Let's clarify how the **Mean Value Theorem (MVT)** applies here: ### **Mean Value Theorem (MVT) Recap** If a function \( f \) is: - **Continuous** on a closed interval \([a, b]\) - **Differentiable** on the open interval \((a, b)\) Then **there is at least one \( c \) in \((a, b)\) such that:** \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] --- ### **Applying to the Graph** - Here, \( a = 0 \) and \( b = 12 \). - The **secant line** from \( (0, f(0)) \) to \( (12, f(12)) \) has slope: \[ m_{sec} = \frac{f(12) - f(0)}{12 - 0} \] (This is the slope of the long dashed line connecting the endpoints.) #### **What does MVT mean here?** You are looking for all \( x \) in \( (0, 12) \) where the **tangent** to the curve is **parallel** to this secant line — that is, where the instantaneous slope equals the average rate of change. #### **From the Graph** - The dashed lines tangent to the curve at certain points are shown. - The points where these dashed tangents are **parallel to the endpoint secant** are the solutions. #### **Reading the graph:** - The graph shows dashed tangent lines at \( x \approx 2 \) and \( x \approx 9 \) that appear parallel to the secant from \( (0, f(0)) \) to \( (12, f(12)) \). - Therefore, **the values of \( x \) that satisfy the MVT are**: \[ x \approx 2, \quad x \approx 9 \] (You may also include any other point in \( (0, 12) \) where a tangent is parallel to the endpoint secant.) --- ## **Summary** **All values of \( x \) in \( (0, 12) \) where the tangent line is parallel to the secant line from \( (0, f(0)) \) to \( (12, f(12)) \) satisfy the Mean Value Theorem. According to the graph, these are approximately \( x = 2 \) and \( x = 9 \).** Let me know if you'd like a more detailed explanation or algebraic setup!