Problem 18 - 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 \(g(x)\) $$ \begin{array}{l}{f(x)=x^{2}-1} \\ {g(x)=\frac{x}{x+1}}\end{array} $$

Step-by-Step Solution

Answer

(f+g)(x)=\frac{x^3+x^2-1}{x+1}, (f-g)(x)=\frac{x^3+x^2-x-1}{x+1}, (f\cdot g)(x)=\frac{x^3-x}{x+1}, \left(\frac{f}{g}\right)(x)=x^2+x-1-\frac{1}{x}.

Step 1: Addition Function

To find \((f+g)(x)\), simply add the two functions: \[ (f+g)(x) = f(x) + g(x). \] Substitute the given functions: \[ f(x) = x^2 - 1 \text{ and } g(x) = \frac{x}{x+1}. \] Thus, \[(f+g)(x) = (x^2 - 1) + \frac{x}{x+1}.\] To combine, the common denominator should be considered. The numerator becomes \[(x^2 - 1)(x+1) + x.\] Therefore, \[(f+g)(x) = \frac{x^3 + x^2 - x - 1 + x}{x+1} = \frac{x^3 + x^2 - 1}{x+1}.\]

Step 2: Subtraction Function

To find \((f-g)(x)\), subtract the functions: \[ (f-g)(x) = f(x) - g(x). \] Substitute the given functions: \[ f(x) = x^2 - 1 \text{ and } g(x) = \frac{x}{x+1}. \] Therefore, \[(f-g)(x) = (x^2 - 1) - \frac{x}{x+1}.\] The common denominator of \(g(x)\) should be used. Thus, the expression becomes \[(x^2 - 1)(x+1) - x.\] Simplifying, \[(f-g)(x) = \frac{x^3 + x^2 - x - 1 - x}{x+1} = \frac{x^3 + x^2 - x - 1}{x+1}.\]

Step 3: Multiplication Function

To find \((f \cdot g)(x)\), multiply the two functions: \[ (f \cdot g)(x) = f(x) \cdot g(x). \] Substitute the given functions: \[ f(x) = x^2 - 1 \text{ and } g(x) = \frac{x}{x+1}. \] Hence, \[(f \cdot g)(x) = (x^2 - 1) \cdot \frac{x}{x+1}.\] Distribute \(x\): \[(f \cdot g)(x) = \frac{x(x^2 - 1)}{x+1} = \frac{x^3 - x}{x+1}.\]

Step 4: Division Function

To find \(\left(\frac{f}{g}\right)(x)\), divide \(f(x)\) by \(g(x)\): \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}. \] Substitute the given functions: \[ f(x) = x^2 - 1 \text{ and } g(x) = \frac{x}{x+1}. \] This gives \[\left(\frac{f}{g}\right)(x) = \frac{x^2 - 1}{\frac{x}{x+1}}.\] To simplify, multiply by the reciprocal of \(g(x)\): \[\left(\frac{f}{g}\right)(x) = (x^2 - 1) \cdot \frac{x+1}{x} = \frac{(x^2 - 1)(x+1)}{x}.\] Expand the numerator: \[\left(\frac{f}{g}\right)(x) = \frac{x^3 + x^2 - x - 1}{x}.\] Simplifying further results in \[\left(\frac{f}{g}\right)(x) = x^2 + x - 1 - \frac{1}{x}.\]

Step 5: Final Result

All operations have been performed: \[(f+g)(x) = \frac{x^3 + x^2 - 1}{x+1}, \] \[(f-g)(x) = \frac{x^3 + x^2 - x - 1}{x+1}, \] \[(f \cdot g)(x) = \frac{x^3 - x}{x+1}, \] \[\left(\frac{f}{g}\right)(x) = x^2 + x - 1 - \frac{1}{x}.\]

Key Concepts

Addition of Functions

When you add two functions together, it is represented as \((f+g)(x)\). This involves finding the sum of the two functions' expressions. In simple terms, if you have functions \(f(x)\) and \(g(x)\), adding them means combining their expressions:

• First, substitute each function's expression: for \(f(x)=x^2-1\) and \(g(x)=\frac{x}{x+1}\), you have \((f+g)(x) = (x^2 - 1) + \frac{x}{x+1}\).
• To add these terms, you should have a common denominator. Here, since \(g(x)\) is a fraction, find a common denominator to neatly add the expressions.
• The numerator becomes \((x^2 - 1)(x+1) + x\), and the result is \(\frac{x^3 + x^2 - 1}{x+1}\).

This result shows the new function after addition which is crucial in function operations.

Subtraction of Functions

Subtraction of functions is like addition, but this time, we're finding the difference between two functions. It's represented as \((f-g)(x)\) and involves subtracting one function from another, using the given expressions:

• Given \(f(x)=x^2-1\) and \(g(x)=\frac{x}{x+1}\), perform the operation \((f-g)(x) = (x^2 - 1) - \frac{x}{x+1}\).
• To subtract, use the common denominator approach used in addition, considering the fraction \(\frac{x}{x+1}\).
• The expression becomes \((x^2 - 1)(x+1) - x\), resulting in \(\frac{x^3 + x^2 - x - 1}{x+1}\).

This subtraction concept is useful when determining how functions differ as parts of broader mathematical tasks.

Multiplication of Functions

Multiplication of functions involves the product of each function's expressions, represented by \((f \cdot g)(x)\). This operation gives a new, multiplied function:

• Substitute the functions: \(f(x)=x^2-1\) and \(g(x)=\frac{x}{x+1}\).
• Multiply the two expressions: \((f \cdot g)(x) = (x^2 - 1) \cdot \frac{x}{x+1}\).
• When multiplied, it expands to \((x^3 - x)\), leading to \(\frac{x^3 - x}{x+1}\).

This multiplication formula helps when determining the influence of one function scaling another.

Division of Functions

Division of functions, noted as \(\left(\frac{f}{g}\right)(x)\), involves dividing one function by another, typically converting division into multiplication by a reciprocal:

• Take the given \(f(x)=x^2-1\) and \(g(x)=\frac{x}{x+1}\).
• Convert this into multiplying by the reciprocal: \(\left(\frac{f}{g}\right)(x) = \frac{x^2 - 1}{\frac{x}{x+1}}\) simplifies to \((x^2 - 1) \cdot \frac{x+1}{x}\).
• After expanding, it simplifies to \(\frac{x^3 + x^2 - x - 1}{x}\), which can be further simplified to \(x^2 + x - 1 - \frac{1}{x}\).

Understanding division in functions aids in scenarios where relative magnitude or change of functions is important.