https://file.aiquickdraw.com/mathbot/file/1728259196552riaxcie3.jpg

Let's work through the problem step by step, using the graph of the functions \( f(x) \) (the blue curve) and \( g(x) \) (the red line) to answer the parts (a) through (f). I'll extract the values of \( f(x) \) and \( g(x) \) from the graph at the required points.

### Values from the graph:
- \( f(2) = 4 \), \( g(2) = 2 \)
- \( f(4) = 0 \), \( g(4) = 4 \)
- \( f(6) = 2 \), \( g(6) = 6 \)

### Now, let's solve each part:

(a) \( (f + g)(2) \)
\[
(f + g)(2) = f(2) + g(2) = 4 + 2 = 6
\]

(b) \( (f + g)(4) \)
\[
(f + g)(4) = f(4) + g(4) = 0 + 4 = 4
\]

(c) \( (f - g)(6) \)
\[
(f - g)(6) = f(6) - g(6) = 2 - 6 = -4
\]

(d) \( (g - f)(6) \)
\[
(g - f)(6) = g(6) - f(6) = 6 - 2 = 4
\]

(e) \( (f \cdot g)(2) \)
\[
(f \cdot g)(2) = f(2) \cdot g(2) = 4 \cdot 2 = 8
\]

(f) \( \left( \frac{f}{g} \right)(4) \)
\[
\left( \frac{f}{g} \right)(4) = \frac{f(4)}{g(4)} = \frac{0}{4} = 0
\]

### Answers:
- (a) 6
- (b) 4
- (c) -4
- (d) 4
- (e) 8
- (f) 0

Would you like any additional details or explanations? 

Here are 5 related questions to deepen your understanding:
1. How would the values change if the graph were shifted vertically?
2. What does the product \( (f \cdot g)(x) \) represent graphically?
3. How would you calculate \( f(x) + g(x) \) at any arbitrary point?
4. How does the graph of \( f(x) \) differ from \( g(x) \) in terms of its general behavior?
5. Could there be points where \( f(x) = g(x) \) based on the graph?

**Tip:** Always check the graph carefully to ensure correct values at specific points—this ensures the accuracy of your calculations!

The graph of two functions, f and g, is illustrated. Use the graph to answer parts (a) through (f), which involve evaluating expressions such as (f + g)(2), (f - g)(6), and more.

Learn how to evaluate expressions like (f + g)(2), (f - g)(6), and more using the graph of functions f(x) and g(x). This problem covers core concepts in algebra and function evaluation, making it suitable for high school students in Grades 9-12.

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