Let's carefully evaluate the given expressions step by step. The provided functions are:

1. $h(x) = \log_3(x + 3) - 2$
2. $g(x)$ from the graph.
3. $f(x)$ from the table.

A. $(g \cdot h^{-1})(1)$

1. $g \cdot h^{-1}(1)$ means we first find $h^{-1}(1)$ (inverse of $h$ at 1) and then substitute into $g$.

Step 1: Solve $h(x) = 1$

The equation for $h(x)$ is: $\log_3(x + 3) - 2 = 1$ Add 2 to both sides: $\log_3(x + 3) = 3$ Rewriting in exponential form: $x + 3 = 3^3 = 27 \implies x = 24$

Thus, $h^{-1}(1) = 24$.

Step 2: Find $g(24)$

From the graph of $g(x)$, there is no information about $g(24)$. Without more details on $g$, we cannot determine $g(24)$.

B. $(h + f^{-1})(0)$

This involves $h(0)$ and $f^{-1}(0)$:

Step 1: Find $h(0)$

Substitute $x = 0$ into $h(x)$: $h(0) = \log_3(0 + 3) - 2 = \log_3(3) - 2$ Since $\log_3(3) = 1$, we have: $h(0) = 1 - 2 = -1$

Step 2: Find $f^{-1}(0)$

From the table, $f(x) = 0$ when $x = 3$. Therefore: $f^{-1}(0) = 3$

Step 3: Add $h(0)$ and $f^{-1}(0)$

$h(0) + f^{-1}(0) = -1 + 3 = 2$

C. $(f \circ g^{-1} \circ h)(0)$

Here we evaluate $h(0)$, then $g^{-1}$, and finally $f$.

Step 1: Find $h(0)$

From part B: $h(0) = -1$

Step 2: Find $g^{-1}(-1)$

From the graph of $g(x)$, $g(x) = -1$ when $x = 2$. Thus: $g^{-1}(-1) = 2$

Step 3: Find $f(2)$

From the table of $f(x)$, $f(2) = -1$. Therefore: $f(g^{-1}(h(0))) = f(2) = -1$

Final Answers:

A. $g \cdot h^{-1}(1)$: Undefined due to lack of information on $g(24)$.
B. $h + f^{-1}(0) = 2$
C. $f \circ g^{-1} \circ h(0) = -1$

Do you want detailed explanations for any step? Here are 5 related questions to explore:

1. How do we calculate the inverse of a logarithmic function like $h(x) = \log_3(x + 3) - 2$?
2. How can we read values of $g(x)$ and its inverse $g^{-1}(x)$ from a graph?
3. How do we interpret $f^{-1}(x)$ from a table of values?
4. What does composition of functions, such as $(f \circ g^{-1} \circ h)(x)$, mean mathematically?
5. What is the domain and range of $h(x)$ based on its equation?

Tip:

When solving for inverse functions, remember to isolate the variable step by step and rewrite logarithmic or exponential equations clearly!