The function $f$ is defined by $$f(x) = \frac{2x + 3}{x - 2} \quad x \in \mathbb{R}, x \neq 2$$ Find $f^{-1}$ - AQA - A-Level Maths Pure - Question 13 - 2020 - Paper 1

To find $ff(y)$, substitute $f(y)$ into itself:

1. Start with $f(y)$:
   $f(y) = \frac{2y + 3}{y - 2}$
2. Substitute $f(y)$ into $f(x)$ to get:
   $ff(y) = f\left(\frac{2y + 3}{y - 2}\right)$
3. Solve accordingly to rewrite this expression.

To find $fg(y)$, first compute $f(g(y))$:

1. Start with $g(y)$:
   $g(y) = \frac{2y^2 - 5y}{2}$
2. Substitute $g(y)$ into $f$:
   $fg(y) = f(g(y)) = f\left(\frac{2y^2 - 5y}{2}\right)$
3. Compute this expression accordingly.

To find the range of $g(x)$:

1. Locate the vertex of the quadratic $g(x) = \frac{2x^2 - 5x}{2}$ within the interval $0 \leq x \leq 4$.
2. Calculate the maximum and minimum values by evaluating $g(0)$ and $g(4)$:
   $g(0) = 0, \quad g(4) = 6$
3. Use the vertex formula $x = -\frac{b}{2a}$ to find:
   $x = \frac{5}{4}$
4. Determine $g(1.25)$:
   $g(1.25) = -1.5625$
5. Therefore, the range is:
   $\{y: -1.5625 \leq y \leq 6\}$.

To identify if $g$ has an inverse:

1. Investigate whether $g$ is one-to-one.
2. Confirm this by checking if $g(0) = g(2.5)$.
3. Since $g(0) = 0$ and $g(2.5) = 0$ indicate that $g$ is not one-to-one.
4. Therefore, $g$ does not have an inverse.

To prove that
$gf(x) = \frac{48 + 29x - 2x^2}{2x^2 - 8x + 8}$

1. Substitute $f(x)$ into $g(x)$:
   $gf(x) = g(f(x))$
2. Solve and reduce the fractions until obtaining the required result.

To find $a$:

1. Identify the conditions where $fg(x)$ is undefined, specifically where the denominator equals 0.
2. Set $2x^2 - 5x - 4 = 0$ and apply the quadratic formula:
   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
3. Consequently, solve for $x$ to determine $a$:
   $a = \frac{5 \pm \sqrt{(5)^2 - 4(2)(-4)}}{2(2)} = \frac{5 \pm \sqrt{41}}{4}$