Let's solve problem 27, where the given functions are:

$f(x) = x^2 - 4$

$g(x) = -x^2 + 4$

We will solve the following conditions:

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

Set $f(x) = 0$:

$x^2 - 4 = 0$

Factorize:

$(x - 2)(x + 2) = 0$

So, the solutions are:

$x = \pm 2$

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

Set $g(x) = 0$:

$- x^2 + 4 = 0$

Rearrange:

$x^2 = 4$

Take the square root:

$x = \pm 2$

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

Set $f(x) = g(x)$:

$x^2 - 4 = -x^2 + 4$

Rearrange:

$x^2 + x^2 = 4 + 4$

$2x^2 = 8$

$x^2 = 4$

$x = \pm 2$

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

Solve:

$x^2 - 4 > 0$

Factor:

$(x - 2)(x + 2) > 0$

Find sign changes using test points:

• For $x < -2$, both factors are negative, so product is positive.
• For $-2 < x < 2$, one factor is negative and one is positive, so product is negative.
• For $x > 2$, both factors are positive, so product is positive.

Thus, the solution is:

$x \in (-\infty, -2) \cup (2, \infty)$

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

Solve:

$- x^2 + 4 \leq 0$

Rearrange:

$x^2 \geq 4$

Take square root:

$|x| \geq 2$

Thus, the solution is:

$x \in (-\infty, -2] \cup [2, \infty)$

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

Solve:

$x^2 - 4 > -x^2 + 4$

Rearrange:

$x^2 + x^2 > 8$

$2x^2 > 8$

$x^2 > 4$

$|x| > 2$

Thus, the solution is:

$x \in (-\infty, -2) \cup (2, \infty)$

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

Solve:

$x^2 - 4 \geq 1$

$x^2 \geq 5$

$|x| \geq \sqrt{5}$

Thus, the solution is:

$x \in (-\infty, -\sqrt{5}] \cup [\sqrt{5}, \infty)$

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Would you like more details or a graph for any of these solutions?

Related Questions

1. How do you determine intervals for quadratic inequalities?
2. What is the graphical interpretation of $f(x) = g(x)$?
3. How do you test intervals for inequality solutions?
4. Why does the function $f(x) = x^2 - 4$ have the same roots as $g(x) = -x^2 + 4$?
5. How would the solutions change if we used different functions?

Tip

For quadratic inequalities, always test points in each interval determined by the critical values!