The task you provided involves matching the graphs of functions with their corresponding characteristics based on the first and second derivatives of the functions.

Here is a breakdown of how to approach this:

Characteristics Interpretation:

• f'(x) < 0: The function is decreasing.
• f'(x) > 0: The function is increasing.
• f''(x) < 0: The function is concave down (the graph is curving downward).
• f''(x) > 0: The function is concave up (the graph is curving upward).

Analysis of Graphs:

1. Graph 1:

   • The graph starts from the top-left and slopes downward to the right.
   • The function is decreasing, but as it continues, the concavity appears to be upward (concave up).
   • This matches Г: $f'(x) < 0$ and $f''(x) > 0$.

2. Graph 2:

   • This graph starts at the bottom-left and increases with downward concavity.
   • The function is increasing, but concave down.
   • This matches А: $f'(x) > 0$ and $f''(x) < 0$.

3. Graph 3:

   • This graph has a flat section, starting from negative slope and curving downward.
   • The function is decreasing and concave down.
   • This matches Б: $f'(x) < 0$ and $f''(x) < 0$.

4. Graph 4:

   • This graph starts from negative slope and gradually turns upward.
   • The function is increasing and concave up.
   • This matches В: $f'(x) > 0$ and $f''(x) > 0$.

Final Matching:

• Graph 1 → Г
• Graph 2 → А
• Graph 3 → Б
• Graph 4 → В

Let me know if you need any further clarification or details! Here are some related questions to explore:

1. How do the first and second derivatives influence the shape of the graph?
2. What is the relationship between concavity and inflection points in a graph?
3. How can you determine the maximum or minimum points of a function from its derivatives?
4. How does the sign of the first derivative indicate the behavior of the graph?
5. What are the implications of a zero second derivative at a point on the graph?

Tip: Understanding the interaction between the first and second derivatives helps in sketching and analyzing the function's graph more effectively.