For a function f, the Newton divided difference table is...

For a function f, the Newton divided difference table is given by: x0 = 0.0 f[x0] = A x1 = 0.2 f[x1] = B f[x0,x1] = 20 x2 = 0.4 f[x2] = C f[x1,x2] = D f[x0,x1,x2] = E x3 = 0.6 f[x3] = 41 f[x2,x3] = 65 f[x1,x2,x3] = 75 f[x0,x1,x2,x3] = F First, find the values of missing entries A, B, C, D, E, F in the table and construct the third degree Newton's interpolating polynomial. Then, use the constructed third degree polynomial to approximate f(0.5). What is the resulting approximate value of f(0.5)? 33.5625 34.5 35.4375 36.375 None of the above

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00:01 Hi, in this question, this is the given table.

00:03 First we have to evaluate f of 0 .3.

00:09 So, for that purpose first we have to find f of x2, x3 which is equal to f of x3 minus f of x2 divided by x3 minus x2.

00:23 On substituting the corresponding values in this, then we get here f of x2, x3 is 120 which is equal to 54 minus c divided by 0 .6 minus 0 .4.

00:40 On further simplifying we get c equals 30.

00:44 Next we have to find f of x1, x2, x3 which is equal to f of x2, x3 minus f of x1, x2 divided by x3 minus x1.

01:01 On substituting the corresponding values then we get 187 .5 equals 120 minus t divided by 0 .6 minus 0 .2.

01:12 On further simplifying we get d equals 45.

01:17 Next we have to find f of x1, x2 which is equal to f of x2 minus f of x1 divided by x2 minus x1.

01:29 On substituting the corresponding value by using the table then we get c minus b divided by 0 .4 minus 0 .2 equals 30 minus b divided by 0 .2.

01:45 So which is equal to 30 minus b divided by 0 .2 equals 45.

02:01 On further simplifying we get b equals 21.

02:04 Next we have to evaluate f of x0, x1 which is equal to f of x1 minus f of x0 divided by x1 minus x0.

02:18 On substituting the corresponding values in this then we get 30 minus b minus a divided by 0 .2 minus 0.

02:27 Therefore 30 equals b as 21 minus a as unknown divided by 0 .2...

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