Problem 14 - Precalculus : Building Concepts and Connections

Question

In Exercises \(7-16,\) for the given functions \(f\) and \(g,\) find each composite function and identify its domain. (a) \((f+g)(x)\) (b) \((f-g)(x)\) (c) \((f g)(x)\) (d) \(\left(\frac{f}{g}\right)(x)\) $$f(x)=2 x-1 ; g(x)=\sqrt{x}$$

Step-by-Step Solution

Answer

The composite functions (a) \((f+g)(x) = 2x - 1 + \sqrt{x}\), (b) \((f-g)(x) = 2x - 1 - \sqrt{x}\), (c) \((f g)(x) = (2x - 1)*\sqrt{x}\), and (d) \(\left(\frac{f}{g}\right)(x) = \frac{2x - 1}{\sqrt{x}}\) if \(g(x) ≠ 0\), have domains (a, b, c) \(x \geq 0\) and (d) \(x > 0\) respectively.

Step 1: Formation of the Composite Functions

To form the composite functions based on the given instructions, simply perform operations on \(f(x)\) and \(g(x)\) as follows:(a) \((f+g)(x) = f(x) + g(x) = 2x - 1 + \sqrt{x}\)(b) \((f-g)(x) = f(x) - g(x) = 2x - 1 - \sqrt{x}\)(c) \((f g)(x) = f(x) * g(x) = (2x - 1)*\sqrt{x}\)(d) \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{2x - 1}{\sqrt{x}}\) if \(g(x) ≠ 0\)\

Step 2: Identification of the Domains

Identify the domain for each composite function as follows:(a) For \((f+g)(x)\), since the square root function is defined for all non-negative values only, the domain is \(x \geq 0\).(b) For \((f-g)(x)\), the domain is also \(x \geq 0\) for the same reason as above.(c) For \((f g)(x)\), the domain is \(x \geq 0\) because multiplication will not affect the domain.(d) For \(\left(\frac{f}{g}\right)(x)\), we exclude the values for which \(g(x) = 0\) from the domain of \(g(x)\). Therefore, in this case, the domain is \(x > 0\).

Key Concepts

Composite Functions Algebra

When dealing with composite functions in algebra, it is akin to combining the outputs of two or more functions into a single expression. Imagine we have two functions, say f and g. With these functions, we can create new functions by performing standard arithmetic operations such as addition, subtraction, multiplication, or division.

For example, given the functions f(x)=2x-1 and g(x)=\text{\(\backslash\text{sqrt}\)}{x}, we can form:

• The sum (f+g)(x) = f(x) + g(x), resulting in 2x - 1 + \text{\(\backslash\text{sqrt}\)}{x}.
• The difference (f-g)(x) = f(x) - g(x), yielding 2x - 1 - \text{\(\backslash\text{sqrt}\)}{x}.
• The product (fg)(x) = f(x) * g(x), which becomes (2x - 1)*\text{\(\backslash\text{sqrt}\)}{x}.
• The quotient \text{\(\backslash\text{frac}\)}{f}{g}\text{\(\backslash\)}(x) = \text{\(\backslash\text{frac}\)}{f(x)}{g(x)}, simplifying to \text{\(\backslash\text{frac}\)}{2x - 1}{\text{\(\backslash\text{sqrt}\)}{x}} provided that \text{\(\backslash\text{sqrt}\)}{x} eq 0.

This process doesn't alter the individual functions f and g, but rather generates entirely new functions that describe different mathematical relationships. It's important to remember that although the arithmetic is straightforward, the resultant composite function may have a range of values or a domain that is shaped by the domains of the original functions.

Function Operations Precalculus

In precalculus, function operations involve adding, subtracting, multiplying, and dividing functions in a manner comparable to that of numbers, but with extra care given to the domains of the functions. The domain of a function is the set of all possible input values (usually 'x' values) for which the function is defined.

Considering Composite Functions

When operating with composite functions, understanding how the domain impacts each operation is paramount. For instance:

• For the sum or difference, \text{\((f\text{\pm}g)(x)\text{\)}}, if one function includes a square root, such as \text{\(\text{\(\backslash\text{sqrt}\)}{x}\text{\)}}, we must ensure that 'x' is greater than or equal to zero for the square root to be real.
• For the product, the resulting domain is usually the intersection of the original domains, but one must still look out for restrictions. If a square root is involved, the argument under the square root must be non-negative.
• For the quotient, we must exclude any 'x' values that result in division by zero, making the rational expression undefined. So in the division of two functions where the denominator includes a square root, 'x' must be strictly greater than zero, avoiding the undefined expression at 'x = 0'.

These observations are key when determining the domain of the resulting function after performing operations, which directly impacts how we interpret and graph these composite functions.

Domain of a Function

The domain of a function is one of the fundamental concepts in algebra and precalculus. It refers to all the possible inputs, or 'x' values, that a function can accept to produce real and defined outputs. When it comes to composite functions, determining the domain can be slightly more complex, considering the domains of the individual functions involved.

For the given example with \text{\(f(x)=2x-1\text{\)}} and \text{\(g(x)=\text{\(\backslash\text{sqrt}\)}{x}\text{\)}}, let's explore how the domain is established.

• In the case of \text{\((f+g)(x)\text{\)} and \((f-g)(x)\text{\)}}, the presence of \text{\(\sqrt{x}\text{\)}} means we accept only non-negative values of 'x', hence the domain is \text{\([0, \backslash\text{infty})\text{\)}}.
• For \text{\((fg)(x)\text{\)}}, even though the functions are multiplied, the square root in \text{\(g(x)\text{\)}} still requires 'x' to be non-negative, so the domain remains \text{\([0, \backslash\text{infty})\text{\)}}.
• However, when looking at \text{\(\text{\(\frac{f}{g}\text{\)}}(x)\text{\)}}, since \text{\(g(x)\text{\)}} is in the denominator, 'x' must be strictly positive to avoid division by zero. Thus, the domain is \text{\((0, \backslash\text{infty})\text{\)}}.

It's imperative for students to understand that the domain is not just about finding the inputs where the function 'works', but also about ensuring that each part of a composite function is defined and real over its full range of application.