The functions $f$ and $g$ are defined by $$egin{align*} f: & x ightarrow 1 - 2x^3, \, x \, ext{in } \, extbf{R} \\ g: & x ightarrow \frac{3}{x} - 4, \, x > 0, \, x \, \text{in } \, extbf{R} \\ \\ ext{(a) Find the inverse function } f^{-1} - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 6

To find the composite function $gf(x)$, we substitute $f(x)$ into $g(x)$. First, we start with:

$f(x) = 1 - 2x^3$

Substituting $f(x)$ into $g(x)$ gives:

$g(f(x)) = g(1 - 2x^3) = \frac{3}{1 - 2x^3} - 4$

This simplifies to:

$g(f(x)) = \frac{3 - 4(1 - 2x^3)}{1 - 2x^3} = \frac{3 - 4 + 8x^3}{1 - 2x^3} = \frac{8x^3 - 1}{1 - 2x^3}$

Hence, we confirm that:

$gf: x \rightarrow \frac{8x^2 - 1}{1 - 2x^2}$.

To solve for $gf(x) = 0$, we set the numerator equal to zero:

$8x^2 - 1 = 0$

Solving for $x$, we get:

$8x^2 = 1$

$x^2 = \frac{1}{8}$

$x = \pm \frac{1}{2\sqrt{2}}$

Since the function $g$ is defined for $x > 0$, we take:

$x = \frac{1}{2\sqrt{2}}$.

To find stationary points, we differentiate $gf(x)$:

$\frac{dy}{dx} = \frac{(1 - 2x^2)(16x) - (8x^2 - 1)(-4x)}{(1 - 2x^2)^2}$

Setting the numerator equal to zero for stationary points gives:

$18x^2 = 0$

Thus, $x = 0$. Now substituting back to find $y$:

$gf(0) = \frac{8(0)^2 - 1}{1 - 2(0)^2} = -1$

Therefore, the coordinates of the stationary point are:

$(0, -1)$.