## mahmit2012 Q3 If f(x) is a differentiable at x=1, find a,b. f(x)=ArcCos(sqrt(x))/sqrt(1-x) for 0<=x<1 ax+b for x=>1 one year ago one year ago

1. mahmit2012

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2. mahmit2012

It is not easy to find out.!

3. experimentX
4. mahmit2012

5. experimentX

the left hand derivative must be equal to right hand derivative at x=1

6. experimentX

for b, the function must be continuous ... so equate left hand limit to right hand limit at x=1

7. niravshah08

the right hand derivative is just "a" and left hand derivative can be found using quotient rule

8. mahmit2012

I know but how can you reach to -1/6 ?

9. experimentX

wolfram is pretty handy ... :D

10. mahmit2012

yeap,but I didn't find any ways in there.

11. niravshah08

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12. experimentX

$a = \left ( \frac{d}{dx} \left ( \frac{\cos^{-1}(\sqrt{x})}{\sqrt{1-x}} \right ) \right )_{x=1}$

13. mahmit2012

Ok find it out?it gets so complexity.Doesn't it?

14. experimentX

$\lim_{x->1}ax+b = \lim_{x->1}\frac{\cos^{-1}(\sqrt{x})}{\sqrt{1-x}}$

15. experimentX

i prefer to find out by using wolfram first .. then use hand to verity it.

16. experimentX
17. mahmit2012

in right hand side the limit is going to be 1 .then a+b=1 but give me a proof for a=-1/6

18. asnaseer

you can find the proof using wolfram - click show steps to see how it is done: http://www.wolframalpha.com/input/?i=d%2Fdx+Cos^-1%28sqrt%28x%29%29%2Fsqrt%281-x%29+

19. experimentX

$\left ( \frac{d}{dx}(ax+b) \right )_{x=1} = a = \left ( \frac{d}{dx} \left ( \frac{\cos^{-1}(\sqrt{x})}{\sqrt{1-x}} \right ) \right )_{x=1}$ http://www.wolframalpha.com/input/?i=d%28Cos^-1%28sqrt%28x%29%29%2Fsqrt%281-x%29%29%2Fdx+%2C+x%3D1

20. mahmit2012

please show me -1/6 . I didn't find it in wolfram...!Did you see any?

21. asnaseer

The link I pasted above shows how to find the derivative. Then just plug x=1 into this to get your answer.

22. asnaseer

sorry - I meant take the limit as x tends to 1

23. mahmit2012

Ok copy past the answer and find it out for x=1.

24. asnaseer

you can find the derivation of this limit here: http://www.wolframalpha.com/input/?i=limit+%281%2F%28%28-1+%2B+x%29+Sqrt[x]%29+%2B+ArcCos[Sqrt[x]]%2F%281+-+x%29^%283%2F2%29%29%2F2+as+x+tends+to+1 again, just click show steps to see it

25. mahmit2012

That was wrong site you posted.

26. experimentX

Jeez what am i getting instead of indeterminate http://www.wolframalpha.com/input/?i=1%2F2+%281%2F%28%28-1%2Bx%29+sqrt%28x%29%29%2B%28cos^%28-1%29%28sqrt%28x%29%29%29%2F%281-x%29^%283%2F2%29%29+where+x%3D1

27. asnaseer

experimentX: thats interesting - looks like wolfram is automagically using the limit as x tends to 1 to calculate this :)

28. mahmit2012

I have never found it yet.so could you please copy past the way (lim x->1)

29. mahmit2012

only i looked the answer -1/6 and derivation of function.

30. experimentX

it seems the case http://www.wolframalpha.com/input/?i=lim+x-%3E1++1%2F2%281%2F%28%28-1%2Bx%29+sqrt%28x%29%29%2B%28cos^%28-1%29%28sqrt%28x%29%29%29%2F%281-x%29^%283%2F2%29%29 but limit of derivative ... sounds fisht @mahmit2012 click Show Steps .. below "=" icon

31. mahmit2012

there is no calculation i chose step but there is only answer.

32. mahmit2012

I accepted the conclusion but show me how?

33. experimentX

Oh ... this is a bit fishy ... the slope of f(x) will be a at x=1 (left hand slope as well as slope) While on the other side, the right hand slope tends to this value 'a' ... i think limiting the value of the derivative will just do fine. You can find the derivative of arccos... right??

34. experimentX

$\frac{d}{dx} \left ( \frac{\cos^{-1}(\sqrt{x})}{\sqrt{1-x}} \right )$ I mean find this first.

35. mahmit2012

ok find out this limit|dw:1337458049743:dw|

36. experimentX

yeah, that would be -1/6

37. experimentX

using L'hospital's rule

38. mahmit2012

ok show me.i really appreciate

39. experimentX

are you having trouble with limits?

40. experimentX

which part .. first or last??

41. mahmit2012

no but this is not easy to find out.check it out.

42. experimentX

Ah sorry ... we need only right hand limit

43. mahmit2012

both of them to each other you should take limit,because each goes to infinity !

44. experimentX

ah .. yes ... looks like i'm nuts

45. experimentX

It seems to be the case of $$\infty - \infty$$ so we can apply L"hospital's rule $\frac{1}{(-1+x) \sqrt x} = -\frac{1}{(1-x) \sqrt x}$ make them a fraction ... and use L'hospital's rule

46. mahmit2012

im waiting yet.

47. experimentX

$\frac{\sqrt x \cos^{-1}(\sqrt x) - \sqrt{1-x} }{(1-x)^{3/2} \sqrt x }$ Use L'hospital rule ... Since now both numerators are denominators are zero ... you can use L'Hospital rule http://www.wolframalpha.com/input/?i=lim+x-%3E1++%28sqrt%281-x%29++-+sqrt%28x%29+cos^-1%28sqrt%28x%29%29%29

48. mahmit2012

yes it is going to 0/0 so what is the rest?

49. mahmit2012

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50. mahmit2012

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51. mahmit2012

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52. experimentX

Ahh ... looks like you got it!! good night!!

53. mahmit2012

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