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) The linear interpolation polynomial $P_{1}(x)$ using the nodes $x_{0}=-1$ and $x_{1}=0$ is given by:<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 values, we get:<br /><br />$P_{1}(x) = \frac{(x - 0)}{(-1 - 0)}(-1)^{3} + \frac{(x - (-1))}{(0 - (-1))}(0)^{3} = -x - 1$<br /><br />(b) The quadratic interpolation polynomial $P_{2}(x)$ using the nodes $x_{0}=-1$, $x_{1}=0$, and $x_{2}=1$ is given by:<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 values, we get:<br /><br />$P_{2}(x) = \frac{(x - 0)(x - 1)}{(-1 - 0)(-1 - 1)}(-1)^{3} + \frac{(x - (-1))(x - 1)}{(0 - (-1))(0 - 1)}(0)^{3} + \frac{(x - (-1))(x - 0)}{(1 - (-1))(1 - 0)}(1)^{3} = x^{2} - x$<br /><br />(c) The cubic interpolation polynomial $P_{3}(x)$ using the nodes $x_{0}=-1$, $x_{1}=0$, $x_{2}=1$, and $x_{3}=2$ is given by:<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 values, we get:<br /><br />$P_{3}(x) = \frac{(x - 0)(x - 1)(x - 2)}{(-1 - 0)(-1 - 1)(-1 - 2)}(-1)^{3} + \frac{(x - (-1))(x - 1)(x - 2)}{(0 - (-1))(0 - 1)(0 - 2)}(0)^{3} + \frac{(x - (-1))(x - 0)(x - 2)}{(1 - (-1))(1 - 0)(1 - 2)}(1)^{3} + \frac{(x - (-1))(x - 0)(x - 1)}{(2 - (-1))(2 - 0)(2 - 1)}(2)^{3} = x^{3} - 3x^{2} + 3x - 1$<br /><br />(d) The linear interpolation polynomial $P_{1}(x)$ using the nodes $x_{0}=1$ and $x_{1}=2$ is given by:<br /><br />$P_{1}(x) = \frac{(x - x_{1})}{(x_{0} - x_{1})}f(x_{0}) + \frac{(x