Problem Breakdown:

Definitions:

• $f(x) = 5x - 2$ for $x > 1$
• $g(x) = 4x^2 - 9$ for $x > 0$

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Part (i) State the range of $g(x)$:

$g(x) = 4x^2 - 9$, for $x > 0$:

1. The smallest value of $x$ is $x = 0^+$, where $g(x) \to -9$ (approaches $-9$ as $x \to 0^+$).
2. As $x \to \infty$, $g(x) \to \infty$.

Thus, the range of $g(x)$ is: $g(x) > -9, \quad \text{or } \ (-9, \infty).$

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Part (ii) Find the domain of $gf(x)$:

The composition $gf(x) = g(f(x)) = 4(f(x))^2 - 9$.

1. Domain of $f(x)$: $x > 1$.
2. $f(x) = 5x - 2$: Since $x > 1$, $f(x) > 5(1) - 2 = 3$, so $f(x) > 3$.
3. Domain of $g(f(x))$: $g(x)$ is defined for $x > 0$. For $gf(x)$, this requires $f(x) > 0$, which is already satisfied because $f(x) > 3$.

Thus, the domain of $gf(x)$ is: $x > 1.$

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Part (iii) Find the exact solutions of $gf(x) = 4$:

1. Substitute $gf(x) = g(f(x)) = 4(f(x))^2 - 9$: $4(f(x))^2 - 9 = 4.$

2. Simplify: $4(f(x))^2 = 13 \quad \Rightarrow \quad (f(x))^2 = \frac{13}{4}.$

3. Solve for $f(x)$: $f(x) = \pm \sqrt{\frac{13}{4}} = \pm \frac{\sqrt{13}}{2}.$

4. Check which solutions are valid:

   • $f(x) = 5x - 2 > 3$, so only $f(x) = \frac{\sqrt{13}}{2}$ is valid.

5. Solve $5x - 2 = \frac{\sqrt{13}}{2}$: $5x = 2 + \frac{\sqrt{13}}{2} \quad \Rightarrow \quad x = \frac{2 + \frac{\sqrt{13}}{2}}{5}.$ Simplify: $x = \frac{4 + \sqrt{13}}{10}.$

Thus, the exact solution is: $x = \frac{4 + \sqrt{13}}{10}.$

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Recap of Answers:

1. Range of $g(x)$: $(-9, \infty)$.
2. Domain of $gf(x)$: $x > 1$.
3. Exact solution of $gf(x) = 4$: $x = \frac{4 + \sqrt{13}}{10}$.

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Do you want further clarifications? 😊
Here are 5 related questions for exploration:

1. What is the range of $gf(x)$ based on this problem?
2. How does the domain of $g(x)$ influence compositions like $gf(x)$?
3. What would happen if $g(x)$ had a domain of $x \geq 0$ instead of $x > 0$?
4. How would $gf(x)$ behave if $f(x)$ were a quadratic function?
5. What are the graphical interpretations of the functions $f(x)$, $g(x)$, and $gf(x)$?

Tip: Always check the compatibility of domains and ranges when composing functions. It ensures the result is well-defined.