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

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

Mathematics
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|dw:1440370564478:dw|
I really need help on this
\(\cos(3\pi)\)

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how'd u get that?
the function is \(\cos(\pi x)\)
\(f'(x) = \lim_{h\rightarrow 0}\dfrac{\cos(\pi(x+h))-\cos(\pi x)}{h}\) Now plug in \(2\) and \(\cos(2\pi)\) reduces to \(1\)
can u show me the steps to how you got that equation?
The 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 \(\cos(\pi*x)\).
so what do you do after that?
well, since we figured out the function is \(\cos(\pi x)\) we simply plug in \(3\) and get \[\cos(3\pi)=-1\]
what do you do with the h in the denominator?
Nothing, we don't need to actually run the limit, this question is not about the derivative, but the function itself.
we looked at the derivative (the thing with the h) to get the function that we needed.
ooh ok. why is the answer -1?
\(\cos(3\pi)=-1\)
It asks for \(f(3)\) not \(f'(3)\)
I'm not understanding why it's negative
you lost me there
what do you mean?
nvm i didn't read the question correctly i thought you were supposed to find the limit
can you help me with another question?
Do you understand this one?
i guess you are suppose to think that if \[f'(2)=\lim_{h\to 0}\frac{\cos(\pi(2+h))-1}{h}\] then \[f(x)=\sin(\pi x)\]
i don't
you mean \(\cos(\pi x)\)?
yeah that
what are you confused about @satellite73 ? seems pretty basic intro calc question to me.
maybe I am confused lol
knowing the derivative does not mean you know the function
if f(x)=2+6x-x^2, for what values of x is f'(x) positive?
\((\cos(\pi *2))'=\lim_{h\rightarrow 0}\dfrac{\cos(\pi(2+h))-\cos(\pi*2)}{h}\implies f(x) = \cos(\pi x)\)
lol
hehe, I cant think of what else they are asking...
that "plus k" is a killer
can u guys help me with my other question?
\[f(x)=2+6x-x^2\\ f'(x)=6-2x\] solve \[6-2x>0\]
x<3?
yup
thank you! :)
yw @zzr0ck3r i think the first question was just ill posed i am sure you answer was what they wanted \(\cos(3x)\)
yeah , 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 :)
lol rereading my own remark makes me cringe
it is not you, it is whoever wrote the stupid question

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