finding c: calc
posted by david on .
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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