Problem 20 - Calculus for Biology and Medicine

Question

Compute \(f(c+h)-f(c)\) at the indicated point. $$ f(x)=\frac{1}{x} ; c=-2 $$

Step-by-Step Solution

Answer

\( f(c+h) - f(c) = \frac{h}{2(-2+h)} \).

Step 1: Understand the Function

Given the function \( f(x) = \frac{1}{x} \), we need to compute the expression \( f(c+h) - f(c) \) specifically at the point \( c = -2 \). This means we'll substitute \( c = -2 \) in our function.

Step 2: Substitute for \( f(c) \)

First, we calculate \( f(c) \) with \( c = -2 \). So, we find \( f(-2) = \frac{1}{-2} = -\frac{1}{2} \).

Step 3: Substitute for \( f(c+h) \)

Next, substitute \( c+h \) into the function: \( f(c+h) = f(-2 + h) = \frac{1}{-2 + h} \).

Step 4: Compute the Difference

The next step is to compute the difference \( f(c+h) - f(c) \). Start by substituting the expressions we found: \( f(c+h) - f(c) = \frac{1}{-2 + h} + \frac{1}{2} \).

Step 5: Simplify the Expression

The expression \( \frac{1}{-2+h} + \frac{1}{2} \) can be simplified by combining it over a common denominator, which is \( 2(-2+h) \). This yields:$$f(c+h) - f(c) = \frac{2 + (-2+h)}{2(-2+h)} = \frac{h}{2(-2+h)}.$$

Key Concepts

Difference Quotient

The difference quotient is a foundational concept in calculus that helps us understand how functions vary. It's often represented as \( \frac{f(x+h) - f(x)}{h} \), which provides an approximate rate of change of the function at \( x \) when \( h \) is very small. In essence, the difference quotient is the average rate of change and forms the basis for the derivative—the instantaneous rate of change.

• To compute this, we evaluate the function at two points: one at \( x \) and the other at \( x + h \).
• In this exercise, we're interested in the numerator \( f(c+h) - f(c) \) of the difference quotient, particularly at the point \( c = -2 \).
• The process helps us gauge the slope of the function between two points, giving insight into its behavior.

Understanding the difference quotient is critical because it allows us to explore the growth trends and slopes of complex functions using basic algebra.

Rational Functions

Rational functions are ratios of polynomial functions, similar in behavior to the function \( f(x) = \frac{1}{x} \) used in this exercise. They are composed of a numerator and a denominator which are both polynomials.

• In our context, \( f(x) = \frac{1}{x} \) is one of the simplest rational functions with a polynomial numerator of degree zero and a denominator of degree one.
• Rational functions are vital in calculus because their analysis involves polynomial division and examination of asymptotic behavior.
• As seen in this problem, such functions can display interesting properties near their vertical asymptotes (where the denominator approaches zero).

Working with rational functions involves determining behavior as inputs approach values that make the denominator zero, exploring horizontal asymptotes, and simplifying complex expressions.

Function Evaluation

Function evaluation involves substituting values into a given expression to determine the function's output. This practice is crucial in solving many types of problems in calculus, such as the one presented in this exercise.

• Evaluating \( f(cd) \) means substituting \( c = -2 \) in \( f(x) = \frac{1}{x} \) to find \( f(-2) = -\frac{1}{2} \).
• We also evaluate \( f(c+h) \) as \( \frac{1}{-2+h} \), showing how the function changes with small increments \( h \).
• The evaluation process gives us direct insight into the function's behavior at specific points, helping us understand more intricate details like slopes and rates of change.

Mastering function evaluation is key for moving forward with more complex calculus tasks, as it provides the necessary outputs that form the basis for deeper analysis such as derivative computations.