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

- amy0799

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

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- schrodinger

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- amy0799

|dw:1440370564478:dw|

- amy0799

I really need help on this

- zzr0ck3r

\(\cos(3\pi)\)

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## More answers

- amy0799

how'd u get that?

- zzr0ck3r

the function is \(\cos(\pi x)\)

- 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

can u show me the steps to how you got that equation?

- 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

so what do you do after that?

- zzr0ck3r

well, since we figured out the function is \(\cos(\pi x)\) we simply plug in \(3\) and get
\[\cos(3\pi)=-1\]

- amy0799

what do you do with the h in the denominator?

- zzr0ck3r

Nothing, we don't need to actually run the limit, this question is not about the derivative, but the function itself.

- zzr0ck3r

we looked at the derivative (the thing with the h) to get the function that we needed.

- amy0799

ooh ok. why is the answer -1?

- zzr0ck3r

\(\cos(3\pi)=-1\)

- zzr0ck3r

It asks for \(f(3)\) not \(f'(3)\)

- amy0799

I'm not understanding why it's negative

- anonymous

you lost me there

- amy0799

what do you mean?

- anonymous

nvm i didn't read the question correctly
i thought you were supposed to find the limit

- amy0799

can you help me with another question?

- zzr0ck3r

Do you understand this one?

- 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

i don't

- zzr0ck3r

you mean \(\cos(\pi x)\)?

- anonymous

yeah that

- zzr0ck3r

what are you confused about @satellite73 ? seems pretty basic intro calc question to me.

- zzr0ck3r

maybe I am confused lol

- anonymous

knowing the derivative does not mean you know the function

- amy0799

if f(x)=2+6x-x^2, for what values of x is f'(x) positive?

- zzr0ck3r

\((\cos(\pi *2))'=\lim_{h\rightarrow 0}\dfrac{\cos(\pi(2+h))-\cos(\pi*2)}{h}\implies f(x) = \cos(\pi x)\)

- anonymous

lol

- zzr0ck3r

hehe, I cant think of what else they are asking...

- anonymous

that "plus k" is a killer

- amy0799

can u guys help me with my other question?

- anonymous

\[f(x)=2+6x-x^2\\
f'(x)=6-2x\] solve
\[6-2x>0\]

- amy0799

x<3?

- anonymous

yup

- amy0799

thank you! :)

- anonymous

yw
@zzr0ck3r i think the first question was just ill posed
i am sure you answer was what they wanted \(\cos(3x)\)

- 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

lol rereading my own remark makes me cringe

- anonymous

it is not you, it is whoever wrote the stupid question

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