Problem 20 - Calculus for Biology and Medicine

Question

Compute \(f(c+h)-f(c)\) at the indicated point. Your answers will contain \(h\) as an unknown variable. \(f(x)=\frac{1}{x} ; c=-2\)

Step-by-Step Solution

Answer

The expression is \(\frac{h}{2(-2+h)}\).

Step 1: Substitute for f(c)

First, calculate the function value at the point given by substituting the value of \(c\) into the function \(f(x)\). Here, \(c = -2\), so we find \(f(-2)\). The function is \(f(x) = \frac{1}{x}\). Thus, \(f(-2) = \frac{1}{-2} = -\frac{1}{2}\).

Step 2: Substitute for f(c+h)

Next, substitute \(c+h\) into the function \(f(x)\) to find \(f(c+h)\). The function is \(f(x) = \frac{1}{x}\), so \(f(c+h) = \frac{1}{c+h}\). Substitute \(c = -2\) to obtain \(f(-2+h) = \frac{1}{-2+h}\).

Step 3: Subtract f(c) from f(c+h)

Now, compute \(f(c+h) - f(c)\) using the results from the first two steps. Substitute \(f(c+h) = \frac{1}{-2+h}\) and \(f(c) = -\frac{1}{2}\) into the expression: \[f(c+h) - f(c) = \frac{1}{-2+h} - \left(-\frac{1}{2}\right) = \frac{1}{-2+h} + \frac{1}{2}.\]

Step 4: Combine into a single fraction

To simplify \(\frac{1}{-2+h} + \frac{1}{2}\), find a common denominator, which is \((-2+h) \, (2)\). So, rewrite the terms: \[\frac{1}{-2+h} = \frac{2}{(2)(-2+h)}, \quad \frac{1}{2} = \frac{-2+h}{2(-2+h)}.\]Combine them:\[f(c+h) - f(c) = \frac{2}{(2)(-2+h)} + \frac{-2+h}{2(-2+h)} = \frac{2 + (-2+h)}{2(-2+h)} = \frac{h}{2(-2+h)}.\]

Key Concepts

Function evaluation

Function evaluation is a core topic in calculus, especially when dealing with rational functions like \( f(x) = \frac{1}{x} \). To evaluate a function, you simply plug a specific value into the given function. In this exercise, you start with \( c = -2 \). For the function \( f(x) = \frac{1}{x} \), the goal is to find \( f(-2) \).

• Plug \( -2 \) into the function: \( f(-2) = \frac{1}{-2} \).
• The result is \( -\frac{1}{2} \). This is because dividing 1 by -2 gives us -0.5, or \( -\frac{1}{2} \) in fraction form.

Evaluating functions helps understand how they behave at specific points. This is particularly useful in biology for calculating specific rates like growth or decay. Remember, you substitute the entire expression or number into every instance of \( x \) in your function, making sure your calculations are accurate.

Difference quotient

The difference quotient is an essential tool for understanding the rate of change of a function. It is used to approximate the derivative and involves two primary tasks: evaluating the function at \( f(c) \) and \( f(c+h) \). Here, \( h \) is a small increment that is typically approaching zero.To find \( f(c+h) - f(c) \), follow these steps:

• First, find \( f(c+h) \) by substituting \( c+h \) into the function. For our function, this means \( f(-2+h) = \frac{1}{-2+h} \).
• Next, use your previously computed \( f(-2) = -\frac{1}{2} \).
• Subtract the results to find the difference quotient: \( f(c+h) - f(c) = \frac{1}{-2+h} + \frac{1}{2} \).

The difference quotient provides a measure of how quickly a function's value changes as \( x \) changes by a small amount \( h \). This helps in identifying the slopes of the secant lines on the graph of the function, hinting at the larger concept of derivatives.

Rational functions

Rational functions, such as \( f(x) = \frac{1}{x} \), are defined as the ratio of two polynomials. These functions can have several properties like asymptotes and domains restricted by division by zero.One key feature of rational functions is that they are undefined where the denominator equals zero. For instance, in our function, \( f(x) = \frac{1}{x} \), it is undefined at \( x = 0 \). This makes it crucial to consider where rational functions break or how they behave near these points.

• The concept of rational functions also involves simplifying them by finding a common denominator. In this problem, you need a common denominator to simplify \( \frac{1}{-2+h} + \frac{1}{2} \).
• The combined denominator here is \( 2(-2+h) \), which helps to write a single fraction in simplified form: \( \frac{h}{2(-2+h)} \).

These simplifications aid in further calculations or integrations, which are pivotal in mathematical modeling for biology, such as understanding enzyme kinetics or population dynamics.