1. **Stating the problem:** Given functions $f(x) = x + 3$ and $g(f(x)) = 2x + 5$, find the expressions for $f^{-1}(x)$, $g(x)$, $g(f^{-1}(x))$, and the value of $k$ such that $f(g(k)) = 3k - 4$. 2. **Find the inverse of $f(x)$:** Given $f(x) = x + 3$, to find $f^{-1}(x)$, swap $x$ and $y$ and solve for $y$: $$y = x + 3$$ $$x = y + 3$$ $$y = x - 3$$ Thus, $$f^{-1}(x) = x - 3$$. 3. **Find $g(x)$:** Given $g(f(x)) = 2x + 5$ and $f(x) = x + 3$, substitute $f(x)$: $$g(x + 3) = 2x + 5$$ Let $y = x + 3$, then $x = y - 3$: $$g(y) = 2(y - 3) + 5 = 2y - 6 + 5 = 2y - 1$$ Therefore, $$g(x) = 2x - 1$$. 4. **Find $g(f^{-1}(x))$:** Substitute $f^{-1}(x) = x - 3$ into $g(x)$: $$g(f^{-1}(x)) = g(x - 3) = 2(x - 3) - 1 = 2x - 6 - 1 = 2x - 7$$. 5. **Find $k$ such that $f(g(k)) = 3k - 4$:** First, find $f(g(k))$: $$g(k) = 2k - 1$$ $$f(g(k)) = f(2k - 1) = (2k - 1) + 3 = 2k + 2$$ Set equal to $3k - 4$: $$2k + 2 = 3k - 4$$ Solve for $k$: $$2 + 4 = 3k - 2k$$ $$6 = k$$ **Final answers:** - $$f^{-1}(x) = x - 3$$ - $$g(x) = 2x - 1$$ - $$g(f^{-1}(x)) = 2x - 7$$ - $$k = 6$$