5(8)(5 points) Extra Problem.Sometimes, it is important to solve the equation f(x)=0 given some function f, like the one in the figure below (blue graph). However, some equations simply don't have a solution in a "closed form", for example if f(x)=exx, the equation exx=0 is unsolvable. So we need to approximate the solution. One of the best methods for solving such equations (approximating the solution) is Newton-Raphson method. We choose a point x0(initial guess for the root of the equation f(x)=0) and compute f(x0). Then we draw a tangent line at (x0,f(x0)) and we claim that the point where that tangent line intersects the x-axis (the x-intercept) is a better approximation for the solution of f(x)=0. We'll call this point x1. We continue the process by computing the tangent line at x1 and thus getting closer and closer to the root, hopefully.Let's make it more mathematical. After choosing x0, the slope of the tangent line isf'(x0)=f(x0)-0x0-x1So we can compute x1 :x1=x0-f(x0)f'(x0)and in generalxn1=xn-f(xn)f'(xn)(a) Let f(x)=exx, compute f'(x)(b) Using x0=-1, compute x1 using the formula above(c) In a similar fashion, compute x2,x3,x4. The more computations you make the more precise your solution is.(d) Compute f(x4) and confirm that f(x4)~~0(1 point) Use implict differerfiation to find an equation of the tangent line to the carve 2xy33xy=40 at the poirt (8,1)The equation defines the tangent line to the curve at the point (8,1)