1. Find Lagrange polynomials that approximate f(x)=x^3 (a) Find the linear interpolation polynomial P_(1)(x) using the nodes x_(0)=-1 and x_(1)=0. (b) Find the quadratic interpolation polynomial P_(2)(x) using the nodes x_(0)=-1 x_(1)=0, and x_(2)=1 (c) Find the cubic interpolation polynomial P_(3)(x) using the nodes x_(0)=-1,x_(1)=0, x_(2)=1 , and x_(3)=2. (d) Find the linear interpolation polynomial P_(1)(x) using the nodes x_(0)=1 and x_(1)=2. (e) Find the quadratic interpolation polynomial P_(2)(x) using the nodes x_(0)=0, x_(1)=1 , and x_(2)=2.

(a) To find the linear interpolation polynomial $P_{1}(x)$ using the nodes $x_{0}=-1$ and $x_{1}=0$, we can use the formula for the Lagrange polynomial:<br /><br />$P_{1}(x) = \frac{(x-x_{1})}{(x_{0}-x_{1})}f(x_{0}) + \frac{(x-x_{0})}{(x_{1}-x_{0})}f(x_{1})$<br /><br />Substituting the given values, we have:<br /><br />$P_{1}(x) = \frac{(x-0)}{(-1-0)}x^{3} + \frac{(x+1)}{(0+1)}x^{3} = x^{3} + x^{3} = 2x^{3}$<br /><br />(b) To find the quadratic interpolation polynomial $P_{2}(x)$ using the nodes $x_{0}=-1$, $x_{1}=0$, and $x_{2}=1$, we can use the formula for the Lagrange polynomial:<br /><br />$P_{2}(x) = \frac{(x-x_{1})(x-x_{2})}{(x_{0}-x_{1})(x_{0}-x_{2})}f(x_{0}) + \frac{(x-x_{0})(x-x_{2})}{(x_{1}-x_{0})(x_{1}-x_{2})}f(x_{1}) + \frac{(x-x_{0})(x-x_{1})}{(x_{2}-x_{0})(x_{2}-x_{1})}f(x_{2})$<br /><br />Substituting the given values, we have:<br /><br />$P_{2}(x) = \frac{(x-0)(x-1)}{(-1-0)(-1-1)}x^{3} + \frac{(x+1)(x-1)}{(0+1)(0-1)}x^{3} + \frac{(x+1)(x-0)}{(1+1)(1-0)}x^{3} = \frac{1}{2}x^{2} + \frac{1}{2}x^{2} + \frac{1}{2}x^{2} = x^{2}$<br /><br />(c) To find the cubic interpolation polynomial $P_{3}(x)$ using the nodes $x_{0}=-1$, $x_{1}=0$, $x_{2}=1$, and $x_{3}=2$, we can use the formula for the Lagrange polynomial:<br /><br />$P_{3}(x) = \frac{(x-x_{1})(x-x_{2})(x-x_{3})}{(x_{0}-x_{1})(x_{0}-x_{2})(x_{0}-x_{3})}f(x_{0}) + \frac{(x-x_{0})(x-x_{2})(x-x_{3})}{(x_{1}-x_{0})(x_{1}-x_{2})(x_{1}-x_{3})}f(x_{1}) + \frac{(x-x_{0})(x-x_{1})(x-x_{3})}{(x_{2}-x_{0})(x_{2}-x_{1})(x_{2}-x_{3})}f(x_{2}) + \frac{(x-x_{0})(x-x_{1})(x-x_{2})}{(x_{3}-x_{0})(x_{3}-x_{1})(x_{3}-x_{2})}f(x_{3})$<br /><br />Substituting the given values, we have:<br /><br />$P_{3}(x) = \frac{(x-0)(x-1)(x-2)}{(-1-0)(-1-1)(-1-2)}x^{3} + \frac{(x+1)(x-1)(x-2)}{(0+1)(0-1)(0-2)}x^{3} + \frac{(x+1)(x-0)(x-2)}{(1+1)(1-0)(1-2)}x^{3} + \frac{(x+1)(x-0)(x-1)}{(2+1)(2-0)(2-1)}x^{3} = \frac{1}{6}x^{3} + \frac{1}{6}x^{3} + \frac{1}{6}x^{3} + \frac{1}{6}x^{3} = x^{3}$<br /><br />(d) To find the linear interpolation polynomial $P_{1}(x)$ using the nodes $x_{0}=1$ and $x_{1}=2$, we can use the formula for the Lagrange polynomial:<br /><br />$P_{1}(x) = \frac{(x-x_{1})}{(x_{