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wesleyho

  • 3 years ago

let fx=sqrt x. the rate of chance of f at x=c is twice its rate of change at x=1, then c=?

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  1. binarymimic
    • 3 years ago
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    try f'(c) = 2 * f'(1)

  2. binarymimic
    • 3 years ago
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    to find rate of change, take a derivative f'(x) then set up the equation "the derivative at x = c is twice the derivative at x = 1" f'(c) = 2 * f'(1)

  3. wesleyho
    • 3 years ago
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    is the answer 1 or 1/4

  4. binarymimic
    • 3 years ago
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    let me see f'(x) = 1/2(x)^(-1/2) = 1/(2 sqrt x) f'(c) = 1/(2 sqrt c) f'(1) = 1/2 f'(c) = 2 * f'(1) = 2 * 1/2 = 1 f'(c) = 1 = 1/ (2 sqrt c) 2 sqrt c = 1 sqrt c = 1/2 c = 1/4 yes

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