Given the functions f(x)=x2−2x−3 and g(x)=x−3, where x ∈ R. Determine the value(s) of x, for which: (a) f(x) > g(x) (b) f(x).g(x) 0

$$x^2 - 2x - x - 3 + 3 > 0$$ $$x^2 - 3x > 0$$

$$x = 0 \quad \text{or} \quad x - 3 = 0 \implies x = 3$$

The critical points are $x = 0$ and $x = 3$.

We test values in the intervals $(-\infty, 0)$, $(0, 3)$, and $(3, \infty)$.

• For $x < 0$ (e.g., $x = -1$): $(-1)(-1 - 3) = (-1)(-4) = 4 > 0$. This interval is a solution.
• For $0 < x < 3$ (e.g., $x = 1$): $(1)(1 - 3) = (1)(-2) = -2 < 0$. This interval is not a solution.
• For $x > 3$ (e.g., $x = 4$): $(4)(4 - 3) = (4)(1) = 4 > 0$. This interval is a solution.

The solution is $x < 0$ or $x > 3$.

$$\boxed{x < 0 \quad \text{or} \quad x > 3}$$

$$(x - 3)(x + 1)(x - 3) \leq 0$$ $$(x - 3)^2 (x + 1) \leq 0$$

$$x - 3 = 0 \implies x = 3$$ $$x + 1 = 0 \implies x = -1$$

The critical points are $x = -1$ and $x = 3$.

We consider the expression $E = (x - 3)^2 (x + 1)$.
Since $(x - 3)^2$ is always non-negative (i.e., $(x - 3)^2 \geq 0$), the sign of $E$ depends on the sign of $(x + 1)$, unless $(x-3)^2 = 0$.

• If $x = 3$, then $E = (3 - 3)^2 (3 + 1) = 0 \cdot 4 = 0$. So $x = 3$ is a solution.
• If $x \neq 3$, then $(x - 3)^2 > 0$. For $E \leq 0$, we must have $(x + 1) \leq 0$.

$$x + 1 \leq 0 \implies x \leq -1$$

Combining these, the solution is $x \leq -1$ or $x = 3$.

The solution is $x \leq -1$ or $x = 3$.

$$\boxed{x \leq -1 \quad \text{or} \quad x = 3}$$