Answer only the parts written in blue in the image below PROBLEM FOUR(a) Find the value of x when y = 0.3 by applying Lagrange's formula inversely.0.40.60.8y 0.3683 | 0.3332 | 0.2897(b) The following table gives the value of the elliptical integraldeF(6) = , 1- }sin²efor certain values of 0.Find the values of 0 if F(0) =0.3887250230F(0) | 0.3706 | 0.4068| 0.4433 |210(c) Use Lagrange multipliers to obtain the interpolating polynomial which passes through allthe following points.-1 0 |1| 2-1 |0| 15f(x) 0Hence use your polynomial to approximate f'(x), f" (x), f"(x), f'(1), f"(1), f" (1).(d) Using Lagrange's formula, find a polynomial which passes through the points(0, – 12), (1,0), (3,6), (4, 12)Mathematics can smile!(e) Using Lagrange's interpolation formula, find the value of y corresponding to x= 10 fromthe following table:511y= f(x) | 12 | 131416Sxy2(f) Given that u = , &r,& and & denote the errors in x,y and z respectively such thatx= y=z=1 and & = &, = & =0.001, then find the relative maximum error in u.(g) Show that the relative error in quotient of two quantities x and y may be expressed asdifference between the relative ernand y.Sxy2(h) Given that u= , Ɛ, ɛ, and ɛ denote the errors in x,y and z respectively such thatx=y=z=2 and ɛ = ɛy = & = 0.003, then find the relative maximum error in u.(i) The number x= 37.46235 is rounded off to 4 significant figures.Compute the percentage error in x.(j) Obtain an interpolating polynomial using Newton Backward Difference Formula for1 20 15-1f(x)-1

Name: Answer only the parts written in blue in the image below PROBLEM FOUR(
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Description: Answer only the parts written in blue in the image below PROBLEM FOUR(a) Find the value of x when y = 0.3 by applying Lagrange's formula inversely.0.40.60.8y 0.3683 | 0.3332 | 0.2897(b) The following table gives the value of the elliptical integralde