Question

For the given functions fand g. find the following. For parts \((a)-(d),\) also find the domain. (a) \((f+g)(x)\) (b) \((f-g)(x)\) (c) \((f \cdot g)(x)\) (d) \(\left(\frac{f}{g}\right)(x)\) (e) \((f+g)\) (3) (f) \((f-g)\) (4) (g) \((f \cdot g)\) ( 2 ) \((h)\left(\frac{f}{g}\right)(1)\) \(f(x)=x-1 ; \quad g(x)=2 x^{2}\)

Short answer

(f+g)(x) = 2x^2 + x - 1; (f-g)(x) = -2x^2 + x - 1; (f·g)(x) = 2x^3 - 2x^2; (f/g)(x) = (x-1)/(2x^2), Domain: x≠0: (f+g)(3) = 20; (f-g)(4) = -29; (f·g)(2) = 8; (f/g)(1) = 0

Step-by-step solution

- Understanding the Functions

We are given two functions: \( f(x) = x - 1 \) \( g(x) = 2x^2 \).

- (f+g)(x) and its Domain

To find \((f+g)(x)\), add the functions: \( (f+g)(x) = f(x) + g(x) \) Substituting the given functions: \( (f+g)(x) = (x - 1) + 2x^2 \) \( (f+g)(x) = 2x^2 + x - 1 \) The domain for both functions is all real numbers, so the domain for \( (f+g)(x) \) is also all real numbers.

- (f-g)(x) and its Domain

To find \((f-g)(x)\), subtract the functions: \( (f-g)(x) = f(x) - g(x) \) Substituting the given functions: \( (f-g)(x) = (x - 1) - 2x^2 \) \( (f-g)(x) = -2x^2 + x - 1 \) The domain for \( (f-g)(x) \) is all real numbers.

- (f·g)(x) and its Domain

To find \((f·g)(x)\), multiply the functions: \( (f \times g)(x) = f(x) \times g(x) \) Substituting the given functions: \( (f \times g)(x) = (x - 1) \times 2x^2 \) \( (f \times g)(x) = 2x^3 - 2x^2 \) The domain for \( (f \times g)(x) \) is all real numbers.

- (f/g)(x) and its Domain

To find \((f/g)(x)\), divide the functions: \( (f/g)(x) = \frac{f(x)}{g(x)} \) Substituting the given functions, we get: \( (f/g)(x) = \frac{x - 1}{2x^2} \) To find the domain, solve for when the denominator equals zero: \( 2x^2 = 0 \implies x = 0 \). So the domain for \( (f/g)(x) \) is all real numbers except \( x = 0 \).

- (f+g)(3)

Substitute \( x = 3 \) into \( (f+g)(x) \): \( (f+g)(3) = 2(3)^2 + 3 - 1 \) \( (f+g)(3) = 18 + 3 - 1 = 20 \)

- (f-g)(4)

Substitute \( x = 4 \) into \( (f-g)(x) \): \( (f-g)(4) = -2(4)^2 + 4 - 1 \) \( (f-g)(4) = -32 + 4 - 1 = -29 \)

- (f·g)(2)

Substitute \( x = 2 \) into \( (f \times g)(x) \): \( (f \times g)(2) = 2(2)^3 - 2(2)^2 \) \( (f \times g)(2) = 16 - 8 = 8 \)

- (f/g)(1)

Substitute \( x = 1 \) into \( (f/g)(x) \): \( (f/g)(1) = \frac{1 - 1}{2(1)^2} = \frac{0}{2} = 0 \)

Key concepts

Function Addition

Function addition involves finding the sum of two functions. For functions f(x) and g(x), function addition is represented as \( (f+g)(x) = f(x) + g(x) \).
To demonstrate, let's use the given functions. For \( f(x) = x - 1 \) and \( g(x) = 2x^2 \), the sum is \( (f+g)(x) = (x - 1) + 2x^2 \).
Simplifying this, we get \( 2x^2 + x - 1 \).
It's also crucial to find the domain. Since both functions f and g have the domain of all real numbers, the domain of \( (f+g)(x) \) is also all real numbers. So, \( (f+g)(x) = 2x^2 + x - 1 \) with a domain of all real numbers.

Function Subtraction

Function subtraction involves finding the difference of two functions. Represented as \( (f-g)(x) = f(x) - g(x) \), it subtracts the values of two functions at each point.
Using the given functions, \( f(x) = x - 1 \) and \( g(x) = 2x^2 \), the subtraction is \( (f - g) (x) = (x - 1) - 2x^2 \).
Simplifying, this results in \( -2x^2 + x - 1 \).
Like addition, the domain for \( (f-g)(x) \) is all real numbers since both f and g are defined for all real numbers. Therefore, \( (f-g)(x) = -2x^2 + x - 1 \) has a domain of all real numbers.

Function Multiplication

For function multiplication, we multiply two functions. This operation is denoted by \( (f \times g)(x) = f(x) \times g(x) \). It multiplies the values of the functions f and g at each point.
With the functions \( f(x) = x - 1 \) and \( g(x) = 2x^2 \), it results in \( (f \times g)(x) = (x - 1) \times 2x^2 \).
Simplifying this, we get \( 2x^3 - 2x^2 \).
The domain for \( (f \times g)(x) \) remains all real numbers, as neither function restricts the domain. Thus, the function \( (f \times g)(x) = 2x^3 - 2x^2 \) is defined for all real numbers.

Function Division

Function division involves dividing one function by another. It's expressed as \( (f/g)(x) = \frac{f(x)}{g(x)} \).
Using f and g, \( f(x) = x - 1 \) and \( g(x) = 2x^2 \), the division is \( (f/g)(x) = \frac{x-1}{2x^2} \).
To determine the domain of this function, we need to ensure g(x) ≠ 0 to avoid division by zero. Here, g(x) = 2x^2, which is zero when \( x = 0 \). Therefore, the domain for \( (f/g)(x) \) is all real numbers except \( x = 0 \).
So, \( (f/g)(x) = \frac{x-1}{2x^2} \) with a domain of all real numbers except zero.

Domain of Functions

The domain of a function is the set of all possible input values (x-values) that the function can accept.
For the given operations:

• Addition and subtraction: If f and g are defined for all real numbers, so are \( (f+g) \) and \( (f-g) \) with domains being all real numbers.
• Multiplication: If both f and g are defined for all real numbers, then \( (f \times g) \) also has a domain of all real numbers.
• Division: Extra caution is needed. The function \( g(x) \) must not equal zero. For our case, g(x) = 2x^2, so the domain excludes x = 0.

Understanding the domain ensures the functions are used properly within their defined scopes. When performing operations on functions, always consider their domains to avoid undefined values.