Question 3.1.3: Linear Approximation to Some Cube Roots Use a linear approxi...

Here we are approximating values of the function $f(x)=\sqrt[3]{x}=x^{1 / 3}$. So, $f^{\prime}(x)=\frac{1}{3} x^{-2 / 3}$. The closest number to any of 8.02, 8.07 or 8.15 whose cube root we know exactly is 8. So, we write

$f(8.02)=f(8)+[f(8.02)-f(8)]$ Add and substract f(8).

$=f(8)+\Delta y$. (1.5)

From (1.4), we have

$\Delta y=f\left(x_1\right)-f\left(x_0\right) \approx f^{\prime}\left(x_0\right) \Delta x=d y$ (1.4)

$=\left(\frac{1}{3}\right) 8^{-2 / 3}(8.02-8)=\frac{1}{600}$. Since Δx = 8.02 − 8. (1.6)

Using (1.5) and (1.6), we get

$f(8.02) \approx f(8)+d y=2+\frac{1}{600} \approx 2.0016667$,

while your calculator accurately returns $\sqrt[3]{8.02} \approx 2.0016653$. Similarly, we get

and $f(8.15) \approx f(8)+\frac{1}{3} 8^{-2 / 3}(8.15-8) \approx 2.0125$,

while your calculator returns $\sqrt[3]{8.07} \approx 2.005816 \text { and } \sqrt[3]{8.15} \approx 2.01242$. In the margin, we show a table with the error in using the linear approximation to approximate $\sqrt[3]{x}$. Note how the error grows large as x gets farther from 8.

To approximate $\sqrt[3]{25.2}$, observe that 8 is not the closest number to 25.2 whose cube root we know exactly. Since 25.2 is much closer to 27 than to 8, we write

$f(25.2)=f(27)+\Delta y \approx f(27)+d y=3+d y$.

In this case,

and we have $f(25.2) \approx 3+d y=3-\frac{1}{15} \approx 2.9333333$,

compared to the value of 2.931794, produced by your calculator. In Figure 3.5, you can clearly see that the farther the value of x gets from the point of tangency, the worse the approximation tends to be.