Answer :
Let's break down each part of the problem step by step.
### (i) Functional Values and Operations
Functions Given:
- [tex]\( f(x) = 2x + 1 \)[/tex]
- [tex]\( g(x) = 3x + 1 \)[/tex]
Definitions for Operations:
- Addition of functions: [tex]\( (f + g)(x) = f(x) + g(x) \)[/tex]
- Multiplication of functions: [tex]\( (f \cdot g)(x) = f(x) \cdot g(x) \)[/tex]
- Composition of functions [tex]\( f \circ g \)[/tex] and [tex]\( g \circ f \)[/tex]:
- [tex]\( (f \circ g)(x) = f(g(x)) \)[/tex]
- [tex]\( (g \circ f)(x) = g(f(x)) \)[/tex]
#### 1. Addition: [tex]\(\bar{g}(x) = (f(x) + g(x))\)[/tex]
[tex]\[ \bar{g}(x) = f(x) + g(x) \][/tex]
[tex]\[ = (2x + 1) + (3x + 1) \][/tex]
[tex]\[ = 2x + 3x + 1 + 1 \][/tex]
[tex]\[ = 5x + 2 \][/tex]
So, [tex]\(\bar{g}(x) = 5x + 2\)[/tex].
#### 2. Multiplication: [tex]\((fg)(x) = (f(x) \cdot g(x))\)[/tex]
[tex]\[ (fg)(x) = (2x + 1) \cdot (3x + 1) \][/tex]
[tex]\[ = 2x \cdot 3x + 2x \cdot 1 + 1 \cdot 3x + 1 \cdot 1 \][/tex]
[tex]\[ = 6x^2 + 2x + 3x + 1 \][/tex]
[tex]\[ = 6x^2 + 5x + 1 \][/tex]
So, [tex]\((fg)(x) = 6x^2 + 5x + 1\)[/tex].
#### 3. Composition [tex]\( f \circ g \)[/tex] (i.e., [tex]\( f(g(x)) \)[/tex]):
[tex]\[ (f \circ g)(x) = f(g(x)) = f(3x + 1) \][/tex]
Substitute [tex]\( 3x + 1 \)[/tex] into [tex]\( f(x) = 2x + 1 \)[/tex]:
[tex]\[ = 2(3x + 1) + 1 \][/tex]
[tex]\[ = 6x + 2 + 1 \][/tex]
[tex]\[ = 6x + 3 \][/tex]
So, [tex]\( (f \circ g)(x) = 6x + 3 \)[/tex].
#### 4. Composition [tex]\( g \circ f \)[/tex] (i.e., [tex]\( g(f(x)) \)[/tex]):
[tex]\[ (g \circ f)(x) = g(f(x)) = g(2x + 1) \][/tex]
Substitute [tex]\( 2x + 1 \)[/tex] into [tex]\( g(x) = 3x + 1 \)[/tex]:
[tex]\[ = 3(2x + 1) + 1 \][/tex]
[tex]\[ = 6x + 3 + 1 \][/tex]
[tex]\[ = 6x + 4 \][/tex]
So, [tex]\( (g \circ f)(x) = 6x + 4 \)[/tex].
### (ii) Equality of [tex]\( fg \)[/tex], [tex]\( f \circ g \)[/tex], and [tex]\( g \circ f \)[/tex]
Let's compare the expressions we derived:
- [tex]\( (fg)(x) = 6x^2 + 5x + 1 \)[/tex]
- [tex]\( (f \circ g)(x) = 6x + 3 \)[/tex]
- [tex]\( (g \circ f)(x) = 6x + 4 \)[/tex]
Clearly,
[tex]\[ (fg)(x) \neq (f \circ g)(x) \][/tex]
[tex]\[ (f \circ g)(x) \neq (g \circ f)(x) \][/tex]
[tex]\[ (fg)(x) \neq (g \circ f)(x) \][/tex]
Therefore, [tex]\( fg \)[/tex], [tex]\( f \circ g \)[/tex], and [tex]\( g \circ f \)[/tex] are not equal.
