Problem 15 - Algebra 2

Question

Find \((f+g)(x),(f-g)(x),(f \cdot g)(x),\) and \(\left(\frac{f}{g}\right)\) for each \(f(x)\) and \(g(x)\) $$ \begin{array}{l}{f(x)=2 x-3} \\ {g(x)=4 x+9}\end{array} $$

Step-by-Step Solution

Answer

\((f+g)(x)=6x+6,\ (f-g)(x)=-2x-12,\ (f \cdot g)(x)=8x^2+6x-27,\ \left(\frac{f}{g}\right)(x)=\frac{2x-3}{4x+9}\).

Step 1: Find (f+g)(x)

To find \((f+g)(x)\), add the given functions \(f(x) = 2x - 3\) and \(g(x) = 4x + 9\). Combine like terms: \((2x) + (4x) = 6x\) and \((-3) + 9 = 6\). Thus, \((f+g)(x) = 6x + 6\).

Step 2: Find (f-g)(x)

To find \((f-g)(x)\), subtract \(g(x)\) from \(f(x)\). Start with \(f(x) = 2x - 3\) and \(g(x) = 4x + 9\) and perform the subtraction: \((2x) - (4x) = -2x\) and \((-3) - 9 = -12\). So, \((f-g)(x) = -2x - 12\).

Step 3: Find (f⋅g)(x)

To find \((f \cdot g)(x)\), multiply \(f(x)\) and \(g(x)\). Use the formula \((a-b)(c+d) = ac + ad - bc - bd\) to expand: \((2x)(4x) = 8x^2\), \((2x)(9) = 18x\), \((-3)(4x) = -12x\), and \((-3)(9) = -27\). Combine the terms: \(8x^2 + (18x - 12x) - 27 = 8x^2 + 6x - 27\). Thus, \((f \cdot g)(x) = 8x^2 + 6x - 27\).

Step 4: Find (f/g)(x)

To find \(\left(\frac{f}{g}\right)(x)\), divide \(f(x)\) by \(g(x)\), giving \(\frac{2x - 3}{4x + 9}\). No further simplification is possible as the numerator and denominator have no common factors. Thus, \(\left(\frac{f}{g}\right)(x) = \frac{2x - 3}{4x + 9}\).

Key Concepts

Addition of Functions

To add two functions, simply combine their corresponding terms. Let's break it down with our example where we need to find \((f+g)(x)\). The functions given are \(f(x) = 2x - 3\) and \(g(x) = 4x + 9\). To perform the addition, we add the terms with the same variable components:

• Combine the \(x\) terms: \((2x) + (4x) = 6x\).
• Combine the constant terms: \((-3) + 9 = 6\).

So, the sum of these functions results in \((f+g)(x) = 6x + 6\).
This operation shows how each term contributes to the total, and it's as straightforward as adding numbers!

Subtraction of Functions

Subtraction of functions involves deducting one function's output from another. For the function \((f-g)(x)\), we will work with \(f(x) = 2x - 3\) and \(g(x) = 4x + 9\). Start by subtracting the terms:

• Subtract the \(x\) terms: \((2x) - (4x) = -2x\).
• Subtract the constant terms: \((-3) - 9 = -12\).

Combining these results gives us \((f-g)(x) = -2x - 12\).
It's like balancing scales by taking away from one side. Always ensure to change signs as needed when subtracting each component.

Multiplication of Functions

Multiplying functions can be thought of as expanding combinations of terms, including both variable and constant parts. For \((f \cdot g)(x)\), using \(f(x) = 2x - 3\) and \(g(x) = 4x + 9\), follow these steps:

• Multiply the first terms: \((2x)(4x) = 8x^2\).
• Cross-multiply the middle terms: \((2x)(9) = 18x\) and \((-3)(4x) = -12x\).
• Multiply the last terms: \((-3)(9) = -27\).

Next, add all the results together. The middle terms \(18x\) and \(-12x\) combine:\(8x^2 + (18x - 12x) - 27 = 8x^2 + 6x - 27\). Thus, \((f \cdot g)(x) = 8x^2 + 6x - 27\).
This embodies distributing and collecting like terms for simplification!

Division of Functions

Dividing functions involves placing one function's expression over another's. For \(\left(\frac{f}{g}\right)(x)\), divide \(f(x) = 2x - 3\) by \(g(x) = 4x + 9\):\[\left(\frac{2x - 3}{4x + 9}\right)\]This represents the quotient of two functions. Each function maintains its entirety, creating a rational expression.
Note that simplification isn't always possible if the numerator and denominator have no common factors.
Just remember that division requires extra care about domain restrictions, ensuring the denominator never equals zero to keep the function well-defined!