Let's work through the questions based on the function definitions given in the image:

Function $f(x) = x^2 - 3x$:

(i) Find the set of values of $x$ for which $f(x) > 4$.

To solve $x^2 - 3x > 4$:

1. Rearrange the inequality: $x^2 - 3x - 4 > 0$
2. Factorize the quadratic expression: $(x - 4)(x + 1) > 0$
3. Solve for the critical points: $x = 4 \quad \text{or} \quad x = -1$
4. Now test intervals to find where the inequality holds:
   • For $x < -1$, say $x = -2$, the expression $(x - 4)(x + 1)$ becomes negative.
   • For $-1 < x < 4$, say $x = 0$, the expression is also negative.
   • For $x > 4$, say $x = 5$, the expression is positive.

Thus, the solution for $f(x) > 4$ is: $x \in (-\infty, -1) \cup (4, \infty)$

(ii) Express $f(x)$ in the form $(x - a)^2 - b$, stating the values of $a$ and $b$.

To complete the square for $f(x) = x^2 - 3x$:

1. Take half the coefficient of $x$ (which is $-3$), divide by 2, and square it: $\left(\frac{-3}{2}\right)^2 = \frac{9}{4}$
2. Rewrite the expression: $f(x) = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4}$ So, $a = \frac{3}{2}$ and $b = \frac{9}{4}$.

(iii) Write down the range of $f$.

Since $f(x) = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4}$, the minimum value of the squared term is 0 (which occurs at $x = \frac{3}{2}$).

Thus, the minimum value of $f(x)$ is: $f\left(\frac{3}{2}\right) = -\frac{9}{4}$ Therefore, the range of $f$ is: $f(x) \in \left[-\frac{9}{4}, \infty\right)$

(iv) State, with a reason, whether $f$ has an inverse.

For a function to have an inverse, it must be bijective (both one-to-one and onto). However, the function $f(x) = x^2 - 3x$ is a quadratic, and quadratics are not one-to-one because they fail the horizontal line test (they are symmetric).

Thus, $f(x)$ does not have an inverse because it is not one-to-one.

Function $g(x) = x - 3\sqrt{x}$ for $x \geq 0$:

(v) Solve the equation $g(x) = 10$.

To solve $x - 3\sqrt{x} = 10$, let $y = \sqrt{x}$, so $x = y^2$. Substituting into the equation:

$y^2 - 3y = 10$ Rearrange the equation: $y^2 - 3y - 10 = 0$ Solve the quadratic equation: $y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)} = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm \sqrt{49}}{2} = \frac{3 \pm 7}{2}$ Thus, $y = 5$ or $y = -2$. Since $y = \sqrt{x}$, $y \geq 0$, so we discard $y = -2$.

Therefore, $y = 5$, which gives $x = 5^2 = 25$.

Thus, the solution is: $x = 25$

Do you want further details or have any questions?

Here are 5 related questions for you:

1. How do we determine the domain of a function based on its equation?
2. What are the conditions for a function to have an inverse?
3. How can we solve inequalities involving quadratic expressions?
4. Can completing the square help in solving quadratic equations? How?
5. How do we find the critical points of a function using derivatives?

Tip:

When solving quadratic inequalities, always test the intervals between the roots to determine the sign of the expression.