Let's solve the given problem step by step.

The function given is:

$f(x) = \frac{3x + 1}{x^2 - 1}.$

Part (a): Find the value of $f(0)$.

To calculate $f(0)$, substitute $x = 0$ into the function:

$f(0) = \frac{3(0) + 1}{0^2 - 1} = \frac{1}{-1} = -1.$

So, $f(0) = -1$.

Part (b): Find $f(t) + f(-t)$.

Substitute $t$ and $-t$ into the function.

1. For $f(t)$: $f(t) = \frac{3t + 1}{t^2 - 1}.$

2. For $f(-t)$: Substitute $x = -t$ into $f(x)$: $f(-t) = \frac{3(-t) + 1}{(-t)^2 - 1} = \frac{-3t + 1}{t^2 - 1}.$

3. Add $f(t)$ and $f(-t)$: $f(t) + f(-t) = \frac{3t + 1}{t^2 - 1} + \frac{-3t + 1}{t^2 - 1}.$

Since the denominators are the same, combine the numerators: $f(t) + f(-t) = \frac{(3t + 1) + (-3t + 1)}{t^2 - 1}.$

Simplify the numerator: $(3t + 1) + (-3t + 1) = 0 + 2 = 2.$

Thus: $f(t) + f(-t) = \frac{2}{t^2 - 1}.$

Final Answers:

(a) $f(0) = -1$.
(b) $f(t) + f(-t) = \frac{2}{t^2 - 1}$.

Would you like further explanation or a step-by-step breakdown?

Related Questions:

1. What happens to $f(x)$ as $x \to \pm \infty$?
2. For what values of $x$ is $f(x)$ undefined?
3. How would you find the derivative $f'(x)$ of the function $f(x)$?
4. Can $f(x)$ be expressed as a sum of partial fractions?
5. What is the behavior of $f(x)$ at $x = 1$ and $x = -1$ (vertical asymptotes)?

Tip:

When working with rational functions, always check for undefined points by examining the denominator for roots!