Problem 44 - College Algebra Essentials

Question

Find \(f+g, f-g,\) fg, and \(\frac{f}{g} .\) Determine the domain for each function. $$f(x)=\frac{3 x+1}{x^{2}-25}, g(x)=\frac{2 x-4}{x^{2}-25}$$

Step-by-Step Solution

Answer

Function sum: \(f+g = \frac{5x-3}{x^{2}-25}\), Function difference: \(f-g = \frac{x+5}{x^{2}-25}\), Function product: \(fg = \frac{6x^{2}-10x-4}{x^{4}-50x^{2}+625}\), Function division: \( \frac{f}{g} = \frac{(3x+1)}{2x-4} \) and their domain is \(x|- \infty < x < -5 or -5 < x < 5 or 5 < x < \infty\)

Step 1: Find \(f+g\)

To find \(f+g\) simply add the two functions together: \[f(x) + g(x) = \frac{3x+1}{x^{2}-25} + \frac{2x-4}{x^{2}-25} = \frac{(3x+1+(2x-4))}{x^{2}-25} = \frac{5x-3}{x^{2}-25}\].

Step 2: Find \(f-g\)

To find \(f-g\) subtract the second function from the first: \[f(x) - g(x) = \frac{3x+1}{x^{2}-25} - \frac{2x-4}{x^{2}-25} = \frac{(3x+1-(2x-4))}{x^{2}-25} = \frac{x+5}{x^{2}-25}\].

Step 3: Find \(fg\)

To find the product of functions \(f\) and \(g\), simply multiply the two functions: \( f(x) \cdot g(x) = (\frac{3x+1}{x^{2}-25}) \cdot (\frac{2x-4}{x^{2}-25}) = \frac{(3x+1)(2x-4)}{(x^{2}-25)^2} = \frac{6x^{2}-10x-4}{x^{4}-50x^{2}+625}\).

Step 4: Find \(\frac{f}{g}\)

To find \(\frac{f}{g}\), divide function \(f\) by function \(g\): \( \frac{f(x)}{g(x)} = \frac{(3x+1)}{2x-4} \).

Step 5: Find the domain

The domain of a function is the set of all real numbers 'x' for which the function is defined. Here, the functions are undefined where the denominator equals zero. The denominator in each function equals zero where \(x^{2}-25=0\) or where \(x=\pm 5.\) Thus, the domain of each function is \({x|- \infty < x < -5 or -5 < x < 5 or 5 < x < \infty}\).

Key Concepts

Function Addition

When we talk about adding two functions, like in the example with functions f and g, we're simply combining their outputs at each point in their domains. It's analogous to adding two numbers, only that the numbers in question are, in fact, the outputs of our functions for each input value x. Here's how you do it:
You add the numerators directly because the denominators are the same. In our example, f plus g gave us
\(\frac{{3x+1}}{x^2-25} + \frac{{2x-4}}{x^2-25} = \frac{{(3x+1)+(2x-4)}}{x^2-25} = \frac{{5x-3}}{x^2-25}\).
Remember to simplify whenever possible to obtain the most reduced form of the function result.

Function Subtraction

Subtracting one function from another works by taking the output of the second function away from the output of the first for each input x. Just as with regular subtraction, we aim to find the difference between the two values.
In the provided problem, that process resulted in:
\(f(x) - g(x) = \frac{3x+1}{x^2-25} - \frac{2x-4}{x^2-25} = \frac{(3x+1)-(2x-4)}{x^2-25}= \frac{x+5}{x^2-25}\).
Just like with function addition, it's crucial to simplify the resulting expression to make its interpretation and further operations easier.

Function Multiplication

Multiplying functions involves finding the product of their outputs for each value in the domain. In our case, we multiply fractions. Just as with numerical fractions, we multiply the numerators together and the denominators together.
Here's what we got with functions f(x) and g(x):
\(f(x) \cdot g(x) = \left(\frac{{3x+1}}{{x^2-25}}\right) \cdot \left(\frac{{2x-4}}{{x^2-25}}\right) = \frac{{(3x+1)(2x-4)}}{{(x^2-25)^2}} = \frac{{6x^2-10x-4}}{{x^4-50x^2+625}}\).
As usual, simplification of the result is highly recommended if the numerator and the denominator have common factors that can be canceled out.

Function Division

To divide one function by another, you take the ratio of their outputs for each input x. Dividing functions is like dividing numbers—it's finding how many times the denominator function, g(x), is contained within the numerator function, f(x).
In our example, division yielded:
\(\frac{f(x)}{g(x)} = \frac{\frac{3x+1}{x^2-25}}{\frac{2x-4}{x^2-25}} = \frac{(3x+1)}{(2x-4)}\).
Notice that when the denominators are the same, they cancel out. Still, it's always good practice to check if the result can be reduced further by simplifying any complex fractions.

Domain of a Function

The domain of a function is the complete set of possible values of the independent variable, which in most cases is x. In simple terms, it's all the values you can put into a function and get a valid output.
For rational functions, which are fractions where the numerator and/or the denominator are polynomials, the domain is all real numbers except those that would make the denominator equal to zero (since dividing by zero is undefined).
Looking at the functions f(x) and g(x) you provided, we exclude the values where the denominator \(x^2-25\) becomes zero. Factoring it, we get \((x-5)(x+5)\). This tells us that x cannot be 5 or -5. Therefore, the domain for both functions is all real numbers except 5 and -5, which can be expressed as: \({ x | x eq 5 \text{ and } x eq -5 }\). Understanding the domain is vital for correctly graphing the function and for knowing what inputs make sense to use in real-world applications.