Problem 15 - Precalculus

Question

Use the given function \(f\) to find and simplify the following: \- \(f(3)\) \- \(f(4 x)\) \- \(f(x-4)\) \- \(f(-1)\) \- \(4 f(x)\) -\(f(x)-4\) \- \(f\left(\frac{3}{2}\right)\) \- \(f(-x)\) \- \(f\left(x^{2}\right)\) $$f(x)=\frac{x}{x-1}$$

Step-by-Step Solution

Answer

1. \( \frac{3}{2} \), 2. \( \frac{4x}{4x-1} \), 3. \( \frac{x-4}{x-5} \), 4. \( \frac{1}{2} \), 5. \( \frac{4x}{x-1} \), 6. \( \frac{-3x+4}{x-1} \), 7. 3, 8. \( \frac{x}{x+1} \), 9. \( \frac{x^2}{x^2-1} \).

Step 1: Find f(3)

To find \( f(3) \), substitute \( x = 3 \) into the function \( f(x) = \frac{x}{x-1} \): \[ f(3) = \frac{3}{3-1} = \frac{3}{2} \] Thus, \( f(3) = \frac{3}{2} \).

Step 2: Find f(4x)

To find \( f(4x) \), substitute \( x = 4x \) into the function: \[ f(4x) = \frac{4x}{4x - 1} \] The expression is already simplified. So, \( f(4x) = \frac{4x}{4x - 1} \).

Step 3: Find f(x-4)

To find \( f(x-4) \), substitute \( x = x - 4 \) into the function: \[ f(x-4) = \frac{x-4}{(x-4) - 1} = \frac{x-4}{x-5} \] Therefore, \( f(x-4) = \frac{x-4}{x-5} \).

Step 4: Find f(-1)

To find \( f(-1) \), substitute \( x = -1 \) into the function: \[ f(-1) = \frac{-1}{-1-1} = \frac{-1}{-2} = \frac{1}{2} \] So, \( f(-1) = \frac{1}{2} \).

Step 5: Find 4f(x)

To find \( 4f(x) \), multiply the function by 4: \[ 4f(x) = 4 \cdot \frac{x}{x-1} = \frac{4x}{x-1} \] Thus, \( 4f(x) = \frac{4x}{x-1} \).

Step 6: Find f(x) - 4

To find \( f(x) - 4 \), subtract 4 from the function: \[ f(x) - 4 = \frac{x}{x-1} - 4 \]Convert 4 to a fraction with the same denominator:\[ f(x) - 4 = \frac{x}{x-1} - \frac{4(x-1)}{x-1} = \frac{x}{x-1} - \frac{4x - 4}{x-1} \] Combine the fractions:\[ f(x) - 4 = \frac{x - (4x - 4)}{x-1} = \frac{x - 4x + 4}{x-1} = \frac{-3x + 4}{x-1} \] Therefore, \( f(x) - 4 = \frac{-3x + 4}{x-1} \).

Step 7: Find f(3/2)

To find \( f\left(\frac{3}{2}\right) \), substitute \( x = \frac{3}{2} \) into the function: \[ f\left(\frac{3}{2}\right) = \frac{\frac{3}{2}}{\frac{3}{2} - 1} = \frac{\frac{3}{2}}{\frac{1}{2}} = \frac{3}{2} \times 2 = 3 \] Thus, \( f\left(\frac{3}{2}\right) = 3 \).

Step 8: Find f(-x)

To find \( f(-x) \), substitute \( x = -x \) into the function: \[ f(-x) = \frac{-x}{-x-1} = \frac{x}{x+1} \] Therefore, \( f(-x) = \frac{x}{x+1} \).

Step 9: Find f(x^2)

To find \( f(x^2) \), substitute \( x = x^2 \) into the function: \[ f(x^2) = \frac{x^2}{x^2-1} \] The expression is already simplified. So, \( f(x^2) = \frac{x^2}{x^2-1} \).

Key Concepts

Understanding Function Evaluation

Evaluating a function means finding the function's output for a given input value. When you evaluate a function like \(f(x)=\frac{x}{x-1}\), you replace the variable \(x\) with the specific number or expression you're interested in. For example, if you want to find \(f(3)\), you substitute 3 into the function, resulting in \(f(3)=\frac{3}{3-1}=\frac{3}{2}\). Whenever you're faced with a new input, you just plug it into the function and perform the arithmetic operations accordingly.

• Identify the variable in the function.
• Substitute the given value for the variable.
• Simplify to get the answer.

Substituting different inputs will yield different outputs, making function evaluation a core process in understanding how functions behave.

Concept of Function Substitution

Function substitution involves replacing the variable in a function with another expression or number. This is a frequent math operation used to simplify or evaluate different forms of a function. For example, in the function \(f(x) = \frac{x}{x-1}\), replacing \(x\) with \(4x\) gives you \(f(4x) = \frac{4x}{4x-1}\). Here, instead of a single number, an expression \(4x\) is substituted for \(x\).

If you substitute \(x\) with \(x-4\), you find \(f(x-4) = \frac{x-4}{x-5}\). The same principle applies regardless of what the substitution expression is.

• Replace \(x\) with the new expression.
• Reorganize the expression if needed.
• Simplify to find the result.

Function substitution allows us to explore the function's behavior when the input value is altered.

Mastering Fraction Simplification

Simplifying fractions is essential when working with rational functions. Simplification means writing the fraction in its simplest form, which might involve simplifying the numerator or the denominator or both. For instance, in \(f(x) - 4 = \frac{x}{x-1} - 4\), converting 4 to the fraction \(\frac{4(x-1)}{x-1}\) with a common denominator and then subtracting allows us to simplify the result to \(\frac{-3x + 4}{x-1}\).

This simplification involves:

• Finding a common denominator.
• Subtracting or adding fractions carefully.
• Re-organizing and simplifying the results.

Simplified fractions are more manageable and easier to interpret, making them valuable for deeper mathematical insights.

Exploring Polynomial Expressions

Polynomial expressions are mathematical expressions that involve variables raised to whole number powers, such as \(x^2\) or \(4x\). When dealing with rational functions, polynomial expressions often appear in either the numerator, the denominator, or sometimes both. With the function \(f(x) = \frac{x}{x-1}\), substituting \(x^2\) for \(x\) gives us another polynomial in the form of \(f(x^2) = \frac{x^2}{x^2-1}\).

Working with polynomial expressions involves a few steps:

• Identify the polynomials in the function.
• Manipulate them according to the rules of algebra.
• Simplify if possible.

These manipulations help in understanding how altering a function input with polynomials can impact the function's form and complexity.