A particle is moving along the x-axis, starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 ms⁻¹ and an acceleration given by $$\dot{x} = 2x^3 + 2x.$$ (i) Show that \( \dot{x} = x^4 + 1 - HSC - SSCE Mathematics Extension 1 - Question 5 - 2004 - Paper 1

To find an expression for x in terms of t, we need to integrate ( \dot{x} = x^4 + 1 ):

1. We recognize that: $\dot{x} = \frac{dx}{dt} = x^4 + 1.$
2. Rearranging gives: $\frac{dx}{x^4 + 1} = dt.$
3. Integrating both sides leads us to: $\int \frac{dx}{x^4 + 1} = t + C'.$
4. The left side can be tackled using standard integral techniques (this might require partial fractions or a trigonometric substitution depending on familiarity) resulting in: $t = \text{integral solution from the left side} + C'.$
5. To solve for x in terms of t, substitute back the limits to find C' using initial conditions, and express x as a function of t based on the integrated result.

To sketch the inverse function y = f⁻¹(x), we note that the graph of an inverse function is a reflection of the original function across the line y = x.

1. Identify key features of f(x) to locate points of intersection with y = x.
2. For example, if the function decreases from (0, 1) to (1, 0), the inverse will increase from (1, 0) to (0, 1).
3. Use the symmetry around the line y = x to draw the inverse, ensuring that the graph respects the domain and range of the original function.

The domain of f⁻¹(x) corresponds to the range of the function f(x). Given that f(x) = ( \frac{1}{1 + x^2} ), which decreases from 1 (when x = 0) to 0 (as x approaches infinity), the domain of f⁻¹(x) is:

$\text{Domain of } f^{-1}(x): (0, 1].$

To find f⁻¹(x), we need to express x in terms of y from the equation of f(x):

1. Set y = f(x) = ( \frac{1}{1 + x^2} ).
2. Rearranging gives: $y(1 + x^2) = 1 \Rightarrow 1 + x^2 = \frac{1}{y} \Rightarrow x^2 = \frac{1}{y} - 1.$
3. Hence, $x = \sqrt{\frac{1}{y} - 1}.$
4. Thus, substituting y back gives: $f^{-1}(x) = \sqrt{\frac{1}{x} - 1}$

At point P where f(x) and f⁻¹(x) intersect:

1. By definition, at this point, y must equal x. Thus: $f(\alpha) = \alpha.$

2. Given that f(x) = ( \frac{1}{1 + x^2} ), substituting gives: $\frac{1}{1 + \alpha^2} = \alpha.$

3. Hence, this proves that α is indeed a root of the equation.

To apply Newton's method:

1. We start with the equation ( f(x) = \frac{1}{1 + x^2} - x = 0. )
2. First, the first derivative is: $$$$
3. Using the first approximation ( \alpha_1 = 0.5 ), we compute: $\alpha_2 = \alpha_1 - \frac{f(\alpha_1)}{f'(\alpha_1)}.$
4. Compute f(0.5) and f'(0.5). Substitute these values to find the new approximation ( \alpha_2. $$