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April 16th, 2018, 11:58 PM   #1
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Calculus - Lagrange

if f(x) = (x^p)/(sinx)^q when 0 < x < 90
= 0 when x = 0
Then prove that L agrange’s mean value theorem is applicable to f(x) in closed interval [0, x] only when p >q .

Could u please put some hint
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April 17th, 2018, 03:05 AM   #2
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The mean value theorem, applied to interval [0, x] requires that f be differentiable on the open interval (0, x) and continuous on the closed interval [0, x]. Can you show that?

(Does the problem really say "0< x< 90"? That "90" looks like 90 degrees and it is very unusual to use degrees rather than radians in Calculus)
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April 17th, 2018, 04:21 AM   #3
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U are correct this is in radians ( pi/2) . However , could u please help me proving that Lagrange's MVT applies to the above only when p> q
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April 17th, 2018, 05:16 AM   #4
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You should focus on that condition of continuity. What can you tell about the continuity of $f(X)$ at $x=0$?
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April 17th, 2018, 06:18 AM   #5
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What is $\displaystyle \lim_{x\to 0}\frac{x^p}{sin^q(x)}$ for different values of p and q?

Notice that:

if p= q you can write the function as $\displaystyle \left(\frac{x}{sin(x)}\right)^p$.

If p> q, taking r= p- q, you can write the function as $\displaystyle \left(\frac{x}{sin(x)}\right)^q x^r$.

If p< q, taking r= q- r, you can write the function as $\displaystyle \left(\frac{x}{sin(x)}\right)^q\left(\frac{1}{sin( x)}\right)^r$.

Last edited by Country Boy; April 17th, 2018 at 06:42 AM.
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April 17th, 2018, 09:24 PM   #6
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Thanks . Got it .The R.H.L = 0 only when p>q
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April 17th, 2018, 10:51 PM   #7
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I mean R.H.L = 0 ( at x--> 0+)
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