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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

Mathematics
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|dw:1337454011771:dw|
It is not easy to find out.!
a = http://www.wolframalpha.com/input/?i=d%2Fdx+Cos^-1%28sqrt%28x%29%29%2Fsqrt%281-x%29+%2C+x%3D1

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Other answers:

experimentX can you give me a way to your answer?
the left hand derivative must be equal to right hand derivative at x=1
for b, the function must be continuous ... so equate left hand limit to right hand limit at x=1
the right hand derivative is just "a" and left hand derivative can be found using quotient rule
I know but how can you reach to -1/6 ?
wolfram is pretty handy ... :D
yeap,but I didn't find any ways in there.
|dw:1337455871084:dw|
\[ a = \left ( \frac{d}{dx} \left ( \frac{\cos^{-1}(\sqrt{x})}{\sqrt{1-x}} \right ) \right )_{x=1} \]
Ok find it out?it gets so complexity.Doesn't it?
\[ \lim_{x->1}ax+b = \lim_{x->1}\frac{\cos^{-1}(\sqrt{x})}{\sqrt{1-x}}\]
i prefer to find out by using wolfram first .. then use hand to verity it.
http://www.wolframalpha.com/input/?i=lim+x-%3E1+Cos^-1%28sqrt%28x%29%29%2Fsqrt%281-x%29
in right hand side the limit is going to be 1 .then a+b=1 but give me a proof for a=-1/6
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+
\[ \left ( \frac{d}{dx}(ax+b) \right )_{x=1} = a = \left ( \frac{d}{dx} \left ( \frac{\cos^{-1}(\sqrt{x})}{\sqrt{1-x}} \right ) \right )_{x=1}\] http://www.wolframalpha.com/input/?i=d%28Cos^-1%28sqrt%28x%29%29%2Fsqrt%281-x%29%29%2Fdx+%2C+x%3D1
please show me -1/6 . I didn't find it in wolfram...!Did you see any?
The link I pasted above shows how to find the derivative. Then just plug x=1 into this to get your answer.
sorry - I meant take the limit as x tends to 1
Ok copy past the answer and find it out for x=1.
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
That was wrong site you posted.
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
experimentX: thats interesting - looks like wolfram is automagically using the limit as x tends to 1 to calculate this :)
I have never found it yet.so could you please copy past the way (lim x->1)
only i looked the answer -1/6 and derivation of function.
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
there is no calculation i chose step but there is only answer.
I accepted the conclusion but show me how?
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??
\[ \frac{d}{dx} \left ( \frac{\cos^{-1}(\sqrt{x})}{\sqrt{1-x}} \right ) \] I mean find this first.
ok find out this limit|dw:1337458049743:dw|
yeah, that would be -1/6
using L'hospital's rule
ok show me.i really appreciate
are you having trouble with limits?
which part .. first or last??
no but this is not easy to find out.check it out.
Ah sorry ... we need only right hand limit
both of them to each other you should take limit,because each goes to infinity !
ah .. yes ... looks like i'm nuts
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
im waiting yet.
\[ \frac{\sqrt x \cos^{-1}(\sqrt x) - \sqrt{1-x} }{(1-x)^{3/2} \sqrt x }\] 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
yes it is going to 0/0 so what is the rest?
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|dw:1337460652950:dw|
|dw:1337460769484:dw|
Ahh ... looks like you got it!! good night!!
|dw:1337460890186:dw|

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