

6x(sin²(x²)) cos(x²)

Oh, are you saying "well john, we brought a 3 to the front, how in the heck did you get a 6x?" Well, duh! Chain rule on the derivative of cos(x²). Have to times by the derivative of the inside (which is 2x).

Well I hope that helps out you guys that may be getting confused and lost on those type of derivatives.

Now, for something I need help with...

Can someone explain to me the steps and purpose of the First Derivative Test? I mean, I was sick the day we learned it, and I've tried to grasp the concepts of it but I'm not exactly sure on what I'm doing with it or why for that matter. I think maybe it can be used to find maxs and mins, but I'm unsure! Any help would be great guys. :-)

Study study study for the quiz tomorrow. :-)

-John

jessie's 4th post

this week in calculus we learned more about first derivatives and second derivatives. i understand more about first derivatives such as the steps:

Steps are as follows:1. Take the derivative2. Set = 03. Solve or x =>max & min (extrema), horiz tangent4. Set up intervals using step 5. plug in 1st derivative6. To find an absolute max/min plug values from #5 into original function. Check endpoints. i also under stand concavities (concave up/ concave down)1st derivative: Positive slope, Negative slope, horizontal tangent 2nd derivative: Concave up, Concave down, point of inflection. if the function is concave up it is positive and above the axis and if its concave down its negative and below the axis. since i wasnt in class for the stuff on second derivatives i still dont understand that. so if anyone can help me that would be good. i understand how to get the first derivative but not the second. Maximum-original, Minimum-orignial, Extrema. i dont think i was there for that either so i need help with that too. And can someone explain an extrema for me?

4th post

Well this was a very interesting week i must say. Learning how to take the first and second derivative proved to be a challenge to me at first, but i believe i sort of understand how to do it now. If the problem asks you to find the intervals on which the function is increasing or decreasing then you take the first derivative, then set that equal to zero and then solve for x. For example on the homework problem #3, the equation is x^2-6x+8. First you would take the derivative on that, which would be 2x-6. Then you would set that equal to zero 2x-6=0. and then solve for x. which would give you x=3. The only thing i do not understand is what to do after that. Do i plug in numbers from the graph to give me the intervals or do i use the (-infinity, 3) u (3, infinity) as my intervals?

Also, when the homework asks for critical points. I know you have to take the first derivative and then solve for x again. But if x=0, how do u find other critical points or is zero your critical points? I know how to do the easy ones, for example,

x^2-6x, they asks for the critical points and all other possible extrema. So you take the derivative which is 2x-6. Then set equal to zero, then solve for x. Which will give you x=3. So my critical point is 3. Which will give me the intervals (-infinity,3) u (3,infinity). after that we take numbers in between those points to plug into our derivative to find out whether the graph is increasing or decreasing. It will also tell us if the numbers are max's or min's. That part i understand, the number one thing i do not understand is how to justify my answers. For that problem i would say "by using the first derivative test, i took the derivative of the original function and set the derivative equal to zero. Then i solved the function for x. When i found the x point i set up intervals. Then i plugged numbers in between my intervals into the derivative i found using the first derivative test. By plugging these numbers in i found that the first interval was decreasing and the second interval was increasing. I then found that the critical point 3 is a min." did i justify that correctly or did i do something wrong in the problem. I think i did all the tests right but i'm not sure. Can someone help me on my justifications and how to find the critical points and extrema. Thanks :)

Post #4

This week in Calculus we reviewed and learned new things. One thing I am not comvortable with is simplifying the second derivative, but I semi-understand how to do it..which really doesn't make sense.

The thing I am most comfortable with is the second derivative test. This test involves points of inflection and convavity, which is concave up or concave down. When taking the second derivative you must be aware that points of inflection only happen if there is a change in concavity.

To do the second derivative, first, you must take the derivative of the equation such as
6/(x^2 +3) = ( (x^2+3)(0) - [(6)(2x)] ) / (x^2 + 3)^2

Which gives you (-12x) / ( (x^2 +3) ^2) in it's simpliest terms. Next, you must take the derivative of this equation. Which would give you (36(x^2 -1)) / (x^2 +3)^3

This must then be solved to it's simpliest state set equal to zero which is (36(x+1)(x-1)) / (x^2+3)^3

Now, you have found the critical values: X = +/- 1. These are only POTENTIAL points of influction. Next, you must set up intervals (-infinity, -1) u (-1, 1) u (1, infinity)

Now, you must pick a number in each interval and plug it into the second derivative. This will tell you if the interval concaves up or down.

f''(-2) = positive = concave up
f''(0) = negative = concave down
f''(2) = positvie = concave up

Another way to write this is:
at (-infinity, -1) and (1, infinity) = concave up
at (-1, 1) = concave down
points of inflection = x=-1
x=1

These are the points of inflection because this is where the concavity changes. It is very important that you know points of inflection ONLY occur with changes of concavity.

Things I am still not comfortable with is finding average velocity, slope, and instantaneous speed. If anyone can help me understand it would make me pumped for life.