Given two functions $f(x) = x^2 - 1$ and $g(x) = 3x + 1$, we are tasked with finding:

a) $f \circ g$ (also written as $f(g(x))$):

To find $f \circ g$, we substitute $g(x)$ into $f(x)$.

$f(g(x)) = f(3x + 1)$

Now, substitute $3x + 1$ for $x$ in $f(x)$:

$f(3x + 1) = (3x + 1)^2 - 1$

First, expand $(3x + 1)^2$:

$(3x + 1)^2 = 9x^2 + 6x + 1$

Now subtract 1:

$f(g(x)) = 9x^2 + 6x + 1 - 1 = 9x^2 + 6x$

Thus,

$f \circ g = 9x^2 + 6x$

b) $g \circ f$ (also written as $g(f(x))$):

To find $g \circ f$, we substitute $f(x)$ into $g(x)$.

$g(f(x)) = g(x^2 - 1)$

Now, substitute $x^2 - 1$ for $x$ in $g(x)$:

$g(x^2 - 1) = 3(x^2 - 1) + 1$

Expand:

$g(x^2 - 1) = 3x^2 - 3 + 1 = 3x^2 - 2$

Thus,

$g \circ f = 3x^2 - 2$

Final Results:

a) $f \circ g = 9x^2 + 6x$

b) $g \circ f = 3x^2 - 2$

Would you like further details or explanations on any of the steps?

Here are 5 related questions:

1. What is the domain of $f \circ g$?
2. Can you reverse the order of function composition, i.e., does $f \circ g = g \circ f$?
3. How would you compute $f \circ f(x)$ and $g \circ g(x)$?
4. How can you graph $f \circ g$ and $g \circ f$ on the same coordinate plane?
5. What happens if $g(x)$ is not linear—how does that change $f \circ g$?

Tip: When composing functions, always substitute one function into the other carefully, ensuring all operations are applied correctly.