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

- anonymous

- chestercat

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- anonymous

|dw:1337454011771:dw|

- anonymous

It is not easy to find out.!

- 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

experimentX can you give me a way to your answer?

- experimentX

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

- experimentX

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

- anonymous

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

- anonymous

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

- experimentX

wolfram is pretty handy ... :D

- anonymous

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

- anonymous

|dw:1337455871084:dw|

- experimentX

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

- anonymous

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

- experimentX

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

- experimentX

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

- experimentX

http://www.wolframalpha.com/input/?i=lim+x-%3E1+Cos^-1%28sqrt%28x%29%29%2Fsqrt%281-x%29

- 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

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

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

- anonymous

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

- asnaseer

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

- asnaseer

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

- anonymous

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

- 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

That was wrong site you posted.

- 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

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

- anonymous

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

- anonymous

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

- 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

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

- anonymous

I accepted the conclusion but show me how?

- 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

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

- anonymous

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

- experimentX

yeah, that would be -1/6

- experimentX

using L'hospital's rule

- anonymous

ok show me.i really appreciate

- experimentX

are you having trouble with limits?

- experimentX

which part .. first or last??

- anonymous

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

- experimentX

Ah sorry ... we need only right hand limit

- anonymous

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

- experimentX

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

- 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

im waiting yet.

- 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

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

- anonymous

|dw:1337460455981:dw|

- anonymous

|dw:1337460652950:dw|

- anonymous

|dw:1337460769484:dw|

- experimentX

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

- anonymous

|dw:1337460890186:dw|

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