To answer these parts using the given functions $f(x) = x^2 - 4$ and $g(x) = x + 8$, let’s break down each one step by step.

(a) Solve $f(x) = 0$

We are asked to solve $x^2 - 4 = 0$.

1. Factor the quadratic equation: $(x - 2)(x + 2) = 0$
2. Set each factor equal to zero: $x - 2 = 0 \quad \text{or} \quad x + 2 = 0$ $x = 2 \quad \text{or} \quad x = -2$ Thus, the solutions are $x = 2$ and $x = -2$.

(d) Solve $f(x) > 0$

We are asked to solve $x^2 - 4 > 0$.

1. Factor the inequality: $(x - 2)(x + 2) > 0$
2. The critical points are $x = 2$ and $x = -2$. We test intervals:
   • For $x < -2$, choose $x = -3$: $(-3)^2 - 4 = 9 - 4 = 5$, which is positive.
   • For $-2 < x < 2$, choose $x = 0$: $(0)^2 - 4 = -4$, which is negative.
   • For $x > 2$, choose $x = 3$: $(3)^2 - 4 = 9 - 4 = 5$, which is positive.

Therefore, $f(x) > 0$ when $x < -2$ or $x > 2$. In interval notation, the solution is: $x \in (-\infty, -2) \cup (2, \infty)$

(g) Solve $f(x) \geq 1$

We are asked to solve $x^2 - 4 \geq 1$.

1. Add 4 to both sides: $x^2 \geq 5$
2. Solve for $x$: $x \geq \sqrt{5} \quad \text{or} \quad x \leq -\sqrt{5}$ Thus, the solution is $x \in (-\infty, -\sqrt{5}] \cup [\sqrt{5}, \infty)$.

(b) Solve $g(x) = 0$

We are asked to solve $x + 8 = 0$.

1. Subtract 8 from both sides: $x = -8$ Thus, the solution is $x = -8$.

(e) Solve $g(x) \leq 0$

We are asked to solve $x + 8 \leq 0$.

1. Subtract 8 from both sides: $x \leq -8$ Thus, the solution is $x \leq -8$.

(c) Solve $f(x) = g(x)$

We are asked to solve $x^2 - 4 = x + 8$.

1. Rearrange the equation: $x^2 - x - 12 = 0$
2. Factor the quadratic equation: $(x - 4)(x + 3) = 0$
3. Set each factor equal to zero: $x = 4 \quad \text{or} \quad x = -3$ Thus, the solutions are $x = 4$ and $x = -3$.

(f) Solve $f(x) > g(x)$

We are asked to solve $x^2 - 4 > x + 8$.

1. Rearrange the inequality: $x^2 - x - 12 > 0$
2. Factor the quadratic inequality: $(x - 4)(x + 3) > 0$
3. Test intervals:
   • For $x < -3$, choose $x = -4$: $(-4 - 4)(-4 + 3) = (-8)(-1) > 0$.
   • For $-3 < x < 4$, choose $x = 0$: $(0 - 4)(0 + 3) = (-4)(3) < 0$.
   • For $x > 4$, choose $x = 5$: $(5 - 4)(5 + 3) = (1)(8) > 0$.

Thus, $f(x) > g(x)$ when $x < -3$ or $x > 4$. In interval notation, the solution is: $x \in (-\infty, -3) \cup (4, \infty)$

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Would you like more details on any part, or do you have any questions? Here are some related problems to explore:

1. Solve $f(x) = 1$ for $f(x) = x^2 - 4$.
2. Find the range of $g(x) = x + 8$.
3. Solve $f(x) \leq 0$.
4. Determine when $f(x) = g(x) + 5$.
5. Analyze the graph of $f(x)$ and $g(x)$ to visualize the solutions.

Tip: When solving inequalities involving quadratics, always test intervals based on critical points found by solving the equation first.