Find the function h(x) by finding the composite function fg(x), where fg(x) means f(g(x)).
Write the algebraic expression for g(x) inside f(...).

$h (x) = f (g (x)) = f (x + 3)$

Find $f (x + 3)$ by replacing the x's in f(x) with the brackets $(x + 3)$ .

$\begin{array}{rcl} h (x) & = & f (x + 3) \\ = & {(x + 3)}^{2} - 2 (x + 3) \\ = & x^{2} + 6 x + 9 - 2 x - 6 \\ = & x^{2} + 4 x + 3 \end{array}$

substituting x + 3 into f(x) [1]

correct simplified expression [1]

Finding the inverse of h(x) will require completing the square

First, write the function in the form $y = h (x)$ and then swap the x and y

$y = x^{2} + 4 x + 3 x = y^{2} + 4 y + 3$

Rearrange the expression to make $y$ the subject again by completing the square
The formula for completing the square for a quadratic in the form $x^{2} + b x + c$ is ${(x + \frac{b}{2})}^{2} - {(\frac{b}{2})}^{2} + c$ .

Completing the square requires dividing the middle term, 4, by 2

$\begin{array}{rcl} y^{2} + 4 y + 3 & = & {(y + \frac{4}{2})}^{2} - {(\frac{4}{2})}^{2} + 3 \\ = & {(y + 2)}^{2} - 2^{2} + 3 \end{array}$

[1]

Simplify

$\begin{array}{rcl} {(y + 2)}^{2} - 2^{2} + 3 & = & {(y + 2)}^{2} - 4 + 3 \\ = & {(y + 2)}^{2} - 1 \end{array}$

Set equal to $x$ and rearrange to make $y$ the subject

$x = {(y + 2)}^{2} - 1$

Rearrange the expression to make $y$ the subject again

$\begin{array}{rcl} x + 1 & = & {(y + 2)}^{2} \\ y + 2 & = & \pm \sqrt{x + 1} \\ y & = & \pm \sqrt{x + 1} - 2 \\ y & = & - 2 \pm \sqrt{x + 1} \end{array}$

[1]

The domain of h(x) is $x \geq - 2$ so this must be the range of the inverse, taking the positive square root will give $y \geq - 2$
Rewrite using the correct notation for an inverse function

$h^{- 1} (x) = - 2 + \sqrt{x + 1}$ [1]

Functions (Edexcel IGCSE Maths)

Exam Questions