Problem 2 - Intermediate Algebra

Question

For the functions \(f\) and \(g,\) find \(a .(f+g)(x), b .(f-g)(x)\) \(c .(f \cdot g)(x),\) and \(d .\left(\frac{f}{g}\right)(x) .\) $$f(x)=x+4, g(x)=5 x-2$$

Step-by-Step Solution

Answer

(f+g)(x)=6x+2, (f-g)(x)=-4x+6, (f \cdot g)(x)=5x^2+18x-8, (f/g)(x)=\frac{x+4}{5x-2}.

Step 1: Calculate \((f+g)(x)\)

The operation \((f+g)(x)\) means that you need to add the functions \(f(x)\) and \(g(x)\). So we have: \[\begin{align*} (f+g)(x) &= f(x) + g(x) \ &= (x+4) + (5x-2). \end{align*}\] Combine like terms: \[ (f+g)(x) = 6x + 2. \]

Step 2: Calculate \((f-g)(x)\)

The operation \((f-g)(x)\) involves subtracting the function \(g(x)\) from \(f(x)\). This gives us:\[\begin{align*} (f-g)(x) &= f(x) - g(x) \ &= (x+4) - (5x-2). \end{align*}\] Distribute the negative sign and combine like terms:\[ (f-g)(x) = x + 4 - 5x + 2 = -4x + 6. \]

Step 3: Calculate \((f \cdot g)(x)\)

The operation \((f \cdot g)(x)\) denotes the multiplication of the two functions. Calculate as follows:\[\begin{align*} (f \cdot g)(x) &= f(x) \cdot g(x) \ &= (x+4)(5x-2). \end{align*}\] Expand using the distributive property:\[ f \cdot g)(x) = x \cdot 5x + x \cdot (-2) + 4 \cdot 5x + 4 \cdot (-2) = 5x^2 - 2x + 20x - 8 = 5x^2 + 18x - 8. \]

Step 4: Calculate \(\left(\frac{f}{g}\right)(x)\)

For \(\left(\frac{f}{g}\right)(x)\), divide function \(f(x)\) by \(g(x)\):\[\begin{align*} \left(\frac{f}{g}\right)(x) &= \frac{f(x)}{g(x)} \ &= \frac{x+4}{5x-2}. \end{align*}\] This expression cannot be simplified further unless more information about the functions is provided.

Key Concepts

Addition of Functions

When working with the addition of functions, you take two functions and simply add their expressions together. Let’s consider the functions given in the exercise: \( f(x) = x+4 \) and \( g(x) = 5x-2 \). To find \((f+g)(x)\), you need to do the following:

• Identify the expressions for both functions, \( f(x) \) and \( g(x) \).
• Add them together: \((f+g)(x) = (x+4) + (5x-2)\).
• Combine like terms: \( 6x + 2 \).

So, the result of adding these two functions is \( 6x + 2 \). By combining terms, you make the expression easier to understand and work with in further calculations. It’s like collecting all similar objects into one basket, which simplifies the process significantly.

Subtraction of Functions

Subtraction of functions involves taking the expression of one function and subtracting the expression of the second function from it. Using our given functions, \( f(x) = x+4 \) and \( g(x) = 5x-2 \), you can find \((f-g)(x)\) as follows:

• Begin by writing out the subtraction: \( (f-g)(x) = (x+4) - (5x-2) \).
• Distribute the negative sign: \( x + 4 - 5x + 2 \).
• Combine like terms: \( -4x + 6 \).

It’s important to carefully handle the minus sign as it affects all the terms of the second function. On simplifying, you find that the result is \( -4x + 6 \). This step demonstrates how different operations modify relationships between the expressions.

Multiplication of Functions

To multiply functions, we multiply their expressions directly. Given \( f(x) = x+4 \) and \( g(x) = 5x-2 \), let’s see how you perform \((f \cdot g)(x)\):

• Write down how the functions combine: \( (f \cdot g)(x) = (x+4)(5x-2) \).
• Use the distributive property, often remembered as FOIL for binomials: multiply the first terms, the outside terms, the inside terms, and then the last terms.
• Simplify the multiplications: \[ x \cdot 5x + x \cdot (-2) + 4 \cdot 5x + 4 \cdot (-2) = 5x^2 - 2x + 20x - 8 \].
• Combine like terms: \( 5x^2 + 18x - 8 \).

Multiplying functions expands their expressions and introduces new terms, requiring careful simplification and attention to the order of operations. The result, \( 5x^2 + 18x - 8 \), illustrates how complex these expressions can become after multiplication.

Division of Functions

Dividing functions, like the given \( f(x) = x+4 \) and \( g(x) = 5x-2 \), creates a fraction of the two function expressions. To derive \( \left(\frac{f}{g}\right)(x) \):

• Set up the division: \( \left(\frac{f}{g}\right)(x) = \frac{x+4}{5x-2} \).
• Recognize that simplifying this can be limited if the expressions do not have common factors.

In this case, the expression \( \frac{x+4}{5x-2} \) remains as is, because there are no further simplifications possible without additional factors. It’s essential to leave the result in fraction form unless specific values allow further reduction. This concept highlights the intricacy of maintaining balance between numerators and denominators in functional division.