Problem 10 - The Calculus with Analytic Geometry

Question

In Exercises 7 through 12, the functions \(f\) and \(g\) are defined. In each problem define the following functions and determine the domain of the resulting function: (a) \(f+g ;\) (b) \(f-g ;\) (c) \(f \cdot g ;\) (d) \(f / g ;\) (e) \(g / f\); (f) \(f \circ g ;(\mathrm{g}) g \circ f\). $$ f(x)=\sqrt{x-2} ; g(x)=\frac{1}{x} $$

Step-by-Step Solution

Answer

(a), (b), (c), (d), (e), (g \circ f): \(x \gt 2\). (f \circ g): \(0 \lt x \lt \frac{1}{2}\).

Step 1: Define the Functions

Given functions: \(f(x) = \sqrt{x-2}\) \(g(x) = \frac{1}{x}\)

Step 2: Define and Find Domain of \(f+g\)

The function \((f+g)(x) = f(x) + g(x)\) becomes: \((f+g)(x) = \sqrt{x-2} + \frac{1}{x}\). Domain: 1. \(x-2 \geq 0 \Rightarrow x \geq 2\) 2. \(x eq 0\) Combined domain: \(x \gt 2\)

Step 3: Define and Find Domain of \(f-g\)

The function \((f-g)(x) = f(x) - g(x)\) becomes: \((f-g)(x) = \sqrt{x-2} - \frac{1}{x}\). Domain: 1. \(x-2 \geq 0 \Rightarrow x \geq 2\) 2. \(x eq 0\) Combined domain: \(x \gt 2\)

Step 4: Define and Find Domain of \(f \cdot g\)

The function \((f \cdot g)(x) = f(x) \cdot g(x)\) becomes: \((f \cdot g)(x) = \sqrt{x-2} \cdot \frac{1}{x}\). Domain: 1. \(x-2 \geq 0 \Rightarrow x \geq 2\) 2. \(x eq 0\) Combined domain: \(x \gt 2\)

Step 5: Define and Find Domain of \(f/g\)

The function \((f/g)(x) = \frac{f(x)}{g(x)}\) becomes: \((f/g)(x) = \frac{\sqrt{x-2}}{\frac{1}{x}} = x \sqrt{x-2}\). Domain: 1. \(x-2 \geq 0 \Rightarrow x \geq 2\) 2. \(x eq 0\) Combined domain: \(x \gt 2\)

Step 6: Define and Find Domain of \(g/f\)

The function \((g/f)(x) = \frac{g(x)}{f(x)}\) becomes: \((g/f)(x) = \frac{\frac{1}{x}}{\sqrt{x-2}} = \frac{1}{x \sqrt{x-2}}\). Domain: 1. \(x-2 \gt 0 \Rightarrow x \gt 2\) 2. \(x eq 0\) Combined domain: \(x \gt 2\)

Step 7: Define and Find Domain of \(f \circ g\)

The function \((f \circ g)(x) = f(g(x))\) becomes: \((f \circ g)(x) = \sqrt{\frac{1}{x} - 2}\). Domain: 1. \(\frac{1}{x} - 2 \geq 0 \Rightarrow \frac{1}{x} \geq 2 \Rightarrow x \leq \frac{1}{2}\) 2. \(x eq 0\) Combined domain: \(0 \lt x \lt \frac{1}{2}\)

Step 8: Define and Find Domain of \(g \circ f\)

The function \((g \circ f)(x) = g(f(x))\) becomes: \((g \circ f)(x) = \frac{1}{\sqrt{x-2}}\). Domain: 1. \(x-2 \gt 0 \Rightarrow x \gt 2\) Combined domain: \(x \gt 2\)

Key Concepts

function operations

Function operations allow us to combine two or more functions using basic arithmetic operations like addition, subtraction, multiplication, and division. Imagine you have two functions, say \( f(x) \) and \( g(x) \). You can create a new function by adding these two together, subtracting one from the other, multiplying them, or even dividing them.

Here's what that looks like:

• \((f+g)(x) = f(x) + g(x)\)
• \((f-g)(x) = f(x) - g(x)\)
• \((f \cdot g)(x) = f(x) \cdot g(x)\)
• \((f / g)(x) = \frac{f(x)}{g(x)}\)

But performing these operations isn't enough. You also need to consider the domain of the new function. The domain is the set of all possible input values (\(x\)) that produce a valid output. For example, if any function involves division by zero, those values must be excluded from the domain. Likewise, if a function requires taking the square root, the argument must be non-negative.

So, remember to check the rules for combining functions and always be mindful of the resulting domain.

domain of functions

The domain of a function is a critical concept in mathematics. It refers to all the input values (\(x\)) for which the function is defined.

For instance, consider the function \( f(x) = \sqrt{x - 2}\). The square root function requires its argument to be non-negative, so:
\( x - 2 \geq 0 \)
\( x \geq 2\)

This means the domain of \( f(x) \) includes all real numbers greater than or equal to 2.

Now, look at another function, \( g(x) = \frac{1}{x} \). Here, division by zero is undefined, so the domain excludes \( x = 0 \).

When you start combining functions using operations like addition or multiplication, you must take into account the domains of both functions. The resulting domain is usually the intersection of the original domains, excluding any values that do not satisfy both functions.

Being careful with domains ensures you're working with valid inputs, which keeps your calculations accurate.

composite functions

Composite functions are a way to combine two functions to create a new one. If you have two functions, \( f(x) \) and \( g(x) \), the composite function \( f \circ g \) means you first apply \( g \) to \( x \) and then apply \( f \) to the result. This is written as \( (f \circ g)(x) = f(g(x)) \).

Here's an example:
If \( f(x) = \sqrt{x - 2} \) and \( g(x) = \frac{1}{x} \), then:

• \((f \circ g)(x) = f(g(x)) = \sqrt{\frac{1}{x} - 2}\)
• To find the domain of this composite function, first consider the domain of \( g(x) \), which is all real numbers except zero.
• Then, you also have to ensure that the output of \( g(x) \) is a valid input for \( f(x) \). So, \( \frac{1}{x} - 2 \geq 0 \)
• This simplifies to \[ \frac{1}{x} \geq 2 \rightarrow x \leq \frac{1}{2} \]
• Combining these restrictions: \(0 < x < \frac{1}{2}\)

Another example is \( g \circ f \), which is written as \( (g \circ f)(x) = g(f(x)) \). Using the same functions:

• \((g \circ f)(x) = g(\sqrt{x - 2}) = \frac{1}{\sqrt{x - 2}} \)
• The domain of \( f \) is \( x \geq 2 \), and so \( x-2 > 0 \)

Composite functions are very useful but require careful consideration of the domain at every step.