The functions f and g are defined by f : x ↦ 1 - 2x², x ∈ ℝ g : x ↦ 3 - 4, x > 0, x ∈ ℝ (a) Find the inverse function f⭥ - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 6

To find gf(x):

Start with g(x) = 3 - 4/x. Substituting f(x) into g:

g(f(x)) = 3 - 4/(1 - 2x²).

Simplifying this gives:

g(f(x)) = \frac{3(1 - 2x²) - 4}{1 - 2x²} = \frac{8x² - 1}{1 - 2x²}.

Thus, gf : x ↦ \frac{8x² - 1}{1 - 2x²}.

To solve gf(x) = 0:

Set the numerator equal to zero:

8x² - 1 = 0 => 8x² = 1 => x² = \frac{1}{8} => x = \pm \frac{1}{2\sqrt{2}}.

Considering the domain of g, we find: \therefore x = \frac{1}{2\sqrt{2}}.

To find the stationary point, take the derivative of gf:

Using the quotient rule:

dy/dx = \frac{(1 - 2x²)(16x) - (8x² - 1)(-4x)}{(1 - 2x²)²}.

Setting the numerator to zero:

18x³ = 0 => x = 0.

Now substituting x = 0 into gf:

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

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