Problem 44 - Precalculus Essentials

Question

Find \(f+g, f-g,\) fg, and \(\frac{f}{x}\). 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

\(f+g = \frac{5x - 3}{x^{2} - 25}\) (domain: \(x \in R, x \neq \pm 5\)), \(f-g = \frac{x + 5}{x^{2} - 25}\) (domain: \(x \in R, x \neq \pm 5\)), \(fg = \frac{6x^{2} - 10x - 4}{x^{4} - 50x^{2} + 625}\) (domain: \(x \in R, x \neq \pm 5, \pm \sqrt{5}\)), \(\frac{f}{x} = \frac{3x + 1}{x^{3} - 25x}\) (domain: \(x \in R, x \neq 0, \pm 5\))

Step 1: Find f+g and its domain

Combine \(f(x)\) and \(g(x)\) by adding the numerators and keeping the common denominator. Then, simplify the fraction. This gives us: \[(f+g)(x) = \frac{(3x + 1) + (2x - 4)}{x^{2} - 25} = \frac{5x - 3}{x^{2} - 25}\]. The domain of \(f+g\) can be found by setting the denominator equal to zero and solving for \(x\), which gives: \(x^{2} - 25 = 0 \Rightarrow x = \pm 5\). Therefore, the domain of \(f+g\) is \(x \in R, x \neq \pm 5\)

Step 2: Find f-g and its domain

Now, subtract \(g(x)\) from \(f(x)\) by subtracting the numerators and keeping the common denominator. This gives: \[(f-g)(x) = \frac{(3x + 1) - (2x - 4)}{x^{2} - 25} = \frac{x + 5}{x^{2} - 25}\]. The domain of \(f-g\) will be the same as \(f+g\), that is, \(x \in R, x \neq \pm 5\)

Step 3: Find fg and its domain

Multiply \(f(x)\) and \(g(x)\) by multiplying the numerators and denominators to get: \[fg(x) = \frac{(3x + 1) \cdot (2x - 4)}{(x^{2} - 25)^2} = \frac{6x^{2} - 10x - 4}{x^{4} - 50x^{2} + 625}\]. The domain of \(fg\) can be found by setting the denominator equal to zero and solving for \(x\), which gives: \(x^{4} - 50x^{2} + 625 = 0 \Rightarrow x = \pm 5, - \sqrt{5}\). Therefore, the domain of \(fg\) is \(x \in R, x \neq \pm 5, \pm \sqrt{5}\)

Step 4: Find \(\frac{f}{x}\) and its domain

Divide \(f(x)\) by \(x\) to get: \[\frac{f}{x} = \frac{3x + 1}{x(x^{2} - 25)} = \frac{3x + 1}{x^{3} - 25x}\]. The domain of \(\frac{f}{x}\) can be found by setting the denominator equal to zero and solving for \(x\), which gives: \(x^{3} - 25x = 0 \Rightarrow x = 0, \pm 5\). Therefore, the domain of \(\frac{f}{x}\) is \(x \in R, x \neq 0, \pm 5\)

Key Concepts

Function Addition and Domain

When you add two functions, just like adding any other mathematical expressions, you combine the like terms. For the functions given, let's add the numerators while keeping the common denominator.
This process creates a new function, \(f+g\)(x) = \frac{5x - 3}{x^{2} - 25}\. What about the domain of this new function? The domain encompasses all the possible inputs into \(f+g\)(x) that won’t cause any mathematical issues, like division by zero.
To identify the domain, we check where the denominator equals zero since these points are not allowed. For \(f+g\)(x), the denominator \(x^{2} - 25\) equals zero when \(x=\pm 5\). Hence, the domain is all real numbers except \(\pm 5\) or written as \(x \in R, x \eq \pm 5\).
Always ensure the denominators in your combined function do not introduce new restrictions on the domain that weren't present in the original functions.

Function Subtraction and Domain

Subtracting functions follows a similar path to addition. You simply subtract the numerators of the two functions while keeping the denominator the same.
The result for \(f-g\)(x) is \frac{x + 5}{x^{2} - 25}\. Checking the domain involves the same steps: look out for values that make the denominator zero. Just as before, \(x=\pm 5\) are the problematic points.
The domain for \(f-g\)(x) is the same as for \(f+g\)(x) because they share the same denominator. Therefore, it is \(x \in R, x \eq \pm 5\). Different operations on functions can sometimes result in the same domain constraints, based on their shared components.

Function Multiplication and Domain

Multiplying functions entails multiplying their numerators and denominators separately, to form a new function.
After multiplying the given functions, the product is \fg(x) = \frac{6x^{2} - 10x - 4}{x^{4} - 50x^{2} + 625}\. Now, we investigate the domain of this function. Unlike addition or subtraction, the multiplication of the denominators here results in a fourth-degree polynomial, which could introduce new zero points.
For fg, the denominator does not change the set of x values that are problematic; it still remains as \(x=\pm 5\). Therefore, the domain continues to be the set of all real numbers excluding \(\pm 5\), which can be expressed as \(x \in R, x \eq \pm 5\). Multiplying functions hence did not alter the original restrictions on the domain.

Function Division and Domain

Dividing one function by another is the same as multiplying by the reciprocal. Therefore, when we divide \f(x)\ by \(x\), we get \frac{f}{x} = \frac{3x + 1}{x(x^{2} - 25)}-\frac{3x + 1}{x^{3} - 25x}\. Now comes the critical part: finding the domain.
Division introduces a new restriction because now the term \(x\) in the denominator also can't be zero. To find the domain, we factor the denominator and find the values where it's zero, which are \(x=0\) and \(x=\pm 5\).

• The domain excludes these values, which means \(x \in R, x \eq 0, \pm 5\).

Division by \(x\) has thus introduced an additional value that cannot be used in the domain, on top of those found from the original denominator \(x^{2} - 25\).