Problem 45 - Precalculus Essentials

Question

Find \(f+g, f-g,\) fg, and \(\frac{f}{x}\). Determine the domain for each function. $$f(x)=\frac{8 x}{x-2}, g(x)=\frac{6}{x+3}$$

Step-by-Step Solution

Answer

\(f+g = \frac{8x^2+30x-12}{(x-2)(x+3)}\), domain: all real numbers except -3 and 2. \(f-g = \frac{8x^2+18x+12}{(x-2)(x+3)}\), domain: all real numbers except -3 and 2. \(fg = \frac{48x}{(x-2)(x+3)}\), domain: all real numbers except -3 and 2. \(\frac{f}{g} = \frac{4x(x+3)}{3(x-2)}\), domain: all real numbers except -3 and 2.

Step 1: Compute the addition \(f+g\)

By adding the two functions and simplifying we get \(f(x)+g(x) = \frac{8x}{x-2} + \frac{6}{x+3}\). Combine the fractions: \(f(x)+g(x) = \frac{8x(x+3)}{(x-2)(x+3)}+\frac{6(x-2)}{(x-2)(x+3)} = \frac{8x^2+24x+6x-12}{(x-2)(x+3)} = \frac{8x^2+30x-12}{(x-2)(x+3)}\). The domain of \(f+g\) excludes -3 and 2, because with those values the denominator (x-2) (x+3) would be zero. Therefore, the domain of \(f+g\) is all real numbers except -3 and 2.

Step 2: Compute the subtraction \(f-g\)

By subtracting the two functions and simplifying we get \(f(x)-g(x) = \frac{8x}{x-2} - \frac{6}{x+3} = \frac{8x(x+3)}{(x-2)(x+3)} - \frac{6(x-2)}{(x-2)(x+3)} = \frac{8x^2+24x-6x+12}{(x-2)(x+3)} = \frac{8x^2+18x+12}{(x-2)(x+3)}\). The domain of \(f-g\) excludes -3 and 2, where the denominator (x-2)(x+3) would be zero. So the domain of this function is all real numbers except -3 and 2.

Step 3: Compute the multiplication \(fg\)

By multiplying the two functions we get \(f(x)g(x) = \frac{8x}{x-2} * \frac{6}{x+3} = \frac{48x}{(x-2)(x+3)}\). The domain of \(fg\) excludes -3 and 2, where the denominator would be zero. So the domain of this function is all real numbers except -3 and 2.

Step 4: Compute the division \(\frac{f}{g}\)

By dividing \(f\) by \(g\) and simplifying we get \(\frac{f(x)}{g(x)} = \frac{\frac{8x}{x-2}}{\frac{6}{x+3}} = \frac{8x(x+3)}{6(x-2)} = \frac{4x(x+3)}{3(x-2)}\). The domain of \(\frac{f}{g}\) also excludes -3 and 2, because these values would make the denominator (x-2)(x+3) zero. So the domain of this function is all real numbers except -3 and 2.

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial when we're dealing with any type of function operation. It's the set of all possible input values (typically 'x' values) for which the function is defined and will produce real numbers as outputs. When we combine functions arithmetically, like addition or multiplication, we must consider the domains of both functions involved.

For instance, in the exercise with functions f(x) and g(x), their individual domains exclude values that would make the denominators zero because division by zero is undefined. Consequently, the domain of f+g, f-g, fg, and f/g would exclude those specific numbers that cause each original function (or their combination) to be undefined.

In this case, we avoid the values -3 and 2 because they make the denominator zero in one or both of the original functions, f(x) and g(x). It's imperative to spot and exclude such 'trouble' values from the domain to avoid any errors or undefined expressions in computations.

Combining Functions

When it comes to combining functions, we are often asked to perform operations such as addition, subtraction, multiplication, or division on two or more functions.

In the provided exercise, functions f(x) and g(x) are combined through these operations to create new functions. While combining, it's essential to apply arithmetic correctly and look carefully for opportunities to simplify the expressions. It often involves finding a common denominator to combine the functions when they’re rational expressions, as in this exercise.

Simplify Combined Functions

Always ensure to fully simplify the combined function by factoring, reducing any common factors, and then expressing the result in the simplest form. Clearly stating each step of the process helps avoid confusion and ensures that the calculation is correct and the resulting function is as simplified as possible.

Simplifying Rational Expressions

The process of simplifying rational expressions is integral to precalculus and is showcased in the exercise through division and combination of the functions f(x) and g(x). A rational expression is a ratio of two polynomials, and simplifying involves factoring both the numerator and the denominator and then reducing common factors.

It's not just about making the expression look 'neater;' simplifying can also reveal more information about the function's behavior and its domain. In the exercise, after combining the functions, we simplify the resulting expression by combining like terms and ensuring that the expression is in its simplest form.

• Identify common factors in both numerator and denominator.
• Cancel out the common factors, being careful not to remove factors that might result in a loss of critical points or domain restrictions.
• After simplification, re-examine the domain of the new function to ensure that no values are excluded that would make the expression undefined.

By fully understanding these steps, students can tackle any rational expression and simplify it with confidence, keeping in mind the behavior and domain of the functions involved.