Trace or copy the graph of the given function f.

The x y-coordinate plane is given. The curve enters the window in the second quadrant, goes down and right, crosses the negative x-axis, changes direction in the third quadrant, goes up and right, passes through the origin, sharply changes direction in the first quadrant, goes down and right linearly, crosses the positive x-axis, and exits the window in the fourth quadrant.

Sketch the graph of f ′ on the same coordinate axes.

Based on your description, we are given a function \( f(x) \) with a curve that:

1. Enters the second quadrant,
2. Crosses the negative x-axis (meaning \( f(x) = 0 \) at some negative value of \( x \)),
3. Changes direction (suggesting a local minimum or maximum),
4. Passes through the origin (where \( f(0) = 0 \)),
5. Has a sharp change in direction in the first quadrant (possibly an inflection point or a local maximum),
6. Crosses the positive x-axis again (so \( f(x) = 0 \) at some positive value of \( x \)),
7. Finally exits the fourth quadrant with a linear behavior (possibly implying a negative slope).

To sketch the graph of the derivative \( f'(x) \), let's interpret these changes in the function:

### Key Observations for \( f'(x) \):
- **Decreasing in the second quadrant**: This suggests that \( f'(x) \) is negative in this region.
- **Crossing the negative x-axis**: The function crosses the x-axis, meaning there’s a zero at that point, implying that \( f'(x) \) is changing sign around this region (potentially turning positive or less negative).
- **Changing direction in the third quadrant**: This indicates a local minimum for \( f(x) \), so \( f'(x) = 0 \) at this turning point, with a transition from negative to positive values.
- **Passes through the origin**: Since the graph passes through the origin without changing direction drastically, \( f'(x) \) should indicate a smooth transition near \( x = 0 \).
- **Sharp change in the first quadrant**: This likely represents a local maximum for \( f(x) \), so \( f'(x) = 0 \) again at this point, with a switch from positive to negative.
- **Linear behavior and crossing the positive x-axis**: After the sharp turn, the function decreases linearly, meaning \( f'(x) \) is negative and constant after this point.

### Sketching \( f'(x) \):
1. In the second quadrant, \( f'(x) \) is negative.
2. As the curve crosses the negative x-axis and changes direction, \( f'(x) \) reaches 0 (indicating a local minimum).
3. \( f'(x) \) becomes positive in the third quadrant until it reaches a local maximum, where \( f'(x) = 0 \) again.
4. After the sharp turn in the first quadrant, \( f'(x) \) drops sharply to negative, indicating a steep decrease.
5. In the fourth quadrant, the negative linear slope suggests \( f'(x) \) remains negative and constant.

Let me know if you'd like me to sketch this! Here are some related questions to explore this idea further:

1. How do you identify critical points from the graph of a function?
2. What role do inflection points play in understanding the second derivative \( f''(x) \)?
3. How does the sign of the derivative \( f'(x) \) indicate whether the function is increasing or decreasing?
4. Can you explain how to estimate where the derivative might be zero based on the graph of \( f(x) \)?
5. What is the connection between concavity and the second derivative?

**Tip**: When sketching derivatives, always observe where the slope of the original function is zero, as those correspond to zeros in the derivative graph.

Trace or copy the graph of the given function f. The x y-coordinate plane is given. The curve enters the window in the second quadrant, goes down and right, crosses the negative x-axis, changes direction in the third quadrant, goes up and right, passes through the origin, sharply changes direction in the first quadrant, goes down and right linearly, crosses the positive x-axis, and exits the window in the fourth quadrant. Sketch the graph of f′ on the same coordinate axes.

Learn how to trace the graph of a function and sketch the derivative graph based on the function's behavior. This problem involves identifying critical points, local maxima and minima, and interpreting inflection points. Suitable for students in Grades 11-12 studying calculus.

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