The problem asks to determine the function $g(x)$ given that:

1. $(f \circ g)(x) = 2x + 5$ and $f(x) = x + 6$
2. $(f \circ g)(x) = 1 - 4x$ and $f(x) = 2x + 13$

The composition of functions $(f \circ g)(x)$ means that $f(g(x))$ should equal the given expression for each case. Let's solve for $g(x)$ in both cases.

1. Case a

Given: $(f \circ g)(x) = 2x + 5 \quad \text{and} \quad f(x) = x + 6$

1. Substitute $g(x)$ into $f(x)$ so that $f(g(x)) = 2x + 5$. Since $f(x) = x + 6$, replace $x$ with $g(x)$: $f(g(x)) = g(x) + 6$

2. Set up the equation: $g(x) + 6 = 2x + 5$

3. Solve for $g(x)$: $g(x) = 2x + 5 - 6$ $g(x) = 2x - 1$

So, for part (a), $g(x) = 2x - 1$.

2. Case b

Given: $(f \circ g)(x) = 1 - 4x \quad \text{and} \quad f(x) = 2x + 13$

1. Substitute $g(x)$ into $f(x)$ so that $f(g(x)) = 1 - 4x$. Since $f(x) = 2x + 13$, replace $x$ with $g(x)$: $f(g(x)) = 2g(x) + 13$

2. Set up the equation: $2g(x) + 13 = 1 - 4x$

3. Solve for $g(x)$: $2g(x) = 1 - 4x - 13$ $2g(x) = -4x - 12$ $g(x) = -2x - 6$

So, for part (b), $g(x) = -2x - 6$.

Summary

• For part (a), $g(x) = 2x - 1$.
• For part (b), $g(x) = -2x - 6$.

Would you like a more detailed explanation of any of these steps?

Here are some related questions you might find useful to explore:

1. How does function composition work in general?
2. What if $f(x)$ or $g(x)$ were nonlinear functions—how would that affect the solution?
3. How can we check if our solution for $g(x)$ is correct?
4. What are the properties of composite functions?
5. Can we have multiple solutions for $g(x)$ in this type of problem?

Tip: When dealing with function compositions, rewriting $f(g(x))$ explicitly in terms of $g(x)$ can often make the problem clearer and easier to solve.