Question

Determine o valor de f(5) utilizando a interpolação na forma de newton para o quadro a seguir.

Answer 1

$Sabemos\ que: \\\\ f(2)=-15 \ \ \| \ \ f(7)=-8 \ \ \| \ \ f(8)=-3 \\\\ Calculando\ \Delta_1: \\\\ \frac{-8-(-15)}{7-2}= \frac{7}{5} \ \ \| \ \ \frac{-3-(-8)}{8-7}= 5 \\\\ Calculando\ \Delta_2: \\\\ \frac{5- \frac{7}{5}}{8-2}= \frac{18}{5}.\frac{1}{6}= \frac{3}{5}$

$f(x)=f(x_0)+(x-x_0).\Delta_1+(x-x_0).(x-x_1).\Delta_2 \\\\ f(x)=-15+(x-2). \frac{7}{5}+(x-2).( x-7).\frac{3}{5} \\\\ f(x)=-15+ \frac{7x}{5} - \frac{14}{5}+(x^2-9x+14). \frac{3}{5} \\\\ f(x)=(-15- \frac{14}{5})+ \frac{7x}{5}+ \frac{3x^2}{5}- \frac{27x}{5}+ \frac{42}{5} \\\\ f(x)= \frac{3x^2}{5}+( \frac{7x}{5}-\frac{27x}{5})+(-15-\frac{14}{5}+\frac{42}{5}) \\\\ f(x)= \frac{3x^2}{5}-\frac{20x}{5}-\frac{47}{5} \\\\ f(x)= \frac{3x^2-20x-47}{5}$

$Logo: \\\\ f(5)= \frac{3.5^2-20.5-47}{5} \\\\ f(5)=\frac{3.25-100-47}{5} \\\\ f(5)= \frac{75-147}{5} \\\\ f(5)= \frac{-72}{5} \\\\ f(5)=-14,4$

Answer 2

resposta correta!  -14,40