Problem 39 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

The operations of \(f(x)\) and \(g(x)\) yield: \(f+g = \sqrt{x} + x - 4\), \(f-g = \sqrt{x} - x + 4\), \(fg = x\sqrt{x} - 4\sqrt{x}\), and \(f/x = 1/\sqrt{x}\). The domain for each of the four functions is \(x\geq0\), except for \(f/x\), where the domain is \(x>0\).

Step 1: Calculate f+g

Sum function, \(f+g\), is found by adding the two given functions together. Here, \(f+g = \sqrt{x} + (x-4)\.

Step 2: Calculate f-g

The difference function, \(f-g\), is found by subtracting the second function from the first. In this case, \(f-g = \sqrt{x} - (x-4) = \sqrt{x} - x + 4\.

Step 3: Calculate fg

The product function, fg, is found by multiplying the two given functions together. Here, \(fg = \sqrt{x} * (x-4) = x\sqrt{x} - 4\sqrt{x}\.

Step 4: Calculate \(f/x\)

The ratio function, \(f/x\), is found by dividing the first function by x. Thus, \(f/x = \sqrt{x} / x = 1/\sqrt{x}\), for x≠0.

Step 5: Determine the Domain

The domain of f+g, f-g, and fg is all x greater than or equal to 0, because of the square root present in \(f(x)\). The domain of \(f/x\) is all x greater than 0 since x in the denominator cannot be 0.

Key Concepts

Sum of Functions

Understanding the sum of functions is an essential concept in precalculus. Picture it as a friendly meeting between two functions where they decide to combine their values. In mathematical terms, to obtain the sum of two functions, simply add their respective values at each point in their domains. For example, given two functions, say f(x) as \( \sqrt{x} \) and g(x) as \( x-4 \), their sum f+g is determined by \( \sqrt{x} + (x-4) \). When considering their domain, you should reckon with the most restrictive function. Here, since f(x) involves a square root, we only look at x >= 0. The sum of two functions inherits this limitation; hence, the domain for f+g is also x >= 0.

This combined function can now serve as a map that denotes a unique journey through higher or lower values, summing the individual heights of f and g at each point.

Difference of Functions

The difference of functions can be envisioned as an amicable tug-of-war where each function pulls with its respective value, aiming to show its strength. Mathematically, the difference function f-g is derived by subtracting one function from another. Given f(x) as \( \sqrt{x} \) and g(x) as \( x-4 \), their difference f-g will be \( \sqrt{x} - (x-4) \). It's crucial to note that this operation, like the sum, is also bound by the original functions' domains. Since f(x) can only accept non-negative values of x, the domain for the difference function remains x >= 0.

Through this subtraction, we obtain a new mathematical path that depicts the relative positioning of these functions across the x-axis, highlighting the vertical disparity between them.

Product of Functions

The product of functions is akin to a partnership where each function's output is multiplied to achieve a collective result. To calculate the product of two functions, fg, multiply their outputs together for each input within their domains. For our functions, f(x) being \( \sqrt{x} \) and g(x) as \( x-4 \), their product is \( \sqrt{x} \times (x-4) = x\sqrt{x} - 4\sqrt{x} \). The domain of this new function, fg, is still dictated by f, hence we look at x >= 0 again.

This product function offers us a new insight, as it weaves the individual functions together into a complex dance, where each move is influenced by both f and g simultaneously.

Ratio of Functions

The ratio of functions is best imagined as a division of labor between two functions, where the output of one is divided by the output of the other. To find the ratio, divide the function f(x) by x. Considering f(x) to be \( \sqrt{x} \), the expression for this ratio becomes \( \sqrt{x} / x \) which simplifies to \( 1/\sqrt{x} \), and is defined for all values of x except when x = 0, as division by zero is undefined. Thus, the domain of our ratio function f/x comprises all positive values of x.

This ratio gives us a peculiar graph that is distinct from the graphs of f, g, or any of their other operations, reflecting the intrinsic relationship between a function and the identity function x.

Function Domain

The function domain is the set of all possible inputs for which a function is defined. Think of it like the 'guest list' for a function's domain: only certain values are 'invited' or allowed to participate. When we talk about operations involving functions, the domain for each resultant function can become narrower based on restrictions from the original functions. As seen in our examples, the square root in f(x), \( \sqrt{x} \), dictates that we can only work with non-negative values, setting the domain for f as x >= 0.

When performing operations like sum, difference, or product, the most restrictive domain from the initial functions prevails. This ensures that not only is each function 'comfortable' to operate, but also that the new, combined function behaves well and keeps its integrity across its domain.