IsTim
  • IsTim
Find the equation of the tangent to the curve at the given x value.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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IsTim
  • IsTim
|dw:1335261368315:dw|
IsTim
  • IsTim
I got up to this:|dw:1335261401688:dw|
IsTim
  • IsTim
|dw:1335261446849:dw|

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More answers

chaise
  • chaise
Calculate the derivative of the equation, make the equation to equal, solve it and plug in your x value. Are you familiar with the quotient rule in derivatives?
IsTim
  • IsTim
Deriatives is f'(x) right? we're not doing that yet.
IsTim
  • IsTim
Methods we're using is substitution, factoring and the conjugate method.
IsTim
  • IsTim
I'm using the Calculus Phobe videos from Calculus Help (my teacher showed me the website and told me to use that).
chaise
  • chaise
Yes, that's correct. So you're differentiating using the definition? If you take the derivative of the equation y=3/(x^2-1) you can find the slope. if: f(x)=3/(x^2-1) then f(x+h)=3/((x+h)^2-1) dy/dx = (f(x+h)-f(x))/h Can you do the rest?
chaise
  • chaise
dy/dx =lim (h-->0) (f(x+h)-f(x))/h rather.
IsTim
  • IsTim
Yes. That's what I did.
IsTim
  • IsTim
But I can't go any further. |dw:1335262087563:dw| That's the furthest I got up to.
chaise
  • chaise
Expand your brackets and factor out a h.
IsTim
  • IsTim
The bottom one to? I was told not too.
chaise
  • chaise
Not sure - I'm not familiar with the long method, I just use the shortcut :)
Callisto
  • Callisto
Can you take derivatives? Must you use first principle to find it?
IsTim
  • IsTim
Derivatives is our next unit.
IsTim
  • IsTim
I can't isolate or factor out anything. I must had made a mistake beforehand I guess.
IsTim
  • IsTim
Anyone?
Callisto
  • Callisto
I hope you don't mind me posting my solution here
1 Attachment
IsTim
  • IsTim
I guess not, but it's notm much like my work.
Callisto
  • Callisto
Oh shoot... I got some mistakes there! Wait
Callisto
  • Callisto
Once again :(
1 Attachment
Callisto
  • Callisto
Oh my dear!!! The final answer should be -4/3 ... not -2 as I've written :S
Callisto
  • Callisto
Is your question \[\frac{3}{x^2-1}\] or \[\frac{3}{x^2+1}\]?
IsTim
  • IsTim
First one.
IsTim
  • IsTim
Maybe I should upload my work, and maybe you check for problems?
Callisto
  • Callisto
Sure!
IsTim
  • IsTim
IT's c.
IsTim
  • IsTim
Starting from c) at Scan001.
Callisto
  • Callisto
f(a) is not 1
Callisto
  • Callisto
f(a) is 3/ (a^2 -1)
IsTim
  • IsTim
but since x=2, I subbed in a=2 into that equation...
Callisto
  • Callisto
for the denominator, (a+h)-a =h , not a+h -2 You need to find the limit before substituting the numbers.
Callisto
  • Callisto
After you've found the limit, you'll see the unknown x. By then, sub x=2 and you'll get the slope. But NOT when you're taking the limit
IsTim
  • IsTim
I thought the unknown to be m, and that h was still the variable withint the equation?
Callisto
  • Callisto
h is something you need to deal with. Should I upload an example to should how you do first principle for you ?
IsTim
  • IsTim
I've got an example here with me from my teacher.
Callisto
  • Callisto
If you don't need it, it's okay :) Should I keep checking for you or you want to do corrections first?
IsTim
  • IsTim
What should I correct?
Callisto
  • Callisto
Correct starting from the third line For numerator: a -> 3/ (a^2 -1) For denominator: a+h-2 -> h (in your previous step, a+h -a, simplify it , you'll get h)
IsTim
  • IsTim
