Problem related to finding roots using Newton-Raphson method

I am really stuck on problems $9$ and $10$.

Here is what I tried to do:

I can use Lagrange's interpolation formula to get the approximating polynomial, however, in order to apply Newton Raphson's method to find roots, I must get exactly the function $f(x)$.Is it possible to obtain such function $f(x)$ or do I have to consider the approximating polynomial itself as $f(x)$?

Any help would be appreciated...

• I think they are asking you to do it graphically. For example, they provide $x_0$ and want you to use the graph and the Newton-Raphson iteration to find $x_1, x_2, ...$. For example, $$x_1 = x_0 - \dfrac{f(x_0)}{f'(x_0)}$$ – Moo Aug 8 '17 at 17:09
• @Moo:i can understand that.but,in order to find out the iterations ,i would need f(x) .please elaborate a bit more... – Abhishek Shrivastava Aug 8 '17 at 17:11
• What is the function value at $x_0$? Don't you know that value from the grapgh of the function they provide? – Moo Aug 8 '17 at 17:12
• To do it graphically, just sketch a tangent line at each of the iteration points, follow it to its root, then make that your new "$x_0$" and repeat. – Ian Aug 8 '17 at 17:13

• At $x_0=0$ we have $f(x_0)=-0.5$ and $f'(x_0)\approx \Delta y/\Delta x =0.5/0.6=5/6$, thus we get $$x_1 = 0 - (-0.5/(5/6))=0.6$$ Next, $f(x_1)=0$, $f'(x_1) \approx -1/0.2=-5$, leads to $$x_2 = x_1 - f(x_1) /f'(x_1) = 0.6 - 0 = 0.6\ \text{ (again)}$$ and we see that $x_3 =x_2$... – AD. Jan 5 '18 at 8:50
• Also, in the above $\Delta y/\Delta x$ are estimated using a ruler... – AD. Jan 5 '18 at 8:51