Problem 13 - Precalculus with Limits

Question

In Exercises 9-16, find (a) \((f+g)(x)\), (b) \((f-g)(x)\), (c) \((fg)(x)\), and (d) \((f/g)(x). What is the domain of \)f/g\(? \)f(x) = x^2 + 6\(, \)g(x) = \sqrt{1-x}$

Step-by-Step Solution

Answer

The functions (a) \((f+g)(x) = (x^2 + 6) + \sqrt{1-x}\), (b) \((f-g)(x) = (x^2 + 6) - \sqrt{1-x}\), (c) \((fg)(x) = (x^2 + 6) * \sqrt{1-x}\), and (d) \((f/g)(x) = (x^2 + 6) / \sqrt{1-x}\) with the domain of \((f/g)(x)\) is \(x < 1\).

Step 1: Addition of functions

We find the sum \(f+g\) by adding \(f(x)\) and \(g(x)\) together. So, \((f+g)(x) = (x^2 + 6) + \sqrt{1-x}\.

Step 2: Subtraction of functions

We find the difference \(f-g\) by subtracting \(g(x)\) from \(f(x)\). So, \((f-g)(x) = (x^2 + 6) - \sqrt{1-x}\.

Step 3: Multiplication of functions

We find the product \(fg\) by multiplying \(f(x)\) and \(g(x)\). So, \((fg)(x) = (x^2 + 6) * \sqrt{1-x}\.

Step 4: Division of functions

We find fraction \(f/g\) by dividing \(f(x)\) by \(g(x)\) where \(g(x) \neq 0\). So, \((f/g)(x) = (x^2 + 6) / \sqrt{1-x}\.

Step 5: Determining the domain of f/g

The domain of \(f/g\) includes all values of \(x\) such that the denominator is not zero. So we need to solve \( \sqrt{1-x} \neq 0\), which gives us \( x \neq 1\). Also, inside the square root \(1-x \ge 0 \), so \( x \le 1 \). Combining these conditions, we have \(x < 1\) as the domain of \(f/g\).

Key Concepts

Adding Functions

When we talk about adding functions in precalculus, we are essentially referring to the process of combining two functions to create a new one by adding their values for each corresponding input. Imagine you have two functions, like puzzle pieces, and you want to put them together to make a bigger picture. That's what addition does in the realm of functions.

To add functions, simply add the value of one function to the value of the other for the same input value. If we have two functions, say, f(x) and g(x), their sum is noted as (f + g)(x), which equates to f(x) + g(x) for all x in the domains of f and g. For example, if f(x) = x^2 + 6 and g(x) = \( \sqrt{1-x} \), then the sum (f + g)(x) is x^2 + 6 + \sqrt{1-x}.

Subtracting Functions

Subtracting functions is quite similar to adding them, but instead of combining values, we're taking the difference. It's like having a piece of clay (the first function) and carving out a shape from it (the second function).

To subtract one function from another, subtract the value of the second function from the value of the first for each input. If we have functions f(x) and g(x), then their difference is represented as (f - g)(x), which means f(x) - g(x). Given our previous functions, f(x) = x^2 + 6 and g(x) = \( \sqrt{1-x} \), their difference (f - g)(x) will be x^2 + 6 - \sqrt{1-x}.

Multiplying Functions

Moving on to multiplying functions, we're not just adding or subtracting values but scaling one function by another. Think of it as stretching or compressing one function according to the values of the other one.

For multiplication, take each function's value and multiply them together for a given input. With f(x) and g(x), the product is (fg)(x) or f(x) * g(x). In our example with f(x) = x^2 + 6 and g(x) = \( \sqrt{1-x} \), the product (fg)(x) is (x^2 + 6) * \sqrt{1-x}.

Dividing Functions

Dividing functions, on the other hand, involves a bit more caution. It's much like dividing numbers, where you cannot divide by zero. In the context of functions, this means we have to be particularly careful to avoid scenarios where the function we're dividing by, becomes zero.

When dividing f(x) by g(x), we write this as (f/g)(x) or f(x) / g(x), under the condition that g(x) != 0. So, for f(x) = x^2 + 6 and g(x) = \( \sqrt{1-x} \), we get the quotient (f/g)(x) as (x^2 + 6) / \sqrt{1-x}, with the extra precaution that x cannot be such that g(x) is zero.

Function Domain

Lastly, a critical concept in working with functions is understanding the domain. The domain of a function is the set of all possible inputs, or x-values, that the function can accept without resulting in any undefined or non-real values.

For adding, subtracting, and multiplying functions, the domain is generally the intersection of the domains of the individual functions involved. However, for division, we must exclude any values that make the denominator zero. In our example with the quotient (f/g)(x), the domain would be all x-values where g(x) (the square root function) is not zero, and also where the expression under the square root is non-negative. This means x < 1 for the domain of (f/g)(x), ensuring we avoid division by zero and the square root of a negative number.