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

Printable View

- April 8th 2005, 03:30 PMBobber% Percent error
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 - April 11th 2005, 09:38 PMpaultwang
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) - April 12th 2005, 05:26 PMMathGuruPlug it in
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% ?