Let’s solve the given problems step by step:

1. Find $\frac{d}{dx} \big(f(g(x))\big) \big|_{x = -2}$

This requires the chain rule, which states: $\frac{d}{dx} \big(f(g(x))\big) = f'(g(x)) \cdot g'(x)$

Steps:

1. Find $g(-2)$: From the graph, when $x = -2$, $g(x) = 0$.

2. Find $g'(x)$ at $x = -2$: The derivative of $g(x)$ at $x = -2$ is the slope of the line segment of $g(x)$ near $x = -2$.
   Observing the graph:

   • The line segment from $x = -4$ to $x = 0$ has a slope: $g'(x) = \frac{3 - (-3)}{0 - (-4)} = \frac{6}{4} = \frac{3}{2}$ So, $g'(-2) = \frac{3}{2}$.

3. Find $f'(x)$ at $g(x) = 0$: From the graph, when $x = 0$, $f(x)$ is on a horizontal segment, so: $f'(0) = 1 \quad (\text{since the slope of the segment is } 1).$

4. Combine using the chain rule: $\frac{d}{dx} \big(f(g(x))\big) \big|_{x = -2} = f'(g(-2)) \cdot g'(-2) = f'(0) \cdot \frac{3}{2} = 1 \cdot \frac{3}{2} = \frac{3}{2}$

2. Find $\frac{d}{dx} \big(g(x^3)\big) \big|_{x = -2}$

This requires the chain rule, which states: $\frac{d}{dx} \big(g(x^3)\big) = g'(x^3) \cdot \frac{d}{dx}(x^3)$

Steps:

1. Find $x^3$ at $x = -2$: $x^3 = (-2)^3 = -8$

2. Find $g'(-8)$: From the graph, when $x = -8$, $g(x)$ is on a line segment with slope: $g'(x) = \frac{0 - (-4)}{-6 - (-8)} = \frac{4}{2} = 2$ So, $g'(-8) = 2$.

3. Find $\frac{d}{dx}(x^3)$: $\frac{d}{dx}(x^3) = 3x^2$

4. Combine using the chain rule: $\frac{d}{dx} \big(g(x^3)\big) \big|_{x = -2} = g'(x^3) \cdot 3x^2 = g'(-8) \cdot 3(-2)^2 = 2 \cdot 3 \cdot 4 = 24$

Final Answers:

1. $\frac{d}{dx} \big(f(g(x))\big) \big|_{x = -2} = \frac{3}{2}$
2. $\frac{d}{dx} \big(g(x^3)\big) \big|_{x = -2} = 24$

Would you like detailed explanations for any specific part?
Here are 5 related questions to practice:

1. What is the chain rule, and how does it apply to composite functions?
2. How do you compute slopes of piecewise linear functions from graphs?
3. How would you solve $\frac{d}{dx} \big(f(g(x^2))\big)$ at a given $x$?
4. What is the geometric significance of derivatives on graphs?
5. How does $g(x)$ behave as $x \to \infty$ based on the graph?

Tip: Always double-check slopes for accuracy by picking two distinct points on the graph.