Problem 17 - 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. \(f(x)=-2 x+1 ; c=2\)

Step-by-Step Solution

Answer

\(f(c+h) - f(c) = -2h\).

Step 1: Identify the function and points

The function given is \( f(x) = -2x + 1 \). You need to compute \( f(c + h) - f(c) \) where \( c = 2 \).

Step 2: Calculate \( f(c) \)

Substitute \( c = 2 \) into the function \( f(x) = -2x + 1 \). \( f(2) = -2(2) + 1 = -4 + 1 = -3 \).

Step 3: Calculate \( f(c + h) \)

Substitute \( c + h \) into the function. \( f(c + h) = f(2 + h) = -2(2 + h) + 1 = -4 - 2h + 1 = -3 - 2h \).

Step 4: Compute \( f(c+h) - f(c) \)

Subtract \( f(c) \) from \( f(c+h) \). \( f(c+h) - f(c) = (-3 - 2h) - (-3) = -3 - 2h + 3 = -2h \).

Key Concepts

Difference Quotient

The difference quotient is a central concept in calculus. It is often used to find the derivative of a function, which represents the rate of change of the function. For a given function \( f(x) \), the difference quotient is typically given by the formula:

• \( \frac{f(x + h) - f(x)}{h} \)

This formula measures the average rate of change of the function over a small interval \( h \). Since \( h \) is a variable, as it approaches 0, the difference quotient approaches the derivative of \( f(x) \) at that point.
This exercise has you compute only the numerator of the difference quotient, which is \( f(c + h) - f(c) \). This forms the basis for calculating derivatives by giving insight into how the function behaves locally. It's like zooming in closely on a curve to see its slope at a specific point. In the context of the given problem, the focus is this numerator: \( f(c + h) - f(c) = -2h \), simplifying the internal workings of the difference quotient formula.

Function Evaluation

Function evaluation refers to the process of finding the output of a function for a given input. It's key for understanding how functions operate and display behavior at various points.
In the exercise, we evaluated the function \( f(x) = -2x + 1 \) at two different points: \( c \) and \( c + h \).

• At \( c = 2 \), simply plug \( 2 \) into the function: \( f(2) = -2(2) + 1 = -3 \).
• At \( c + h = 2 + h \), you substitute it into the function as: \( f(2 + h) = -2(2 + h) + 1 = -3 - 2h \).

Function evaluation allows us to comprehend what the function outputs for specific inputs. This is vital not just for manual calculations but also for understanding how to shift, stretch, or transform functions.

Linear Functions

Linear functions are a type of function defined by a linear equation, usually of the form \( f(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.
These functions graph as straight lines and have a constant rate of change, making them simple yet powerful tools in algebra and calculus.

• The slope \( m \) tells us how steep the line is; in the exercise, \( m = -2 \) means the line decreases 2 units vertically for each unit increase horizontally.
• The y-intercept \( b \) is where the line crosses the y-axis; for this function, \( b = 1 \).

Linear functions are important because they help build an understanding of more complex functions. By examining their straightforward slope and y-intercept structure, one can predict outputs for new inputs without much calculation. This exercise utilized these properties to determine changes in the function's value.

Limits

Limits in calculus explain how a function behaves as it approaches a certain point. They are foundational for defining both derivatives and integrals. In essence, a limit helps consider what happens to \( f(x) \) as \( x \) gets very close to a certain value.For the difference quotient, understanding limits is crucial because it represents the action of the formula when \( h \) approaches 0:

• If you have \( \frac{f(x + h) - f(x)}{h} \), you study what happens as \( h \to 0 \).
• This allows us to capture the derivative, which is the instantaneous rate of change at \( x \).

In practical terms, limits tackle the concept of approaching but never quite reaching a particular value, helping us handle situations that initially result in indeterminate forms like \( \frac{0}{0} \). They make derivatives possible, as the limit defines the exact slope of a curve at a single point.