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 m/s and an acceleration given by dot{x} = 2x^3 + 2x - HSC - SSCE Mathematics Extension 1 - Question 5 - 2004 - Paper 1

To express x in terms of t, we begin with the fact that:

$\dot{x} = x^4 + 1.$
We can set up the integral:

$\int \frac{dx}{x^4 + 1} = \int dt.$
Using partial fractions or trigonometric substitution, we compute the integral, noting that this integral can be complicated. After integrating and solving the equation:

Assuming we get a workable solution, we can express x as:

$x = f(t).$

The graph of y = f(x), where f(x) = \frac{1}{1+x^2}, is given. On the same axes, the inverse function, y = f^{-1}(x), is also drawn accurately based on the reflection property across the line y = x.

The domain of f^{-1}(x) is determined by considering the range of f(x). Since f(x) = \frac{1}{1+x^2} attains values between 0 (exclusive) and 1 (inclusive), we have:

Domain of f^{-1}(x): (0 < x \leq 1).

To find f^{-1}(x), we start from:

$y = \frac{1}{1 + x^2}.$
Interchanging x and y gives:

The point P where the graphs of y = f(x) and y = f^{-1}(x) meet indicates that at point P, the values of x and y are equal. Therefore, substituting into the equation gives us:

$x = f(x)\quad \Rightarrow \quad x = \frac{1}{1 + x^2},$
which rearranges to:

$x^2 + x - 1 = 0.$
Thus, α must be a root of this equation.

Assuming \alpha = 0.5 as the first approximation:

Using Newton's method, we begin with:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$
Calculating f(0.5) gives us:

$f(0.5) = 0.5 + (0.5)^2 - 1 = -0.25.$
Next, compute f'(x) for our function and evaluate at x = 0.5. We can substitute these values into the formula to obtain a second approximation for α.