Answer :
Sure, let's proceed step by step with the given functions [tex]\(f(x) = \sqrt{x}\)[/tex] and [tex]\(g(x) = x + 3\)[/tex].
### Part a: [tex]\((f \circ g)(x)\)[/tex]
The composition [tex]\((f \circ g)(x)\)[/tex] means applying [tex]\(g(x)\)[/tex] first and then applying [tex]\(f\)[/tex] to the result.
1. First, find [tex]\(g(x)\)[/tex]:
[tex]\[ g(x) = x + 3 \][/tex]
2. Now, apply [tex]\(f\)[/tex] to the result of [tex]\(g(x)\)[/tex]:
[tex]\[ f(g(x)) = f(x + 3) \][/tex]
Since [tex]\(f(x) = \sqrt{x}\)[/tex],
[tex]\[ f(x + 3) = \sqrt{x + 3} \][/tex]
So, [tex]\((f \circ g)(x) = \sqrt{x + 3}\)[/tex].
### Part b: [tex]\((g \circ f)(x)\)[/tex]
The composition [tex]\((g \circ f)(x)\)[/tex] means applying [tex]\(f(x)\)[/tex] first and then applying [tex]\(g\)[/tex] to the result.
1. First, find [tex]\(f(x)\)[/tex]:
[tex]\[ f(x) = \sqrt{x} \][/tex]
2. Now, apply [tex]\(g\)[/tex] to the result of [tex]\(f(x)\)[/tex]:
[tex]\[ g(f(x)) = g(\sqrt{x}) \][/tex]
Since [tex]\(g(x) = x + 3\)[/tex],
[tex]\[ g(\sqrt{x}) = \sqrt{x} + 3 \][/tex]
So, [tex]\((g \circ f)(x) = \sqrt{x} + 3\)[/tex].
### Part c: [tex]\((f \circ g)(6)\)[/tex]
To find [tex]\((f \circ g)(6)\)[/tex], substitute [tex]\(x = 6\)[/tex] into the composition [tex]\((f \circ g)(x)\)[/tex]:
[tex]\[ (f \circ g)(6) = \sqrt{6 + 3} = \sqrt{9} = 3 \][/tex]
### Part d: [tex]\((g \circ f)(6)\)[/tex]
To find [tex]\((g \circ f)(6)\)[/tex], substitute [tex]\(x = 6\)[/tex] into the composition [tex]\((g \circ f)(x)\)[/tex]:
[tex]\[ (g \circ f)(6) = \sqrt{6} + 3 \][/tex]
Since [tex]\(\sqrt{6}\)[/tex] is an irrational number, the expression remains as:
[tex]\[ (g \circ f)(6) = \sqrt{6} + 3 \][/tex]
### Summary of the Solutions
- [tex]\((f \circ g)(x) = \sqrt{x + 3}\)[/tex]
- [tex]\((g \circ f)(x) = \sqrt{x} + 3\)[/tex]
- [tex]\((f \circ g)(6) = 3\)[/tex]
- [tex]\((g \circ f)(6) = \sqrt{6} + 3\)[/tex]
These are the final values for each part of the problem.
### Part a: [tex]\((f \circ g)(x)\)[/tex]
The composition [tex]\((f \circ g)(x)\)[/tex] means applying [tex]\(g(x)\)[/tex] first and then applying [tex]\(f\)[/tex] to the result.
1. First, find [tex]\(g(x)\)[/tex]:
[tex]\[ g(x) = x + 3 \][/tex]
2. Now, apply [tex]\(f\)[/tex] to the result of [tex]\(g(x)\)[/tex]:
[tex]\[ f(g(x)) = f(x + 3) \][/tex]
Since [tex]\(f(x) = \sqrt{x}\)[/tex],
[tex]\[ f(x + 3) = \sqrt{x + 3} \][/tex]
So, [tex]\((f \circ g)(x) = \sqrt{x + 3}\)[/tex].
### Part b: [tex]\((g \circ f)(x)\)[/tex]
The composition [tex]\((g \circ f)(x)\)[/tex] means applying [tex]\(f(x)\)[/tex] first and then applying [tex]\(g\)[/tex] to the result.
1. First, find [tex]\(f(x)\)[/tex]:
[tex]\[ f(x) = \sqrt{x} \][/tex]
2. Now, apply [tex]\(g\)[/tex] to the result of [tex]\(f(x)\)[/tex]:
[tex]\[ g(f(x)) = g(\sqrt{x}) \][/tex]
Since [tex]\(g(x) = x + 3\)[/tex],
[tex]\[ g(\sqrt{x}) = \sqrt{x} + 3 \][/tex]
So, [tex]\((g \circ f)(x) = \sqrt{x} + 3\)[/tex].
### Part c: [tex]\((f \circ g)(6)\)[/tex]
To find [tex]\((f \circ g)(6)\)[/tex], substitute [tex]\(x = 6\)[/tex] into the composition [tex]\((f \circ g)(x)\)[/tex]:
[tex]\[ (f \circ g)(6) = \sqrt{6 + 3} = \sqrt{9} = 3 \][/tex]
### Part d: [tex]\((g \circ f)(6)\)[/tex]
To find [tex]\((g \circ f)(6)\)[/tex], substitute [tex]\(x = 6\)[/tex] into the composition [tex]\((g \circ f)(x)\)[/tex]:
[tex]\[ (g \circ f)(6) = \sqrt{6} + 3 \][/tex]
Since [tex]\(\sqrt{6}\)[/tex] is an irrational number, the expression remains as:
[tex]\[ (g \circ f)(6) = \sqrt{6} + 3 \][/tex]
### Summary of the Solutions
- [tex]\((f \circ g)(x) = \sqrt{x + 3}\)[/tex]
- [tex]\((g \circ f)(x) = \sqrt{x} + 3\)[/tex]
- [tex]\((f \circ g)(6) = 3\)[/tex]
- [tex]\((g \circ f)(6) = \sqrt{6} + 3\)[/tex]
These are the final values for each part of the problem.
