The functions f and g are defined by f: x ↦ 1 - 2x², x ∈ ℝ g: x ↦ 3/(x - 4), x > 0, x ∈ ℝ (a) Find the inverse function f⁻¹ - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 6

To find the composite function gf, we substitute the function g into f:

1. Start with the expression for g: $g(x) = \frac{3}{1 - 2x²} - 4$

2. Substitute g into f: $gf(x) = f(g(x)) = f\left(\frac{3}{1 - 2x²} - 4\right)$

3. Now, first simplify g(x): $g(x) = \frac{3 - 4(1 - 2x²)}{(1 - 2x²)} = \frac{3 - 4 + 8x²}{1 - 2x²} = \frac{8x² - 1}{1 - 2x²}$

Thus, we conclude:

$gf : x \mapsto \frac{8x² - 1}{1 - 2x²}$

To solve for x where gf(x) = 0:

1. Set the equation to zero: $\frac{8x² - 1}{1 - 2x²} = 0$

2. The fraction is zero when the numerator is zero: $8x² - 1 = 0$

3. Rearranging gives: $x² = \frac{1}{8}$

4. Solving for x leads to: $x = \pm \frac{1}{2\sqrt{2}} \approx \pm 0.353$

To find the stationary point, we differentiate gf:

1. Using the quotient rule: $\frac{dy}{dx} = \frac{(1 - 2x²)(16x) - (8x² - 1)(-4x)}{(1 - 2x²)²}$

2. Setting the numerator equal to zero for critical points: $18x^2 = 0$

3. Solving gives: $x = 0$

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

Thus, the stationary point is (0, -1).