Quiz 9 - Newton’s method and antiderivatives

Problem 1

Question

State Newton’s method formula to find the roots of a function \(f(x) = 0\) then use Newton’s method to obtain an approximation for the intersection of \(y = \tan(x)\) and \(y = -1 - x\). Use \(x_0 = 0\) for the initial guess. Find \(x_1\).

Solution

Newton’s method can be written as: \[ x_{n-1} = x_n - \frac{f(x_n)}{f'(x_n)} \] Using this for the intersection we have: \[ f(x) = \tan(x) + 1 + x \] so, \[ \begin{align*} x_1 &= x_0 - \frac{f(x_0)}{f'(x_0)} \\ &= 0 - \frac{\tan(0) + 1 + 0}{\sec^2(0) + 1} \\ &= - \frac{0 + 1}{1 + 1} \\ \Aboxed{x_1 &= -\frac{1}{2}} \end{align*} \]

Problem 2

Question

Find the antiderivatives of the following:

a)

\[ f(x) = 4x^2 + 3 \]

b)

\[ f(x) = \frac{2}{3x^4} + x^2 \]

b)

\[ f(x) = 6 + \frac{4}{3}x \]

Solution

a)

\[ \begin{align*} f(x) &= 4x^2 + 3 \\ \Aboxed{F(x) &= \frac{4}{3}x^3 + 3x} \end{align*} \]

b)

\[ \begin{align*} f(x) &= \frac{2}{3x^4} + x^2 \\ \Aboxed{F(x) &= -\frac{2}{9x^3} + \frac{1}{3}x^3} \end{align*} \]

c)

\[ \begin{align*} f(x) &= 6 + \frac{4}{3}x \\ \Aboxed{F(x) &= 6x + \frac{2}{3}x^2} \end{align*} \]