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.
3 answers
5 trillion
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
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
easy 1billion compounded to the 55th decimal x times the quantum of infinity
so X?
so X?
1 answer