Problem 42 - College Algebra

Question

Let \(f(x)=\frac{1}{x} \cdot\) Find a number \(c\) such that the average rate of change of the function \(f\) on the interval \((1, c)\) is \(-\frac{1}{4}\)

Step-by-Step Solution

Answer

The number \( c \) is \( \frac{4}{3} \).

Step 1: Understand the Average Rate of Change

The average rate of change of a function \( f(x) \) on an interval \((a, b)\) is given by the formula \( \frac{f(b) - f(a)}{b - a} \). Here, \( a = 1 \) and \( b = c \), and we want this rate to be \( -\frac{1}{4} \).

Step 2: Write the Expression for Average Rate of Change

Using the average rate of change formula: \( \frac{f(c) - f(1)}{c - 1} = -\frac{1}{4} \). Since \( f(x) = \frac{1}{x} \), we have \( f(c) = \frac{1}{c} \) and \( f(1) = 1 \). Substitute these into the formula: \( \frac{\frac{1}{c} - 1}{c - 1} = -\frac{1}{4} \).

Step 3: Solve for \(c\)

Solve the equation \( \frac{\frac{1}{c} - 1}{c - 1} = -\frac{1}{4} \). Simplify the numerator: \( \frac{1}{c} - 1 = \frac{1-c}{c} \). Plug this into the equation: \( \frac{\frac{1-c}{c}}{c-1} = -\frac{1}{4} \). Multiply both sides by \( c(c-1) \) to eliminate fractions: \( 4(1-c) = -c \).Simplify: \( 4 - 4c = -c \). Move terms around: \( 4 = 3c \). Divide by 3: \( c = \frac{4}{3} \).

Step 4: Verify the Solution

Substitute \( c = \frac{4}{3} \) back into the rate of change equation to verify: \( \frac{ \frac{1}{\frac{4}{3}} - 1 }{ \frac{4}{3} - 1 } = -\frac{1}{4} \). Simplify the expression: \( \frac{ \frac{3}{4} - 1 }{ \frac{1}{3} } = \frac{-\frac{1}{4}}{ \frac{1}{3} } = -\frac{1}{4} \). The left side equals the right side, confirming \( c = \frac{4}{3} \) is correct.

Key Concepts

Function Notation

Function notation is a way of representing functions in a standard form. It allows for clear communication and makes it easier to work with functions mathematically.
One common representation is the notation \(f(x)\), where \(f\) denotes the function and \(x\) is the independent variable. This notation shows that \(x\) is being input into the function \(f\) and produces an output or result based on the function's rule. For example, if \(f(x) = \frac{1}{x}\), it means the output is the reciprocal of the input \(x\).
Using function notation, you can easily substitute specific values for \(x\) to evaluate the function. For instance, if you want to find \(f(2)\), plug in \(x = 2\), so \(f(2) = \frac{1}{2}\). This simplicity and clarity make function notation a valuable tool in algebra and calculus.

Intervals in Mathematics

Intervals in mathematics are important concepts that refer to the range of values that a variable can take. They are denoted using round or square brackets.

• Round brackets, \((a, b)\), denote an open interval. It includes values between \(a\) and \(b\), but not \(a\) or \(b\) themselves.
• Square brackets, \([a, b]\), show a closed interval. This includes values between and including \(a\) and \(b\).

In the original exercise, the interval \((1, c)\) is an open interval which includes all values between 1 and \(c\) but not the endpoints.
Understanding how to use intervals helps in determining domains and ranges of functions, and also assists in analyzing behavior over specific ranges.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying algebraic expressions to solve equations.
Basic strategies involve:

• Combining like terms.
• Using the distributive property.
• Isolating variables to one side of the equation.

For the exercise, the aim was to simplify the expression \(\frac{\frac{1}{c} - 1}{c - 1} = -\frac{1}{4}\) to solve for \(c\). The manipulation steps involved:

• Rewriting the numerator expression \(\frac{1}{c} - 1\) as \(\frac{1-c}{c}\).
• Multiplying both sides of the equation by \(c(c-1)\) to clear the fractions.
• Rearranging terms and isolating \(c\) to find \(c = \frac{4}{3}\).

Algebraic manipulation is a powerful tool in mathematics, providing methods to simplify complex problems into manageable pieces.