# finding c: calc

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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1. 5 trillion

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posted by Bob
2. 1) f'(x)=1/2 times the square root of x

2) Plug in 1 as we are finding f'(1)

3) Multiply [1/2 times the square root of x] times 2 which equals 1

3) Now equate this to [1/2 times the square root of x]

4) You should get 1/4

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