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

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binarymimicBest ResponseYou've already chosen the best response.2
try f'(c) = 2 * f'(1)
 one year ago

binarymimicBest ResponseYou've already chosen the best response.2
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)
 one year ago

wesleyhoBest ResponseYou've already chosen the best response.0
is the answer 1 or 1/4
 one year ago

binarymimicBest ResponseYou've already chosen the best response.2
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
 one year ago
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