The functions f and g are defined by f: x ↦ 1 - 2x³, x ∈ ℝ g: x ↦ 3/x - 4, x > 0, x ∈ ℝ (a) Find the inverse function f⁻¹ - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 6

To find the composite function $gf(x)$, we first substitute $g(x)$ into $f(x)$:

1. Compute $g(x)$: $g(x) = \frac{3}{x} - 4$ for $x > 0$.

2. Substitute $g(x)$ into $f(x)$: $gf(x) = f(g(x)) = f\left(\frac{3}{x} - 4\right)$

3. Plugging in: $= 1 - 2\left(\frac{3}{x} - 4\right)^3$

4. Simplifying leads to: $gf(x) = \frac{8x^2 - 1}{1 - 2x^2}$.

To solve $gf(x) = 0$:

$\frac{8x^2 - 1}{1 - 2x^2} = 0$

We only need to solve the numerator:

$8x^2 - 1 = 0$ $8x^2 = 1$ $x^2 = \frac{1}{8}$ $x = \frac{1}{2}\sqrt{2}$.

To find the stationary point, we first need to differentiate $gf(x)$:

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

Using the quotient rule:

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

Setting the numerator equal to zero to find critical points:

1. Simplifying the numerator leads to: $18x^4 = 0$ $x = 0$.

2. Substitute back to find $y$: $gf(0) = \frac{8(0)^2 - 1}{1 - 2(0)^2} = -1$.

Thus, the coordinates of the stationary point are $(0, -1)$.