For problem 27, we are asked to solve several inequalities and equations using the given functions $f(x) = -x^2 + 1$ and $g(x) = 4x + 1$.

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

We start by solving for $f(x) = 0$: $f(x) = -x^2 + 1 = 0$ $-x^2 + 1 = 0 \implies x^2 = 1$ $x = \pm 1$ So the solutions are $x = 1$ and $x = -1$.

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

Next, we solve for $g(x) = 0$: $g(x) = 4x + 1 = 0$ $4x = -1 \implies x = -\frac{1}{4}$ Thus, $x = -\frac{1}{4}$.

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

We now solve $f(x) = g(x)$: $-x^2 + 1 = 4x + 1$ Subtract 1 from both sides: $-x^2 = 4x$ Move all terms to one side: $x^2 + 4x = 0$ Factor the equation: $x(x + 4) = 0$ So, $x = 0$ or $x = -4$.

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

We now solve $f(x) > 0$: $-x^2 + 1 > 0$ $-x^2 > -1 \implies x^2 < 1$ So $-1 < x < 1$. The solution is $x \in (-1, 1)$.

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

Next, solve $g(x) \leq 0$: $4x + 1 \leq 0$ $4x \leq -1 \implies x \leq -\frac{1}{4}$ Thus, $x \leq -\frac{1}{4}$.

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

Finally, we solve $f(x) > g(x)$: $-x^2 + 1 > 4x + 1$ Subtract 1 from both sides: $-x^2 > 4x$ Move all terms to one side: $x^2 + 4x < 0$ Factor the inequality: $x(x + 4) < 0$ Using a sign chart or testing intervals, we find that the solution is $-4 < x < 0$.

Would you like more details on any step or a deeper explanation on a particular part?

Here are 5 related questions to expand your understanding:

1. How do we solve quadratic inequalities like $f(x) > g(x)$?
2. What are the steps for solving quadratic equations like $f(x) = 0$?
3. Why do we factor expressions like $x^2 + 4x < 0$?
4. How do sign charts help in solving inequalities?
5. How can we interpret the graphical intersection of $f(x)$ and $g(x)$?

Tip: When solving inequalities, always check the boundary points by substituting them into the inequality.