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

• huan.conchito
In summary, the percent error in f(x) is approximately 1.5%, with a negative error of 1.53% and a positive error of 1.47%. This can be calculated using the formula \Delta f_p=\frac{1}{4}\Delta x_p, where \Delta x_p is the percent error in x. Additionally, it is important to note that the relative error in f(x) is asymmetrical for a symmetrical error in x and can only be accurately calculated if f(x) is monotone increasing for the given domain.
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

Last edited:
anyone know ?

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

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

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.

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.

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.

## 1. What does it mean for x to be accurate to 6%?

When we say that x is accurate to 6%, it means that the value of x falls within a range of 6% above or below the true value. This is also known as the margin of error.

## 2. How does the accuracy of x affect the accuracy of f(x)?

The accuracy of x can have a direct impact on the accuracy of f(x). If x is not accurate, it can lead to errors in the calculations and ultimately affect the accuracy of f(x).

## 3. Can the accuracy of f(x) be improved if x is accurate to 6%?

Yes, the accuracy of f(x) can be improved by ensuring that x is accurate to 6%. This will reduce the margin of error and lead to more precise calculations.

## 4. What factors can affect the accuracy of f(x) when x is accurate to 6%?

Aside from the accuracy of x, other factors that can affect the accuracy of f(x) include measurement errors, rounding errors, and limitations of the mathematical model used to calculate f(x).

## 5. How can we determine the accuracy of f(x) when x is accurate to 6%?

To determine the accuracy of f(x), we can compare the calculated value of f(x) with the actual or expected value. The closer the calculated value is to the expected value, the more accurate f(x) is considered to be.

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