Suppose f(x) = x^(1/4). If x is accurate to within 6%, within what percent is f(x) accurate ?
i know that %percent error = delta F/ f(x) = f'(x)*deltaX / f(x)
what i dont know is how to get delta f out this this information
f(x) = x^(1/4). If x is accurate to within 6%, within what percent is f(x) accurate
x=1
f(x) = 1
x=1.06
f(x) = 1.01467385
Therefore it is accurate to within 1.467% when x=1.
Why is this not exactly 3/(200*1) = 1.5% ?