Consider the following points.

(−1, 3), (0, 0), (1, 1), (4, 52)

(a) Determine the polynomial function of least degree whose graph passes through the given points.

Consider the following points.

(−2, 44), (−1, 0), (0, −14), (1, −16), (2, 0)

(a) Determine the polynomial function of least degree whose graph passes through the given points.

Respuesta :

Answer:

(a) [tex]P(x)=0.4x^{3} +2x^{2} -1.4x[/tex]

(b)[tex]P(x)=x^{4} -x^{3} +5x^{2} -7x-14[/tex]

Step-by-step explanation:

Let´s use Divided Differences Method of Polynomial Interpolation given by this iteration:

[tex]f[x_k,x_k_+_1,...,x_k_+_i]=\frac{f[x_k_+_1,x_k_+_2,...,x_k_+_i]-f[x_k,x_k_+_1,...,x_k_+i_-_1]}{x_k_+_i-x_k}[/tex]

k∈[0,n-i]

Thus the Newton polynomial can be written as:

[tex]P_n_-_1(x)=f[x_0]+f[x_0,x_1](x-x_0)+f[x_o,x_1,x_2](x-x_0)(x-x_1)+...+f[x_n,x_n_-_1,...,x_1](x-x_n)(x-x_n_-_1)...(x-x_1)[/tex]

(a) I attached you the procedure in the first table, using it we have:

[tex]P(x)=3+(-3)(x+1)+(2)(x+1)(x)+0.4(x+1)(x)(x-1)[/tex]

Operate P(x) using the distributive property:

[tex]P(x)=0.4x^{3} +2x^{2} -1.4x[/tex]

(b) I attached you the procedure in the second table, using it we have:

[tex]P(x)=44+(-44)(x+2)+(15)(x+2)(x+1)+(-3)(x+2)(x+1)(x)+(1)(x+2)(x+1)(x)(x-1)[/tex]

Operate P(x) using the distributive property:

[tex]P(x)=x^{4} -x^{3} +5x^{2} -7x-14[/tex]

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