Let's find both $f(g(x))$ and $g(f(x))$ given the functions:

• $f(x) = \ln(x^2 + 4)$
• $g(x) = 2x - 1$

Step 1: Find $f(g(x))$

To find $f(g(x))$, substitute $g(x) = 2x - 1$ into $f(x)$:

$f(g(x)) = f(2x - 1) = \ln((2x - 1)^2 + 4)$

Now, simplify the expression inside the logarithm:

$f(g(x)) = \ln((2x - 1)^2 + 4)$

This is the expression for $f(g(x))$.

Step 2: Find $g(f(x))$

To find $g(f(x))$, substitute $f(x) = \ln(x^2 + 4)$ into $g(x)$:

$g(f(x)) = g(\ln(x^2 + 4)) = 2\ln(x^2 + 4) - 1$

So,

$g(f(x)) = 2\ln(x^2 + 4) - 1$

This is the expression for $g(f(x))$.

Summary:

• $f(g(x)) = \ln((2x - 1)^2 + 4)$
• $g(f(x)) = 2\ln(x^2 + 4) - 1$

Would you like a more detailed explanation of the steps or any further questions?

Related Questions:

1. What happens to $f(g(x))$ if $g(x) = x^2$ instead?
2. How does the derivative of $f(g(x))$ differ from that of $g(f(x))$?
3. How would you find the inverse of $f(x)$?
4. What are the domain restrictions for both composite functions?
5. How can you expand $f(g(x))$ further if the logarithm is required to be in simpler terms?

Tip: Always verify the domain of composite functions to ensure both inner and outer functions are defined.