Posted by **david** on Tuesday, December 5, 2006 at 1:36am.

let f(x)=square root x. if the rate of change of at x=c is twice its rate of change at x=1, then c=?

what???? i don't get get it.

to begin with, i took the derivative if that helps which is 1/(2*sqrtx) .

f'(x) = 1/(2*sqrtx) is correct

f'(x=1) = 1/2

The problem is to find the value of a constant x=c such that the derivative is twice the value at x=1, or 1.

f'(c) = 1 = 1/(2*sqrt c)

2 sqrt c = 1

sqrt c = 1/2

c = 1/4.

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