With what accuracy must the abscissa of the curve xy = 4

With what accuracy must the abscissa of the curve $xy = 4$ be measured at $x = 0.5$ so that the error in the computed ordinate does not exceed $0.1$?

Step 1: The ordinate as a function of the abscissa

\[ y = \frac{4}{x}, \qquad y(0.5) = \frac{4}{0.5} = 8. \]

Step 2: Differentiate

\[ \frac{dy}{dx} = -\frac{4}{x^{2}}, \qquad \bigl|\Delta y\bigr| \approx \frac{4}{x^{2}}\,\bigl|\Delta x\bigr|. \]

Step 3: Evaluate the sensitivity at $x = 0.5$

\[ \frac{4}{\bigl(0.5\bigr)^{2}} = \frac{4}{0.25} = 16. \]

So near this point the ordinate changes sixteen times as fast as the abscissa — an error is magnified, not damped.

Step 4: Impose the requirement

\[ 16\,\bigl|\Delta x\bigr| \le 0.1 \;\Longrightarrow\; \bigl|\Delta x\bigr| \le \frac{0.1}{16} = 0.00625. \]

Step 5: The same thing in relative terms

For $y = 4/x$ we have $\ln y = \ln4-\ln x$, so the relative errors are equal in magnitude: \[\frac{\bigl|\Delta y\bigr|}{y} = \frac{\bigl|\Delta x\bigr|}{x} \;\Longrightarrow\; \frac{0.1}{8} = 0.0125 = \frac{\bigl|\Delta x\bigr|}{0.5} \;\Longrightarrow\; \bigl|\Delta x\bigr| = 0.00625,\]

which confirms the answer by an independent route.

Step 6: Verification

\[x = 0.50625:\quad y = \frac{4}{0.50625} = 7.901235, \qquad \Delta y = -0.098765 \;\;\checkmark\]

\[x = 0.49375:\quad y = \frac{4}{0.49375} = 8.101266, \qquad \Delta y = +0.101266\]

\[\text{Strict bound: } \frac{4}{0.5-\delta} \le 8.1 \iff \delta \le 0.5-\frac{4}{8.1} = 0.006173 \;\;\checkmark\]