Calculus Explanation *asap*

[Gamingphreek]

Tomorrow i have a test in AP Calculus on limits. While the first part was easy, the second part is merely tricky, not hard, just a lot of abstract stuff.

I was wondering if someone could explain how my book got the answer to a problem. I have worked for a while and cannot come up with it:

Problem: lim H->0 for [f(x+h)-f(x)]/h

I am subsituting f(x) for (4/x) in the problem.

You cannot use substitution method as it would give you indeterminate form, so there is some way to simplify it.

Additionally, i remember my teacher saying something that i had never know. Because f(x) is (4/x) that means f(h) is (4/h)?? Can someone attest to this.

Im not trying to get answers and people to do work for me (I am studying not doing HW)... i simply dont understand the problem (more specifically the answer).

Any help would be appreciated.

-Kevin

 

[akubi]

....
(4/(x+h) - 4/x)/h = (-4h/(x(x+h))/h = -4/(x(x+h))

lim h->0 -4/(x(x+h)) = -4/(x(x+0)) = -4/x^2

 

[IAteYourMother]

how old are you?

 

[Gamingphreek]

(4/(x+h) - 4/x)/h = (-4h/(x(x+h))/h = -4/(x(x+h))

Sorry for the stupidity...

Can you go into smaller steps here.

You have to find the commom denominator in order to subtract (4/(x+h) - 4/x) correct? Or is this step not right for this problem.

-Kevin

Edit: I am 17 why...

 

[akubi]

Originally posted by: Gamingphreek

(4/(x+h) - 4/x)/h = (-4h/(x(x+h))/h = -4/(x(x+h))

Sorry for the stupidity...

Can you go into smaller steps here.

You have to find the commom denominator in order to subtract (4/(x+h) - 4/x) correct? Or is this step not right for this problem.

-Kevin

Edit: I am 17 why...

yes you do. common denominator is x(x+h) obviously

 

[aplefka]

The Thread I Will Always Think Of When You Post

Anyway, I feel for you somewhat. I know math can get confusing so good luck on your test. I thought I was gonna fail a test, as in like 50 or thereabouts, and I ended up getting an 86. Just do your best.

 

[TheStu]

so yo havent hit derivatives yet yes?

anyway...

yes you have to use a common denominator to subtract them... in this case it is x(x+h)

 

[compnovice]

Originally posted by: Gamingphreek

(4/(x+h) - 4/x)/h = (-4h/(x(x+h))/h = -4/(x(x+h))

Sorry for the stupidity...

Can you go into smaller steps here.

4/(x+h)-4/x= [4x-{4(x+h)}]/(x+h)(x)=[4x-4x-4h]/(x+h)(x)
=[-4h/(x(x+h)]/h = -4/(x(x+h))

 

[RaWrulez]

I am in Calc I at Troy Univ and we are having a test tomorrow on the same thing. I took AP Calc in High School but that was 6 years ago... ahhaha

Anyway good luck.

 

[bleeb]

Originally posted by: akubi
....
(4/(x+h) - 4/x)/h = (-4h/(x(x+h))/h = -4/(x(x+h))

lim h->0 -4/(x(x+h)) = -4/(x(x+0)) = -4/x^2

 

[Hacp]

Originally posted by: TheStu
so yo havent hit derivatives yet yes?

anyway...

yes you have to use a common denominator to subtract them... in this case it is x(x+h)

We cover derivatives in precal a little.

 

[Gamingphreek]

Originally posted by: aplefka
The Thread I Will Always Think Of When You Post

Anyway, I feel for you somewhat. I know math can get confusing so good luck on your test. I thought I was gonna fail a test, as in like 50 or thereabouts, and I ended up getting an 86. Just do your best.

Lol... that and hitting 90mph on 95 are the extent of my stupidity. No smoking or any of that (Go to parties, but i just talk with my friends and laugh if they get drunk). At any rate, going into that subdivision in the first place wasn't smart, especially without an experienced driver beside me.

OK....back to the problem.

i got the common denominator as x(x+h) as you said which gives me:

(4x-4x-4h)/(x(x+h) .....

Oh dear god (I feel so incredibly retarded for asking for help on that question ), i know exactly what i was doing wrong. Dunno why i crossed something out . I was complicating it too much, probably should have worked it out on a clean sheet of paper too as opposed to my note sheet.

I dont want to seem like i am stupid or anything, but i need 1 more explained. My teacher simply gave the answer but did not explain this one. This one we have to use the squeeze theorem to solve it. The way i understand it is that the squeeze theorem is the value in between the limits of 2 equations.

In this equation (this is the one i can do):

lim X->C f(x)

c=0
4-x^2<=f(x)<=4+x^2

When graphed, there were two parabola and in the center, at the point they met was 4. Which coincedentally was the answer as it was both of the equations limit.

However, the book through a nice equation in there which is only made up of variables.

lim X->C f(x)

c=a
b-|x-a|<=f(x)<=b+|x-a|

Is there a simple way to tell the limit of f(x) here. I believe on this problem there is something simply fundamental that i haven't yet grasped. (For some reason 'b' sticks in my mind as the answer)

Is there anyway you can help me out once more?

