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
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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
Let me guess,
You said,
%percent error = df / f = f' * dx / f
f' = df/dx, thus:
df / f = df/dx * dx / f
Your x can change 6% so dx = 0.06
df/dx = 1/(4*x^(3/4))
error = 1/(4*x^(3/4)) * 0.06 / x^(1/4) = 3/(200*x)
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% ?