### (iii) Domain and Range of Each Operation
Domain:
Since [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are defined from [tex]\( \mathbb{R} \rightarrow \mathbb{R} \)[/tex], meaning they are defined for all real [tex]\( x \)[/tex]:
- The domains of [tex]\( f \)[/tex], [tex]\( g \)[/tex], [tex]\( \bar{g} \)[/tex], [tex]\( (fg) \)[/tex], [tex]\( f \circ g \)[/tex], and [tex]\( g \circ f \)[/tex] are all [tex]\( \mathbb{R} \)[/tex].
Range:
1. Addition [tex]\(\bar{g}(x) = 5x + 2\)[/tex]:
- Domain: [tex]\( \mathbb{R} \)[/tex]
- Range: Since [tex]\( 5x + 2 \)[/tex] is a linear function with coefficients that are non-zero, the range is [tex]\( \mathbb{R} \)[/tex].
2. Multiplication [tex]\((fg)(x) = 6x^2 + 5x + 1\)[/tex]:
- Domain: [tex]\( \mathbb{R} \)[/tex]
- Range: [tex]\( 6x^2 + 5x + 1 \)[/tex] is a quadratic function. As a quadratic function with a positive leading coefficient, its range is [tex]\( [1 - (\text{vertex value}), \infty) \)[/tex]. [tex]\( y \)[/tex] can attain any value greater than or equal to the minimum value of the quadratic function, which is global minimum.
3. Composition [tex]\( (f \circ g)(x) = 6x + 3\)[/tex]:
- Domain: [tex]\( \mathbb{R} \)[/tex]
- Range: Since [tex]\( 6x + 3 \)[/tex] is a linear function, the range is [tex]\( \mathbb{R} \)[/tex].
4. Composition [tex]\( (g \circ f)(x) = 6x + 4\)[/tex]:
- Domain: [tex]\( \mathbb{R} \)[/tex]
- Range: Since [tex]\( 6x + 4 \)[/tex] is a linear function, the range is [tex]\( \mathbb{R} \)[/tex].
Thus, every operation defined over real numbers has a domain of all real numbers, and ranges as noted above.
### (i) Functional Values and Operations
Functions Given:
- [tex]\( f(x) = 2x + 1 \)[/tex]
- [tex]\( g(x) = 3x + 1 \)[/tex]
Definitions for Operations:
- Addition of functions: [tex]\( (f + g)(x) = f(x) + g(x) \)[/tex]
- Multiplication of functions: [tex]\( (f \cdot g)(x) = f(x) \cdot g(x) \)[/tex]
- Composition of functions [tex]\( f \circ g \)[/tex] and [tex]\( g \circ f \)[/tex]:
- [tex]\( (f \circ g)(x) = f(g(x)) \)[/tex]
- [tex]\( (g \circ f)(x) = g(f(x)) \)[/tex]
#### 1. Addition: [tex]\(\bar{g}(x) = (f(x) + g(x))\)[/tex]
[tex]\[ \bar{g}(x) = f(x) + g(x) \][/tex]
[tex]\[ = (2x + 1) + (3x + 1) \][/tex]
[tex]\[ = 2x + 3x + 1 + 1 \][/tex]
[tex]\[ = 5x + 2 \][/tex]
So, [tex]\(\bar{g}(x) = 5x + 2\)[/tex].
#### 2. Multiplication: [tex]\((fg)(x) = (f(x) \cdot g(x))\)[/tex]
[tex]\[ (fg)(x) = (2x + 1) \cdot (3x + 1) \][/tex]
[tex]\[ = 2x \cdot 3x + 2x \cdot 1 + 1 \cdot 3x + 1 \cdot 1 \][/tex]
[tex]\[ = 6x^2 + 2x + 3x + 1 \][/tex]
[tex]\[ = 6x^2 + 5x + 1 \][/tex]
So, [tex]\((fg)(x) = 6x^2 + 5x + 1\)[/tex].
