To approximate \(\sqrt[3]{1511}\) using the function \(g(x) = 15x\), we can observe that \(g(x)\) can be used to find values for \(x\) corresponding to the cube root.
If we are to start with 3, which was an arbitrary choice, the values of interest will be based on \(g(x) = 1511\) by manipulating it as follows:
We want to find \(x\) such that: \[ g(x) = 15x = 1511 \] Solving for \(x\): \[ x = \frac{1511}{15} \approx 100.7333 \]
However, we want to provide the next five x-values starting from 3. If we simply continue to increment \(x\) by 1, we get:
- \(x = 3\)
- \(x = 4\)
- \(x = 5\)
- \(x = 6\)
- \(x = 7\)
Thus, the next five x-values starting from 3 are:
4, 5, 6, 7, and 8.
But if you need to find better approximations for \(g(x) = 1511\) systematically around \(x=3\), perhaps iterating through small increments might be more effective, where the focus is to find the points that yield results closer to 1511 when placed in \(15x\).
If you require a different set or further iterations to achieve this, do let me know!
1 answer