Let's work through the problem together by analyzing the graph and solving each part:

We have two functions, $f(x)$ and $g(x)$, represented by the red and blue curves, respectively. The question involves evaluating composite functions.

(a) Find $(g \circ f)(-1)$

This is the composition of the functions, meaning we first evaluate $f(-1)$, then use that result to find $g$ at that value.

1. Find $f(-1)$: From the graph, when $x = -1$, the value of $f(x)$ (red curve) is $1$. So, $f(-1) = 1$.

2. Now, find $g(1)$: From the graph, when $x = 1$, the value of $g(x)$ (blue curve) is $-2$. Therefore, $g(1) = -2$.

Thus, $(g \circ f)(-1) = g(f(-1)) = g(1) = -2$.

(b) Find $(f \circ g)(4)$

This is also a composition, meaning we first evaluate $g(4)$, then use that result to find $f$ at that value.

1. Find $g(4)$: From the graph, when $x = 4$, the value of $g(x)$ (blue curve) is $-3$. So, $g(4) = -3$.

2. Now, find $f(-3)$: From the graph, when $x = -3$, the value of $f(x)$ (red curve) is $2$. Therefore, $f(-3) = 2$.

Thus, $(f \circ g)(4) = f(g(4)) = f(-3) = 2$.

(c) Find $(f \circ f)(4)$

This composition involves evaluating $f(4)$, then using that result to find $f$ at that value.

1. Find $f(4)$: From the graph, when $x = 4$, the value of $f(x)$ (red curve) is $5$. So, $f(4) = 5$.

2. Now, find $f(5)$: From the graph, when $x = 5$, the value of $f(x)$ (red curve) is also $5$. Therefore, $f(5) = 5$.

Thus, $(f \circ f)(4) = f(f(4)) = f(5) = 5$.

Summary of answers:

• (a) $(g \circ f)(-1) = -2$
• (b) $(f \circ g)(4) = 2$
• (c) $(f \circ f)(4) = 5$

Would you like further explanation on any of the steps or additional details?

Related Questions:

1. What is the value of $(f \circ g)(-1)$?
2. Find $(g \circ g)(2)$.
3. How does the composition of functions differ from basic function evaluation?
4. Can you explain why the graph of a function helps in determining the composition?
5. How would you solve $(f \circ g)(-4)$?

Tip:

When evaluating compositions, always work inside out—start with the innermost function first and use its result for the outer function!