Solution:

Part (a): $(f \circ g)(x) = f(g(x))$

Substitute $g(x)$ into $f$:

$f(g(x)) = f(x^2 - 1) = 2(x^2 - 1) + 3 = 2x^2 - 2 + 3 = 2x^2 + 1$

Part (b): $(g \circ f)(x) = g(f(x))$

Substitute $f(x)$ into $g$:

$g(f(x)) = g(2x + 3) = (2x + 3)^2 - 1$

$= 4x^2 + 12x + 9 - 1 = 4x^2 + 12x + 8$

Part (c): $(f \circ g)(2) = 2(2)^2 + 1 = 2(4) + 1 = 9$

Alternatively: $g(2) = 2^2 - 1 = 3$, then $f(3) = 2(3) + 3 = 9$ ✓

Find the inverse:

Step 1: Replace $f(x)$ with $y$: $y = \frac{2x + 1}{x - 3}$

Step 2: Swap $x$ and $y$: $x = \frac{2y + 1}{y - 3}$

Step 3: Solve for $y$: $x(y - 3) = 2y + 1$ $xy - 3x = 2y + 1$

Step 4: Write as $f^{-1}(x)$: $f^{-1}(x) = \frac{3x + 1}{x - 2}$

Domains and Ranges:

For $f(x) = \frac{2x + 1}{x - 3}$:

• Domain: $x \neq 3$, or $(-\infty, 3) \cup (3, \infty)$
• Range: $y$ (horizontal asymptote), or

For $f^{-1}(x) = \frac{3x + 1}{x - 2}$:

• Domain: $x \neq 2$ (which equals the range of $f$) ✓
• Range: $y \neq 3$ (which equals the domain of ) ✓

Solution:

Let $y = f(x) = \frac{3x - 2}{x + 1}$

To find the inverse, swap $x$ and $y$, then solve for $y$:

$x = \frac{3y - 2}{y + 1}$

$x(y + 1) = 3y - 2$

$xy + x = 3y - 2$

$xy - 3y = -x - 2$

$y(x - 3) = -x - 2$

$y = \frac{-x - 2}{x - 3} = \frac{-(x + 2)}{x - 3}$

Therefore: $f^{-1}(x) = \frac{-x - 2}{x - 3}$ or

Verification: $f(f^{-1}(x)) = f\left(\frac{-x-2}{x-3}\right)$

$= \frac{3\left(\frac{-x-2}{x-3}\right) - 2}{\frac{-x-2}{x-3} + 1}$

$= \frac{\frac{-3x-6}{x-3} - 2}{\frac{-x-2}{x-3} + 1}$

$= \frac{\frac{-3x-6-2(x-3)}{x-3}}{\frac{-x-2+(x-3)}{x-3}}$

$= \frac{-3x-6-2x+6}{-x-2+x-3}$

$= \frac{-5x}{-5} = x$ ✓

Note: The domain restriction $x \geq 2$ makes $f$ one-to-one (right half of parabola).

Complete the square first: $f(x) = x^2 - 4x + 3 = (x - 2)^2 - 4 + 3 = (x - 2)^2 - 1$

With domain $x \geq 2$, the range is $y \geq -1$.

Find the inverse:

Start with $y = (x - 2)^2 - 1$

Swap variables: $x = (y - 2)^2 - 1$

Solve for $y$: $x + 1 = (y - 2)^2$ $\pm\sqrt{x + 1} = y - 2$

Since original domain is $x \geq 2$, the range is $y \geq 2$, so we need the positive square root: $f^{-1}(x) = 2 + \sqrt{x + 1}$

Domain of $f^{-1}$: $x \geq -1$ (the range of $f$)

Find $f^{-1}(8)$: $f^{-1}(8) = 2 + \sqrt{8 + 1} = 2 + \sqrt{9} = 2 + 3 = 5$

Verify: $f(5) = 5^2 - 4(5) + 3 = 25 - 20 + 3 = 8$ ✓

Answer: $f^{-1}(x) = 2 + \sqrt{x + 1}$ for , and