Let's break down each part step-by-step, using the graph and the provided work for verification: --- ### (a) \( g(x) = 2f(x) \), find \( g'(-1) \): #### **Step 1:** Differentiate using the constant multiple rule: \[ g'(x) = 2f'(x) \] #### **Step 2:** Plug in \( x = -1 \): \[ g'(-1) = 2f'(-1) \] #### **Step 3:** From the graph, find \( f'(-1) \): - At \( x = -1 \), the slope of the graph (derivative) is \( 3 \). #### **Step 4:** Substitute: \[ g'(-1) = 2 \times 3 = 6 \] **Final Answer:** \(\boxed{6}\) --- ### (b) \( h(x) = e^{f(x)} \), find \( h'(1) \): #### **Step 1:** Differentiate using the chain rule: \[ h'(x) = e^{f(x)} \cdot f'(x) \] #### **Step 2:** Plug in \( x = 1 \): \[ h'(1) = e^{f(1)} \cdot f'(1) \] #### **Step 3:** From the graph: - \( f(1) = 4 \) (the \( y \)-value at \( x = 1 \)) - \( f'(1) = 0 \) (the slope at \( x = 1 \) is flat) #### **Step 4:** Substitute: \[ h'(1) = e^4 \cdot 0 = 0 \] **Final Answer:** \(\boxed{0}\) --- ### (c) \( k(x) = \frac{1}{f(x)} \), find \( k'(3) \): #### **Step 1:** Differentiate using the quotient rule or chain rule: \[ k(x) = [f(x)]^{-1} \implies k'(x) = -\frac{f'(x)}{[f(x)]^2} \] #### **Step 2:** Plug in \( x = 3 \): \[ k'(3) = -\frac{f'(3)}{[f(3)]^2} \] #### **Step 3:** From the graph: - \( f(3) = 2 \) (the \( y \)-value at \( x = 3 \)) - \( f'(3) = 0 \) (the slope at \( x = 3 \) is flat) #### **Step 4:** Substitute: \[ k'(3) = -\frac{0}{2^2} = 0 \] **Final Answer:** \(\boxed{0}\) --- ## **Summary of Answers** - (a) \(\boxed{6}\) - (b) \(\boxed{0}\) - (c) \(\boxed{0}\) All steps and final answers are verified with the graph and the solutions provided!