Problem 3 - Intermediate Algebra

Question

For the functions \(f\) and \(g,\) find \(a .(f+g)(x), b .(f-g)(x)\) \(c .(f \cdot g)(x),\) and \(d .\left(\frac{f}{g}\right)(x) .\) $$ f(x)=x^{2}+1, g(x)=5 x $$

Step-by-Step Solution

Answer

a.\((f+g)(x)=x^2 + 5x + 1\), b.\((f-g)(x)=x^2 - 5x + 1\), c.\((f \cdot g)(x)=5x^3 + 5x\), d.\(\left(\frac{f}{g}\right)(x)=\frac{x^2 + 1}{5x}\)."

Step 1: Addition of Functions

The sum of the functions \((f+g)(x)\) is given by adding \(f(x)\) and \(g(x)\): \[(f+g)(x) = f(x) + g(x) = x^2 + 1 + 5x.\] Therefore, the expression for \((f+g)(x)\) is \(x^2 + 5x + 1\).

Step 2: Subtraction of Functions

The difference of the functions \((f-g)(x)\) is given by subtracting \(g(x)\) from \(f(x)\): \[(f-g)(x) = f(x) - g(x) = x^2 + 1 - 5x.\] Thus, the expression for \((f-g)(x)\) is \(x^2 - 5x + 1\).

Step 3: Multiplication of Functions

The product of the functions \((f \cdot g)(x)\) is found by multiplying \(f(x)\) by \(g(x)\): \[(f \cdot g)(x) = (x^2 + 1)(5x).\]Expanding this gives:\[f(x) \cdot g(x) = 5x(x^2 + 1) = 5x^3 + 5x.\]Therefore, \((f \cdot g)(x) = 5x^3 + 5x.\)

Step 4: Division of Functions

The division of the functions \(\left(\frac{f}{g}\right)(x)\) is given by dividing \(f(x)\) by \(g(x)\): \[\left(\frac{f}{g}\right)(x) = \frac{x^2 + 1}{5x}.\]This simplifies to \(\frac{x^2 + 1}{5x}\), where \(x eq 0\) to avoid division by zero.

Key Concepts

Addition of Functions

When we talk about the addition of functions, we are essentially combining two functions by adding their respective outputs. In mathematical notation, this is represented as

• \((f+g)(x) = f(x) + g(x)\)

For this exercise, we're given two functions:

• \(f(x) = x^2 + 1\)
• \(g(x) = 5x\)

To find \((f+g)(x)\), we simply add the expressions:

• \(x^2 + 1\) and \(5x\)

This results in the combined function:

• \((f+g)(x) = x^2 + 5x + 1\)

Adding functions helps summarize the effects both functions have on a particular input \(x\). This can be very beneficial in scenarios where multiple processes or changes affect a variable.

Subtraction of Functions

Subtraction of functions involves taking the output of one function and subtracting it from another. The formula looks like this:

• \((f-g)(x) = f(x) - g(x)\)

Using our functions:

• \(f(x) = x^2 + 1\)
• \(g(x) = 5x\)

We subtract the output of \(g(x)\) from \(f(x)\):

• \(x^2 + 1 - 5x\)

This simplifies to:

• \((f-g)(x) = x^2 - 5x + 1\)

Function subtraction is valuable when we want to find how much one quantity surpasses another over a range of inputs, which is often used in comparing differences between two processes in fields like economics and physics.

Multiplication of Functions

Multiplying functions involves finding the product of their outputs at any given \(x\). Mathematically, it is expressed as:

• \((f \cdot g)(x) = f(x) \cdot g(x)\)

For this exercise, the functions provided are:

• \(f(x) = x^2 + 1\)
• \(g(x) = 5x\)

We multiply these to get:

• \((f \cdot g)(x) = (x^2 + 1)(5x)\)

After expanding the expression, we have:

• \((f \cdot g)(x) = 5x^3 + 5x\)

Multiplying functions can be useful in representing complex scenarios where outputs need to interact multiplicatively, common in physics and engineering applications.

Division of Functions

Division of functions is about finding the ratio of their outputs. This is represented as:

• \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}\)

Given the functions:

• \(f(x) = x^2 + 1\)
• \(g(x) = 5x\)

The division is performed as follows:

• \(\left(\frac{f}{g}\right)(x) = \frac{x^2 + 1}{5x}\)

It's important to remember that in division of functions, the denominator function \(g(x)\) must not equal zero, to avoid undefined expressions. In this case, \(xeq 0\). Division of functions helps to determine the relative proportion or change between two different functions, often used in comparison studies in research and finance.