Sorry if I sound ignorant, but that's how I saw it in the example...the h is alone.
Callisto
  • Callisto
Do you mind uploading your example also?
IsTim
  • IsTim
1 Attachment
Callisto
  • Callisto
First, the situation is different . You're given 2 values there, but only 1 here. Second, to be frank, I don't understand what you (or your teacher) is doing. It's not the normal way as I've learnt.. Third, I still insist that you shouldn't sub the value immediately...
IsTim
  • IsTim
Ok. I think I know what you are doing, and what my teacher is doing. You're going from a general formula, while my teacher is using the specific formula. The general formula doesn't sub in the x or a value until the end.
Callisto
  • Callisto
It's good that you notice the difference :)
IsTim
  • IsTim
Just looking through my notes I guess. I've got a test coming up, so I thought reviewing would be best.
Callisto
  • Callisto
Yup :)
Mimi_x3
  • Mimi_x3
well, where do you need help with? This is pretty long..
IsTim
  • IsTim
Maybe I should just leave this question. I tried this for over 2 hours by now.
Mimi_x3
  • Mimi_x3
First, what is the question? \[\frac{1}{x-2} \]?? or \[\frac{3}{x^{2}-1} \]??
IsTim
  • IsTim
Second one, whereas x=2.
IsTim
  • IsTim
Ok. Thanks for the attempt guys. I'll try other problems, and maybe this might make more sense.
Mimi_x3
  • Mimi_x3
wait..i will try and do it first.. can't you take the derivative which is easier?
IsTim
  • IsTim
Others suggested that, but that's for the next unit. I understand a bit on how that works, but I'm not allowed using it here.
Mimi_x3
  • Mimi_x3
Hmm..I might try it but it might take a while since i havent done these in a while...
IsTim
  • IsTim
Ok. Well, I guess you can do so, if yo uhave free time. But most likely you too have something important coming up (immediately).
Mimi_x3
  • Mimi_x3
No, I don't have anything important..i might be able to do it but not sure..lol
Mimi_x3
  • Mimi_x3
I have to review these anyway
apoorvk
  • apoorvk
^^That was mammoth. phew. Istim do you know how to differentiate using chain rule?
Mimi_x3
  • Mimi_x3
lol, i failed...do you remember how to do these apoor? my brain is failling me :(
IsTim
  • IsTim
No, I don't (Sorry, cleaning up house).
apoorvk
  • apoorvk
Hmm. so you want to do this by the first principle of finding derivatives okay.
IsTim
  • IsTim
I am just oh so selective in the method used. That be the end of me.
Mimi_x3
  • Mimi_x3
\[\large\frac{f(x+h)-f(x)}{h} =>\frac{\frac{3}{(x+h)^{2}-1}-\frac{3}{x^{2}-1}}{h} \]
Mimi_x3
  • Mimi_x3
right so far?
IsTim
  • IsTim
I have no idea.
chaise
  • chaise
should be the limit as h approaches zero, but, apart from that, correct. ^_^
Mimi_x3
  • Mimi_x3
\[\large =>\frac{\frac{3}{h^{2}+2hx+x^{2}-1}-\frac{3}{x^{2}-1}}{h} \] \[\large >\frac{\frac{3(x^{2}-1)-3(h^{2}+2hx+x^{2}-1)}{(h^{2}+2hx+x^{2}-1)(x^{2}-1)}}{h} \]
Mimi_x3
  • Mimi_x3
I think that im doing it wrong..i dont think that its this complicated..
chaise
  • chaise
Perhaps you should just use quotient rule.
Mimi_x3
  • Mimi_x3
but it says derivatiive thing is not allowed..
IsTim
  • IsTim
Before using deratives. Referencing Calculus Phobe from Calculus Help. I think I used Conjugate Method.
Mimi_x3
  • Mimi_x3
whaat..im lost..i looked at your example but it got confusing..then i looked at my book its more confusing..using the chain rule is easier.
Mimi_x3
  • Mimi_x3
\[\frac{3}{x^{2}-1} => 3(x^2-1)^{-1}\] \[-3(x^2-1)^{-1}*2x \] \[-6x(x^2-1)^{-1}\] easier i suppose..

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