anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1337454011771:dw|
anonymous
  • anonymous
It is not easy to find out.!
experimentX
  • experimentX
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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More answers

anonymous
  • anonymous
experimentX can you give me a way to your answer?
experimentX
  • experimentX
the left hand derivative must be equal to right hand derivative at x=1
experimentX
  • experimentX
for b, the function must be continuous ... so equate left hand limit to right hand limit at x=1
anonymous
  • anonymous
the right hand derivative is just "a" and left hand derivative can be found using quotient rule
anonymous
  • anonymous
I know but how can you reach to -1/6 ?
experimentX
  • experimentX
wolfram is pretty handy ... :D
anonymous
  • anonymous
yeap,but I didn't find any ways in there.
anonymous
  • anonymous
|dw:1337455871084:dw|
experimentX
  • experimentX
\[ a = \left ( \frac{d}{dx} \left ( \frac{\cos^{-1}(\sqrt{x})}{\sqrt{1-x}} \right ) \right )_{x=1} \]
anonymous
  • anonymous
Ok find it out?it gets so complexity.Doesn't it?
experimentX
  • experimentX
\[ \lim_{x->1}ax+b = \lim_{x->1}\frac{\cos^{-1}(\sqrt{x})}{\sqrt{1-x}}\]
experimentX
  • experimentX
i prefer to find out by using wolfram first .. then use hand to verity it.
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=lim+x-%3E1+Cos^-1%28sqrt%28x%29%29%2Fsqrt%281-x%29
anonymous
  • anonymous
in right hand side the limit is going to be 1 .then a+b=1 but give me a proof for a=-1/6
asnaseer
  • 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+
experimentX
  • 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
anonymous
  • anonymous
please show me -1/6 . I didn't find it in wolfram...!Did you see any?
asnaseer
  • asnaseer
The link I pasted above shows how to find the derivative. Then just plug x=1 into this to get your answer.
asnaseer
  • asnaseer
sorry - I meant take the limit as x tends to 1
anonymous
  • anonymous
Ok copy past the answer and find it out for x=1.
asnaseer
  • 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
anonymous
  • anonymous
That was wrong site you posted.
experimentX
  • 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
asnaseer
  • asnaseer
experimentX: thats interesting - looks like wolfram is automagically using the limit as x tends to 1 to calculate this :)
anonymous
  • anonymous
I have never found it yet.so could you please copy past the way (lim x->1)
anonymous
  • anonymous
only i looked the answer -1/6 and derivation of function.
experimentX
  • 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
anonymous
  • anonymous
there is no calculation i chose step but there is only answer.
anonymous
  • anonymous
I accepted the conclusion but show me how?
experimentX
  • 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??
experimentX
  • experimentX
\[ \frac{d}{dx} \left ( \frac{\cos^{-1}(\sqrt{x})}{\sqrt{1-x}} \right ) \] I mean find this first.
anonymous
  • anonymous
ok find out this limit|dw:1337458049743:dw|
experimentX
  • experimentX
yeah, that would be -1/6
experimentX
  • experimentX
using L'hospital's rule
anonymous
  • anonymous
ok show me.i really appreciate
experimentX
  • experimentX
are you having trouble with limits?
experimentX
  • experimentX
which part .. first or last??
anonymous
  • anonymous
no but this is not easy to find out.check it out.
experimentX
  • experimentX
Ah sorry ... we need only right hand limit
anonymous
  • anonymous
both of them to each other you should take limit,because each goes to infinity !
experimentX
  • experimentX
ah .. yes ... looks like i'm nuts
experimentX
  • 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
anonymous
  • anonymous
im waiting yet.
experimentX
  • 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
anonymous
  • anonymous
yes it is going to 0/0 so what is the rest?
anonymous
  • anonymous
|dw:1337460455981:dw|
anonymous
  • anonymous
|dw:1337460652950:dw|
anonymous
  • anonymous
|dw:1337460769484:dw|
experimentX
  • experimentX
Ahh ... looks like you got it!! good night!!
anonymous
  • anonymous
|dw:1337460890186:dw|

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