This problem is our first introduction to Newton's Method for the solution of nonlinear equations.
The positive solution of is obviously How might one calculate a numerical value of $\sqrt{2}$? The idea of Newton's method is to pick a guess $x_0$, compute the tangent to the graph of $f$ at $\big(x_0,f(x_0)\big)$, and then replace the guess $x_0$ with $x_1$ which is the $x$ intercept of the tangent. (You should draw a picture of this idea.)
Suppose $x_0 = 1.5$. The tangent to the graph of $f$ at the point $(x_0,f(x_0))$ can be written in slope intercept form as $y =$ $x +$ .
The tangent intercepts the $x$-axis at $x_1 =$ . Now let's repeat the process. The tangent to the graph of $f$ at the point $(x_1,f(x_1))$ can be written in slope intercept form as $y =$ $x +$ .
The tangent intercepts the $x$-axis at $x_2 =$ .
Note the accuracy of this approximation:
$\sqrt{2} - x_2 =$ .
Note: To obtain the same result in your last answer as ww you need to compute all intermediate answers to the full accuracy of your calculator. To accomplish this store all intermediate results in your calculator. Do not copy them on paper and then reenter them by hand later. Doing so introduces rounding errors and compromises the accuracy of your calculations and solutions. In fact, you should make a habit of this every time you use your calculator: store intermediate results and avoid having to reenter them. Most calculators keep more digits internally than they can display, so losing accuracy by copying and reentering is inevitable.

You can earn partial credit on this problem.