Question

answer the following true or false: if ( y = 8x^{2} ) and you know the value of ( x ) with an error no more than ( pm e ) then the relative error in using this approximation of ( x ) to approximate the value of ( y ) does not depend on ( x ). false true

Explanation:

Step1: Recall the formula for relative error

The formula for relative error is $\frac{\Delta y}{y}$.

Step2: Differentiate \(y = 8x^{2}\)

Differentiating \(y = 8x^{2}\) with respect to \(x\) gives \(dy=16x dx\).

Step3: Calculate the relative error

The relative error \(\frac{\Delta y}{y}\approx\frac{dy}{y}\). Substituting \(y = 8x^{2}\) and \(dy = 16x dx\) into \(\frac{dy}{y}\), we get \(\frac{16x dx}{8x^{2}}=\frac{2dx}{x}\). Here \(dx\) is the error in \(x\) (i.e., \(dx = E\)). So the relative error is \(\frac{2E}{x}\), which depends on \(x\).

Answer:

False