3 Newton's method 3. For this problem, we define the function g(x)=3+x-1. We want to find a number x such that g(x) = 0. In other words, we want to solve the equation 2³+x-1=0. You may use a calculator for this question. (a) Calculate g(0) and g(1). This guarantees there has to be some number 0<<1 such that g(x)=0. Why? Hint: which theorem we can imply here. (e) Now repeat the process you did in the last three steps, but starting with 22 instead of with r₁. Write the equation of the line tangent to y = g(x) at the point with x-coordinate 22. We will call it L2. Draw L₂ on the picture. Call the new value you obtain 23. Calculate g (23). Is it close enough to zero? Then obtain 24. Is g (24) close enough to 0 ? (b) We are going to make a bunch of successive guesses for the solution to the equation. None of them will be exact, but each one will be better than the previous one. Our first guess is going to be 21 = 1. Write the equation of the line tangent to y= g(x) at the point with 2-coordinate 21. Draw this line. We will call it L₁. Keep track of your data: n In 9(Zn) (e) We are looking for the point of the graph y= g(x) that intersects the r-axis. Since this point is not too far from (21,9 (21)), we can look for the point where the line L₁ intersects the x-axis instead. (Convince yourself that this makes sense!) Calculate this point. Call its 2-coordinate 22. This is our second guess. 2 3 4 5 4. Now you try the whole thing by yourself. Use a similar method to the previous problem to find a solution to 23-2-1=0.This question is for your practice and you don't need to return your work. (d) Calculuate g(x2). Notice that g (22) is not zero yet but it is closer to zero that g(x1). We are improving!