I really dont want this to appear that you guys are doing my work or i dont pay attention in class which results in me asking questions here, but i have no way of asking my teacher before hand. Additionally, she doesn't explain it as well as i would like.

Thanks for everything!!

-Kevin

 

[computeerrgghh]

Heres a hint with fermat's difference quotients... unless your teacher is a major jacka$$, the h in the denominator will always cancel out nicely. So if there is a square root, remember to multiply by the conjugate, etc.

 

[Gamingphreek]

Originally posted by: computeerrgghh
Heres a hint with fermat's difference quotients... unless your teacher is a major jacka$$, the h in the denominator will always cancel out nicely. So if there is a square root, remember to multiply by the conjugate, etc.

If it didn't the limit would not exist then. Or at least it would be in indeterminate form, something we haven't learned to much about yet other than it is 0/0.

I have a Ti-89 Titanium Calculator, so for the problems with numbers in them (unlike the last one i posted) i can find the answer. However, I need to understand how to do this by hand (some problems may require me to show my work).

Just to make sure i know exactly what the conjugate is. Conjugate of sqrt(x) is sqrt(x)/sqrt(x) correct?

As you can see i am really nervous as this is the first real major test in this class.

-Kevin

 

[bonkers325]

seriously? this is easy stuff! do you still need help?

edit: if u still need help, look at this

 

[bonkers325]

if you're still awake i made a quick explanation for you

 

[Gamingphreek]

bonkers thankyou, but i had already figured out that problem (with the aid of others)

I need help on the new problem listed above...

I dont want to seem like i am stupid or anything, but i need 1 more explained. My teacher simply gave the answer but did not explain this one. This one we have to use the squeeze theorem to solve it. The way i understand it is that the squeeze theorem is the value in between the limits of 2 equations.

In this equation (this is the one i can do):

lim X->C f(x)

c=0
4-x^2<=f(x)<=4+x^2

When graphed, there were two parabola and in the center, at the point they met was 4. Which coincedentally was the answer as it was both of the equations limit.

However, the book through a nice equation in there which is only made up of variables.

lim X->C f(x)

c=a
b-|x-a|<=f(x)<=b+|x-a|

Is there a simple way to tell the limit of f(x) here. I believe on this problem there is something simply fundamental that i haven't yet grasped. (For some reason 'b' sticks in my mind as the answer)

 

[bonkers325]

beats me, i got a 65 in AB calc for both semesters, got a 4 on the test... but took it again in college and still got a 30 on the limits section of the test.

but if x approaches a, then f(x) = b from the left and the right

 

[IAteYourMother]

Calc C is a b!tch... predator prey problems... ugh

 

[MBrown]

I hate math but I have to take cal for my EE. I guess I could post in this forum when I start my precal class next semester. Incase your wondering why im going for a EE and I hate math. Well I only hate the math that I will never use for anything in my life but i guess EEing uses a lot of calculus so I better start liking it.

 

[Gamingphreek]

bump.... need an answer in the next 25 mins!!

 

[SynthDude2001]

I could have helped you about two years ago, but I've forgotten much of what I once knew.

 

[DaWhim]

Originally posted by: MBrown
I hate math but I have to take cal for my EE. I guess I could post in this forum when I start my precal class next semester. Incase your wondering why im going for a EE and I hate math. Well I only hate the math that I will never use for anything in my life but i guess EEing uses a lot of calculus so I better start liking it.

a few years down the road, when you realize EE is not your thing, you will want to change major. thus, screwed up.

I had a friend that wanted to do ME, after 1 sem, he dropped it.

 

[Gamingphreek]

Well i took the test. There was only one problem that i was really unsure about. I got the answer via Calculator but was not able to prove it, so probably wont get credit. Other than that everything went pretty well. Ill make sure to post the grade when i get it.

-Kevin

 

[Kyteland]

Originally posted by: Gamingphreek
bonkers thankyou, but i had already figured out that problem (with the aid of others)

I need help on the new problem listed above...

I dont want to seem like i am stupid or anything, but i need 1 more explained. My teacher simply gave the answer but did not explain this one. This one we have to use the squeeze theorem to solve it. The way i understand it is that the squeeze theorem is the value in between the limits of 2 equations.

In this equation (this is the one i can do):

lim X->C f(x)

c=0
4-x^2<=f(x)<=4+x^2

When graphed, there were two parabola and in the center, at the point they met was 4. Which coincedentally was the answer as it was both of the equations limit.

However, the book through a nice equation in there which is only made up of variables.

lim X->C f(x)

c=a
b-|x-a|<=f(x)<=b+|x-a|

Is there a simple way to tell the limit of f(x) here. I believe on this problem there is something simply fundamental that i haven't yet grasped. (For some reason 'b' sticks in my mind as the answer)

If f(x) is a rational or polynomial function and a is in the domain of f then:
lim x->a f(x) = f(a)
This limits law should be in your book somewhere.

You equation is basically asking you to take the limit as x goes to a of f(x). You may not know what f(x) defines, but the answer is f(a). Since lim x->a (b-|x-a|) = b-|a-a| = b and lim x->a (b+|x-a|) = b+|a-a| = b you can use the squeze theorem to prove that lim x->a f(x) = f(a) = b.

Better late than never, eh?

Edit: What was the question you couldn't get?