Conangrayisbaeidn
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Given the functions:
[tex]\[ f(x)=|x| \][/tex]
[tex]\[ g(x)=x+1 \][/tex]

(a) [tex]\[ f \circ g \][/tex]
[tex]\[\square\][/tex]

(b) [tex]\[ g \circ f \][/tex]
[tex]\[\square\][/tex]

Find the domain of each function and each composite function. Enter your answers using interval notation.

Domain of [tex]\( f \)[/tex]:
[tex]\[\square\][/tex]

Domain of [tex]\( g \)[/tex]:
[tex]\[\square\][/tex]

Domain of [tex]\( f \circ g \)[/tex]:
[tex]\[\square\][/tex]

Domain of [tex]\( g \circ f \)[/tex]:
[tex]\[\square\][/tex]

Sagot :

GinnyAnsweridn
Let's start by breaking down each part of the composite functions as well as determining the domains.

### Composite Functions:

#### (a) [tex]\( f \circ g \)[/tex]
The notation [tex]\( f \circ g \)[/tex] means [tex]\( f(g(x)) \)[/tex].

Given [tex]\( g(x) = x + 1 \)[/tex], we need to evaluate [tex]\( f \)[/tex] at [tex]\( g(x) \)[/tex]:
[tex]\[ f(g(x)) = f(x + 1) \][/tex]

Since [tex]\( f(x) = |x| \)[/tex], we substitute [tex]\( x + 1 \)[/tex] into [tex]\( f \)[/tex]:
[tex]\[ f(x + 1) = |x + 1| \][/tex]

Thus:
[tex]\[ f \circ g (x) = |x + 1| \][/tex]

#### (b) [tex]\( g \circ f \)[/tex]
The notation [tex]\( g \circ f \)[/tex] means [tex]\( g(f(x)) \)[/tex].

Given [tex]\( f(x) = |x| \)[/tex], we need to evaluate [tex]\( g \)[/tex] at [tex]\( f(x) \)[/tex]:
[tex]\[ g(f(x)) = g(|x|) \][/tex]

Since [tex]\( g(x) = x + 1 \)[/tex], we substitute [tex]\( |x| \)[/tex] into [tex]\( g \)[/tex]:
[tex]\[ g(|x|) = |x| + 1 \][/tex]

Thus:
[tex]\[ g \circ f (x) = |x| + 1 \][/tex]

### Evaluating Composite Functions at a Given Point:

Evaluating these composite functions at [tex]\( x = 2 \)[/tex]:
- For [tex]\( f \circ g (2) \)[/tex]:
[tex]\[ f(g(2)) = f(2 + 1) = f(3) = |3| = 3 \][/tex]

- For [tex]\( g \circ f (2) \)[/tex]:
[tex]\[ g(f(2)) = g(|2|) = g(2) = 2 + 1 = 3 \][/tex]

So, the evaluations are:
[tex]\[ f \circ g (2) = 3 \][/tex]
[tex]\[ g \circ f (2) = 3 \][/tex]

### Domains:

The domain of a function [tex]\( f \)[/tex] is the set of all [tex]\( x \)[/tex] for which the function is defined.

- Domain of [tex]\( f(x) = |x| \)[/tex]:
- The absolute value function [tex]\( f(x) = |x| \)[/tex] is defined for all real numbers.
- Therefore, the domain of [tex]\( f \)[/tex] is [tex]\( (-\infty, \infty) \)[/tex].

- Domain of [tex]\( g(x) = x + 1 \)[/tex]:
- The linear function [tex]\( g(x) = x + 1 \)[/tex] is also defined for all real numbers.
- Therefore, the domain of [tex]\( g \)[/tex] is [tex]\( (-\infty, \infty) \)[/tex].

- Domain of [tex]\( f \circ g (x) = |x + 1| \)[/tex]:
- Since both [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are defined for all real numbers, [tex]\( f(g(x)) \)[/tex] is defined for all real numbers.
- Therefore, the domain of [tex]\( f \circ g \)[/tex] is [tex]\( (-\infty, \infty) \)[/tex].

- Domain of [tex]\( g \circ f (x) = |x| + 1 \)[/tex]:
- Similarly, since both [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are defined for all real numbers, [tex]\( g(f(x)) \)[/tex] is defined for all real numbers.
- Therefore, the domain of [tex]\( g \circ f \)[/tex] is [tex]\( (-\infty, \infty) \)[/tex].

### Summary:

- The evaluations of the composite functions at [tex]\( x = 2 \)[/tex] are:
[tex]\[ f \circ g (2) = 3 \][/tex]
[tex]\[ g \circ f (2) = 3 \][/tex]

- Domains:
- Domain of [tex]\( f \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
- Domain of [tex]\( g \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
- Domain of [tex]\( f \circ g \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
- Domain of [tex]\( g \circ f \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
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