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clara1223

  • one year ago

Using the Mean Value Theorem find all numbers c on the interval [0,5] that satisfy the theorem for f(x)=3*sqrt(25-x^2). I have already found that (f(5)-f(0))/(5-0)=-3, and that the derivative of the function is (-3x)/sqrt(25-x^2). When I set that equal to -3 and solve for x I get an answer that is not in the interval so I am doing something wrong. Can someone help me?

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  1. clara1223
    • one year ago
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    @zepdrix can you help me?

  2. Loser66
    • one year ago
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    I don't get why you said so, \(\dfrac{-3x}{\sqrt{25-x^2}}=-3\\x = \sqrt{25-x^2}\) \(x^2 = 25 -x^2\\2x^2 = 25\\x=\pm\dfrac{5}{\sqrt2}\) has positive one is on the interval. Am I missing something?

  3. clara1223
    • one year ago
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    Oh I didnt cancel out the -3 in the beginning and somehow I got my math wrong. I got \[\pm \frac{ 5\sqrt{3} }{ \sqrt{2} }\] Thanks!

  4. Loser66
    • one year ago
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    np

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