To solve part (a), we need to find where the derivative of the function $f(x)$, denoted $f'(x)$, equals 0. This occurs at the points where the slope of the function is horizontal (i.e., where the tangent line is flat). These points correspond to the critical points of the function, which are either maximums, minimums, or inflection points.

Looking at the description of the graph:

• A ray starts at $(3, 3)$ and passes through $(1, 2)$.
• A smooth curve begins at $(3, 3)$, falls to a minimum at $(4, 2)$, and then rises again, passing through $(6, 6)$.

Since the derivative is 0 at the minimum point of a curve (because the slope of the tangent line at this point is horizontal), the minimum occurs at $x = 4$.

Thus, the answer is:

• The x-value where $f'(x) = 0$ is $x = 4$.

Answer:

A. The $x$-value where $f'(x) = 0$ is $x = 4$.

Would you like further details, or do you have any questions?

Here are 5 related questions you might find helpful:

1. What does the derivative $f'(x)$ tell us about the shape of a graph?
2. How do we identify critical points of a function?
3. What is the significance of a minimum or maximum in a function’s graph?
4. How do we interpret the derivative in terms of increasing or decreasing behavior?
5. What happens to $f'(x)$ at inflection points?

Tip: Always look for points where the graph has horizontal tangents to find where $f'(x) = 0$.