amy0799
  • amy0799
if f'(2) = lim h-->0 (cos(pi(2+h))-1)/h what is f(3)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amy0799
  • amy0799
|dw:1440370564478:dw|
amy0799
  • amy0799
I really need help on this
zzr0ck3r
  • zzr0ck3r
\(\cos(3\pi)\)

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More answers

amy0799
  • amy0799
how'd u get that?
zzr0ck3r
  • zzr0ck3r
the function is \(\cos(\pi x)\)
zzr0ck3r
  • zzr0ck3r
\(f'(x) = \lim_{h\rightarrow 0}\dfrac{\cos(\pi(x+h))-\cos(\pi x)}{h}\) Now plug in \(2\) and \(\cos(2\pi)\) reduces to \(1\)
amy0799
  • amy0799
can u show me the steps to how you got that equation?
zzr0ck3r
  • zzr0ck3r
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 \(\cos(\pi*x)\).
amy0799
  • amy0799
so what do you do after that?
zzr0ck3r
  • zzr0ck3r
well, since we figured out the function is \(\cos(\pi x)\) we simply plug in \(3\) and get \[\cos(3\pi)=-1\]
amy0799
  • amy0799
what do you do with the h in the denominator?
zzr0ck3r
  • zzr0ck3r
Nothing, we don't need to actually run the limit, this question is not about the derivative, but the function itself.
zzr0ck3r
  • zzr0ck3r
we looked at the derivative (the thing with the h) to get the function that we needed.
amy0799
  • amy0799
ooh ok. why is the answer -1?
zzr0ck3r
  • zzr0ck3r
\(\cos(3\pi)=-1\)
zzr0ck3r
  • zzr0ck3r
It asks for \(f(3)\) not \(f'(3)\)
amy0799
  • amy0799
I'm not understanding why it's negative
anonymous
  • anonymous
you lost me there
amy0799
  • amy0799
what do you mean?
anonymous
  • anonymous
nvm i didn't read the question correctly i thought you were supposed to find the limit
amy0799
  • amy0799
can you help me with another question?
zzr0ck3r
  • zzr0ck3r
Do you understand this one?
anonymous
  • anonymous
i guess you are suppose to think that if \[f'(2)=\lim_{h\to 0}\frac{\cos(\pi(2+h))-1}{h}\] then \[f(x)=\sin(\pi x)\]
anonymous
  • anonymous
i don't
zzr0ck3r
  • zzr0ck3r
you mean \(\cos(\pi x)\)?
anonymous
  • anonymous
yeah that
zzr0ck3r
  • zzr0ck3r
what are you confused about @satellite73 ? seems pretty basic intro calc question to me.
zzr0ck3r
  • zzr0ck3r
maybe I am confused lol
anonymous
  • anonymous
knowing the derivative does not mean you know the function
amy0799
  • amy0799
if f(x)=2+6x-x^2, for what values of x is f'(x) positive?
zzr0ck3r
  • zzr0ck3r
\((\cos(\pi *2))'=\lim_{h\rightarrow 0}\dfrac{\cos(\pi(2+h))-\cos(\pi*2)}{h}\implies f(x) = \cos(\pi x)\)
anonymous
  • anonymous
lol
zzr0ck3r
  • zzr0ck3r
hehe, I cant think of what else they are asking...
anonymous
  • anonymous
that "plus k" is a killer
amy0799
  • amy0799
can u guys help me with my other question?
anonymous
  • anonymous
\[f(x)=2+6x-x^2\\ f'(x)=6-2x\] solve \[6-2x>0\]
amy0799
  • amy0799
x<3?
anonymous
  • anonymous
yup
amy0799
  • amy0799
thank you! :)
anonymous
  • anonymous
yw @zzr0ck3r i think the first question was just ill posed i am sure you answer was what they wanted \(\cos(3x)\)
zzr0ck3r
  • zzr0ck3r
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 :)
zzr0ck3r
  • zzr0ck3r
lol rereading my own remark makes me cringe
anonymous
  • anonymous
it is not you, it is whoever wrote the stupid question

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