#### 3. Composition [tex]\( f \circ g \)[/tex] (i.e., [tex]\( f(g(x)) \)[/tex]):
[tex]\[ (f \circ g)(x) = f(g(x)) = f(3x + 1) \][/tex]
Substitute [tex]\( 3x + 1 \)[/tex] into [tex]\( f(x) = 2x + 1 \)[/tex]:
[tex]\[ = 2(3x + 1) + 1 \][/tex]
[tex]\[ = 6x + 2 + 1 \][/tex]
[tex]\[ = 6x + 3 \][/tex]
So, [tex]\( (f \circ g)(x) = 6x + 3 \)[/tex].
#### 4. Composition [tex]\( g \circ f \)[/tex] (i.e., [tex]\( g(f(x)) \)[/tex]):
[tex]\[ (g \circ f)(x) = g(f(x)) = g(2x + 1) \][/tex]
Substitute [tex]\( 2x + 1 \)[/tex] into [tex]\( g(x) = 3x + 1 \)[/tex]:
[tex]\[ = 3(2x + 1) + 1 \][/tex]
[tex]\[ = 6x + 3 + 1 \][/tex]
[tex]\[ = 6x + 4 \][/tex]
So, [tex]\( (g \circ f)(x) = 6x + 4 \)[/tex].
### (ii) Equality of [tex]\( fg \)[/tex], [tex]\( f \circ g \)[/tex], and [tex]\( g \circ f \)[/tex]
Let's compare the expressions we derived:
- [tex]\( (fg)(x) = 6x^2 + 5x + 1 \)[/tex]
- [tex]\( (f \circ g)(x) = 6x + 3 \)[/tex]
- [tex]\( (g \circ f)(x) = 6x + 4 \)[/tex]
Clearly,
[tex]\[ (fg)(x) \neq (f \circ g)(x) \][/tex]
[tex]\[ (f \circ g)(x) \neq (g \circ f)(x) \][/tex]
[tex]\[ (fg)(x) \neq (g \circ f)(x) \][/tex]
Therefore, [tex]\( fg \)[/tex], [tex]\( f \circ g \)[/tex], and [tex]\( g \circ f \)[/tex] are not equal.
### (iii) Domain and Range of Each Operation
Domain:
Since [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are defined from [tex]\( \mathbb{R} \rightarrow \mathbb{R} \)[/tex], meaning they are defined for all real [tex]\( x \)[/tex]:
- The domains of [tex]\( f \)[/tex], [tex]\( g \)[/tex], [tex]\( \bar{g} \)[/tex], [tex]\( (fg) \)[/tex], [tex]\( f \circ g \)[/tex], and [tex]\( g \circ f \)[/tex] are all [tex]\( \mathbb{R} \)[/tex].
Range:
1. Addition [tex]\(\bar{g}(x) = 5x + 2\)[/tex]:
- Domain: [tex]\( \mathbb{R} \)[/tex]
- Range: Since [tex]\( 5x + 2 \)[/tex] is a linear function with coefficients that are non-zero, the range is [tex]\( \mathbb{R} \)[/tex].
2. Multiplication [tex]\((fg)(x) = 6x^2 + 5x + 1\)[/tex]:
- Domain: [tex]\( \mathbb{R} \)[/tex]
- Range: [tex]\( 6x^2 + 5x + 1 \)[/tex] is a quadratic function. As a quadratic function with a positive leading coefficient, its range is [tex]\( [1 - (\text{vertex value}), \infty) \)[/tex]. [tex]\( y \)[/tex] can attain any value greater than or equal to the minimum value of the quadratic function, which is global minimum.
3. Composition [tex]\( (f \circ g)(x) = 6x + 3\)[/tex]:
- Domain: [tex]\( \mathbb{R} \)[/tex]
- Range: Since [tex]\( 6x + 3 \)[/tex] is a linear function, the range is [tex]\( \mathbb{R} \)[/tex].
4. Composition [tex]\( (g \circ f)(x) = 6x + 4\)[/tex]:
- Domain: [tex]\( \mathbb{R} \)[/tex]
- Range: Since [tex]\( 6x + 4 \)[/tex] is a linear function, the range is [tex]\( \mathbb{R} \)[/tex].
Thus, every operation defined over real numbers has a domain of all real numbers, and ranges as noted above.
