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

1. amy0799

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2. amy0799

I really need help on this

3. zzr0ck3r

$$\cos(3\pi)$$

4. amy0799

how'd u get that?

5. zzr0ck3r

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

6. 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 $$1$$

7. amy0799

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

8. 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 $$\cos(\pi*x)$$.

9. amy0799

so what do you do after that?

10. zzr0ck3r

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

11. amy0799

what do you do with the h in the denominator?

12. zzr0ck3r

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

13. zzr0ck3r

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

14. amy0799

ooh ok. why is the answer -1?

15. zzr0ck3r

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

16. zzr0ck3r

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

17. amy0799

I'm not understanding why it's negative

18. anonymous

you lost me there

19. amy0799

what do you mean?

20. anonymous

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

21. amy0799

can you help me with another question?

22. zzr0ck3r

Do you understand this one?

23. anonymous

i guess you are suppose to think that if $f'(2)=\lim_{h\to 0}\frac{\cos(\pi(2+h))-1}{h}$ then $f(x)=\sin(\pi x)$

24. anonymous

i don't

25. zzr0ck3r

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

26. anonymous

yeah that

27. zzr0ck3r

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

28. zzr0ck3r

maybe I am confused lol

29. anonymous

knowing the derivative does not mean you know the function

30. amy0799

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

31. zzr0ck3r

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

32. anonymous

lol

33. zzr0ck3r

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

34. anonymous

that "plus k" is a killer

35. amy0799

can u guys help me with my other question?

36. anonymous

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

37. amy0799

x<3?

38. anonymous

yup

39. amy0799

thank you! :)

40. anonymous

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

41. 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 :)

42. zzr0ck3r

lol rereading my own remark makes me cringe

43. anonymous

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

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