Accuracy of f(x) when x is Accurate to 6%”

[huan.conchito]

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 don't know is how to get delta f out this this information

 

[huan.conchito]

anyone know ?

 

[michael376071]

I think I ran into this in calc, but not quite sure, I believe you can differentiate dy/dx, then you could plug in the %error for dx, then solve for dy. That probably isn't it but its my best guess at the moment so late at night for me right now :tongue:

 

[SpaceTiger]

$$\Delta f = \frac{df}{dx}|_{x_0}\Delta x$$
$$\Delta f = \frac{1}{4}x_0^{-3/4}\Delta x=\frac{1}{4}\frac{x_0^{1/4}\Delta x_p}{100}$$

where $$\Delta x_p$$ is your percent error in x and x₀ is the value of x at which you're finding the error. To get the percent error in f, you just evaluate:

$$\Delta f_p=100\frac{\Delta f}{f(x_0)}=100\frac{\Delta f}{x_0^{1/4}}=\frac{1}{4}\Delta x_p$$

In other words, it's just a quarter of the percent error in x. If you're just looking for the raw error in f, it depends on the value you're evaluating at and it's given by the second equation above.

 

[michael376071]

Ah that is indeed it, if you used the method I said before, you would actually need to plug in .06x, not just .06, then solve for dy and divide by the function leaving you with 1.5% I believe.

 

[Curious3141]

Honestly, there is no need to use calculus for this, unless a calculator is forbidden. You can get an exact answer much more easily directly with a calculator.

$$f(x) = x^{\frac{1}{4}}$$

We're given that the maximum error in x is 6 %, meaning that x varies from a min of 0.94x to a max of 1.06x

Hence,

$$min(f(x)) = (0.94x)^{\frac{1}{4}}$$ which approximately equals to $$0.9847(x^{\frac{1}{4}}) = 0.9847f(x)$$, meaning a percentage error of - 1.53 %

and

$$max(f(x)) = (1.06x)^{\frac{1}{4}}$$ which approx. $$1.01467f(x)$$, meaning a percentage error of + 1.47 %

So you can see the relative error in f(x) is assymetrical for a symmetrical relative error in x, and goes from a maximum of 1.53 % on the negative side to a max of 1.47 % on the positive side.

To be rigorous, you actually need to establish that f(x) is monotone increasing for the domain in question (true in this case). But it wouldn't work, for instance, in cases where there is a local maximum or minimum to f(x) around the values in consideration. And example of this would be in computing the relative error in $$f(x) = (x-1)^2$$ for a 5 % error in x where x is known to be close to 1.