The function $f$ is defined by $$f(x) = \frac{8x + 5}{2x + 3} \text{ for } x \neq -\frac{3}{2}$$ (a) Find $f^{-1}(\frac{3}{2})$ - Edexcel - A-Level Maths Pure - Question 12 - 2022 - Paper 2

To express $f(x)$ in the form $A + \frac{B}{2x + 3}$, we start again with:

$f(x) = \frac{8x + 5}{2x + 3}$

We can attempt to divide the numerator by the denominator:

Let’s rewrite $8x + 5$:

$8x + 5 = 4(2x + 3) - 7$

Thus, we substitute this back into the function:

$f(x) = \frac{4(2x + 3) - 7}{2x + 3}$

This simplifies to:

$f(x) = 4 - \frac{7}{2x + 3}$

Here we can identify:

$A = 4$ $B = -7$

Thus, we have shown that:

$f(x) = 4 + \frac{-7}{2x + 3}$

To find the range of $g^{-1}(x)$, we first compute the range of $g(r)$ where:

$g(r) = 16 - r^2$

For the domain $0 \leq r \leq 4$, the minimum occurs at $r = 4$:

$g(4) = 16 - 4^2 = 0$

And the maximum occurs at $r = 0$:

$g(0) = 16 - 0^2 = 16$

Thus, the range of $g(r)$ is:

$0 \leq g(r) \leq 16$

Therefore, the range of $g^{-1}(x)$ is (reversing the limits):

$0 \leq g^{-1}(x) \leq 4$

We already have that $0 \leq r \leq 4$. To find the range of $f(g^{-1}(r))$, we evaluate:

$g^{-1}(r) = \sqrt{16 - r}$

Then substituting into $f(x)$ gives us:

$f(g^{-1}(r)) = f(\sqrt{16 - r})$

Using the earlier found form for $f(x)$:

$f(x) = 4 - \frac{7}{2x + 3}$

We can substitute:

$f(g^{-1}(r)) = 4 - \frac{7}{2(\sqrt{16 - r}) + 3}$

Now as $r$ varies from $0$ to $16$, we observe:

• When $r = 0$, we have:

$f(g^{-1}(0)) = f(4) = \frac{8(4) + 5}{2(4) + 3} = \frac{37}{11}$

• When $r = 16$, we find:

$f(g^{-1}(4)) = f(0) = \frac{5}{3}$

Thus, the range of $f(g^{-1}(r))$ will be:

$\left[\frac{5}{3}, \frac{37}{11}\right]$