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amy0799
 one year ago
if f'(2) = lim h>0 (cos(pi(2+h))1)/h
what is f(3)?
amy0799
 one year ago
if f'(2) = lim h>0 (cos(pi(2+h))1)/h what is f(3)?

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amy0799
 one year ago
Best ResponseYou've already chosen the best response.0I really need help on this

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2the function is \(\cos(\pi x)\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\(f'(x) = \lim_{h\rightarrow 0}\dfrac{\cos(\pi(x+h))\cos(\pi x)}{h}\) Now plug in \(2\) and \(\cos(2\pi)\) reduces to \(1\)

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0can u show me the steps to how you got that equation?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2The definition of the derivative of \(f\) is \(f'(x) =\lim_{h\rightarrow 0} \dfrac{f(x+h)f(x)}{h}\) So if we plug in \(2\) we get \(f'(2) =\lim_{h\rightarrow 0} \dfrac{f(2+h)f(2)}{h}\). Compare this to what you had and you see that the function is \(\cos(\pi*x)\).

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0so what do you do after that?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2well, since we figured out the function is \(\cos(\pi x)\) we simply plug in \(3\) and get \[\cos(3\pi)=1\]

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0what do you do with the h in the denominator?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Nothing, we don't need to actually run the limit, this question is not about the derivative, but the function itself.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2we looked at the derivative (the thing with the h) to get the function that we needed.

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0ooh ok. why is the answer 1?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2It asks for \(f(3)\) not \(f'(3)\)

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0I'm not understanding why it's negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nvm i didn't read the question correctly i thought you were supposed to find the limit

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0can you help me with another question?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Do you understand this one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i guess you are suppose to think that if \[f'(2)=\lim_{h\to 0}\frac{\cos(\pi(2+h))1}{h}\] then \[f(x)=\sin(\pi x)\]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2you mean \(\cos(\pi x)\)?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2what are you confused about @satellite73 ? seems pretty basic intro calc question to me.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2maybe I am confused lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0knowing the derivative does not mean you know the function

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0if f(x)=2+6xx^2, for what values of x is f'(x) positive?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\((\cos(\pi *2))'=\lim_{h\rightarrow 0}\dfrac{\cos(\pi(2+h))\cos(\pi*2)}{h}\implies f(x) = \cos(\pi x)\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2hehe, I cant think of what else they are asking...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that "plus k" is a killer

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0can u guys help me with my other question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x)=2+6xx^2\\ f'(x)=62x\] solve \[62x>0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yw @zzr0ck3r i think the first question was just ill posed i am sure you answer was what they wanted \(\cos(3x)\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2yeah , I didn't think of the +k thing, but I am not sure if they introduce them to that problem in intro differentiable clac. Honestly I forgot about the k thing lol. I hate calculus :)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2lol rereading my own remark makes me cringe

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is not you, it is whoever wrote the stupid question
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