1. Using Newton's method of approximation approximate the cube root of 4, assume the root is close to 2, using only 2 iterations. 2. i. What is a (a).linear spline (b)a cubic spline. (6 marks) iii. ii. Construct a linear spline that interpolates the points (1,2),(2,1),(4,4) and (5,3) Study the function below and check if it satisfies conditions for a cubic spline f(x)= ) S_(1)(x)=5+4x^2+x^2+2x^3 on [0,1] S_(2)(x)=12+12(x-1)+7(x-1)^2+(x-1)^3 on [1,2] . Let f(x) be a function with a continuous second derivative on the interval [x_(0),x_(1)] Denote the corresponding function values by y_(0)=f(x_(0)) and y_(1)=f(x_(1)) . Consider the degree 1 interpolating polynomial p(x) through (x_(0),y_(0)) and (x_(1),y_(1)) a. Using the Lagrange formulation, find the interpolating polynomial p(x) at the points (x_(0),y_(0)) and (x_(1),y_(1)) b. State the Trapezoid Formula for numerical integration for the approximation of I=int _(x_(0))^x_(1)f(x)dx and use your result in a) above to show how it is derived. c. The interval [x_(0),x_(1)] can be partitioned into n intervals to give a more accurate results , show how this formula can be extend to cater for this situation. (3+7+5=15 marks)

1. Using Newton's method of approximation to approximate the cube root of 4, we start with an initial guess of 2. We will perform two iterations.<br /><br />Let $f(x) = x^3 - 4$ and $f'(x) = 3x^2$.<br /><br />Iteration 1:<br />$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{2^3 - 4}{3 \cdot 2^2} = 2 - \frac{8 - 4}{12} = 2 - \frac{4}{12} = 2 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3} \approx 1.67$<br /><br />Iteration 2:<br />$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 1.67 - \frac{1.67^3 - 4}{3 \cdot 1.67^2} = 1.67 - \frac{4.41 - 4}{3 \cdot 2.79} = 1.67 - \frac{0.41}{8.37} = 1.67 - 0.049 = 1.62$<br /><br />Therefore, the cube root of 4 is approximately 1.62.<br /><br />2. i. A linear spline is a piecewise linear function that interpolates a set of points. It consists of a set of line segments connected end-to-end, where each line segment is defined by two points.<br /><br />A cubic spline is a piecewise cubic function that interpolates a set of points. It consists of a set of cubic curves connected end-to-end, where each cubic curve is defined by four points.<br /><br />ii. To construct a linear spline that interpolates the points $(1,2)$, $(2,1)$, $(4,4)$, and $(5,3)$, we can use the following steps:<br /><br />1. Arrange the points in order of their x-coordinates: $(1,2)$, $(2,1)$, $(4,4)$, $(5,3)$.<br />2. Choose two points, say $(1,2)$ and $(2,1)$, and find the equation of the line passing through them. The equation of the line is $y = -x + 3$.<br />3. Choose another two points, say $(4,4)$ and $(5,3)$, and find the equation of the line passing through them. The equation of the line is $y = -\frac{1}{3}x + \frac{16}{3}$.<br />4. The linear spline is then given by the piecewise function:<br />$$<br />\begin{cases}<br />y = -x + 3, & \text{if } x \leq 2 \\<br />y = -\frac{1}{3}x + \frac{16}{3}, & \text{if } x > 2<br />\end{cases}<br />$$<br /><br />iii. To check if the function $f(x)$ satisfies the conditions for a cubic spline, we need to check if it is continuous and has a continuous second derivative on the interval $[0,1]$ and $[1,2]$.<br /><br />The function $f(x)$ is defined as:<br />$$<br />f(x) = \begin{cases}<br />S_1(x) = 5 + 4x^2 + x^2 + 2x^3, & \text{on } [0,1] \\<br />S_2(x) = 12 + 12(x-1) + 7(x-1)^2 + (x-1)^3, & \text{on } [1,2]<br />\end{cases}<br />$$<br /><br />To check if $f(x)$ is continuous, we need to check if the left-hand limit and right-hand limit of $f(x)$ are equal at $x=1$. We can do this by evaluating the left-hand limit and right-hand limit of $f(x)$ as $x$ approaches 1.<br /><br />Left-hand limit:<br />$$<br />\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} S_1(x) = \lim_{x \to 1^-} (5 + 4x^2 + x^2 + 2x^3) = 5 + 4(1)^2 + (1)^2 + 2(1)^3 = 12<br />$$<br /><br />Right-hand limit:<br />$$<br />\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} S_2(x) = \lim_{x \to 1^+} (12 + 12(x-1) +