Problem 37 - Calculus for Biology and Medicine

Question

The following limit represents the derivative of a function \(f(x)\) at the point \(x=a\) : $$\lim _{h \rightarrow 0} \frac{\frac{1}{(2+h)^{2}+1}-\frac{1}{5}}{h}$$ Find \(f\) and \(a\).

Step-by-Step Solution

Answer

The function is \(f(x) = \frac{1}{x^2 + 1}\) and the point is \(a = 2\).

Step 1: Recognize the Limit Definition of the Derivative

Recall that the derivative of a function at a point \(x = a\) is given by the limit \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \). The given limit can be compared to this definition to identify parts \(f(a+h)\) and \(f(a)\).

Step 2: Identify f(a+h) in the Numerator

In the expression \( \frac{1}{(2+h)^2 + 1} \), recognize this as \(f(a+h)\). Therefore, \(f(x) = \frac{1}{x^2 + 1}\) because if \( a + h = x \) and \( a = 2 \), then substituting back gives this part of the expression.

Step 3: Identify f(a) in the Numerator

The term \( \frac{1}{5} \) represents \(f(a)\). Substituting \(a = 2\) into \(f(x) = \frac{1}{x^2 + 1}\), we find \(f(2) = \frac{1}{5}\). This implies \(a = 2\).

Step 4: Verify Values of f and a

To verify, substitute \(a = 2\) into \(f(x) = \frac{1}{x^2 + 1}\): \(f(2) = \frac{1}{2^2 + 1} = \frac{1}{5}\). Both parts \(f(a+h)\) and \(f(a)\) match the given expression.

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus, capturing how a function changes at any given point. If you've ever wondered how fast a car accelerates or how the population of a city shifts, derivatives can help you analyze these rates of change. For instance, the derivative of a function at a point tells you the slope of the tangent to the function at that point. Think of a curve and a line that just touches it without crossing it. That line is tangent, and the slope of that line is given by the derivative. When you're given a limit like \[\lim _{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}\] it's a step closer to finding that slope at point \(a\). Understanding and applying the definition of a derivative through limits is crucial for exploring calculus topics further.

Function Analysis Basics

Function analysis involves breaking down a function to understand its behavior and properties. In the context of derivatives, you'll often start by analyzing a function to identify components like \(f(a+h)\) and \(f(a)\). For example, you might be given an expression like \( \frac{1}{(2+h)^2 + 1} \), where you need to match this with \(f(a+h)\). By recognizing patterns, you determine how the function behaves around specific points. Some steps to analyze functions include:

• Identifying the formula or expression involved.
• Calculating initial values such as \(f(2)\) for a given function \(f(x)\).
• Recognizing which terms represent small changes, such as \(h\), in the expression.

Analyzing a function to find its derivative gives you insights into how the function changes immediately at a point, providing a snapshot of its behavior.

Mastering Limit Calculations

Limits are the doorway to understanding change. In calculus, they allow you to find the value a function approaches as the input approaches a certain point. This is especially useful when you're dealing with scenarios where direct substitution doesn't work – like dividing by zero. For calculating derivatives using limits, you begin with:

• Setting up the expression using \(h\), which represents a small change.
• Applying the limit as \(h\) approaches zero to your difference quotient.
• Simplifying the expression if possible to find the slope accurately.

The limit equation \(lim _{h \rightarrow 0} \frac{\frac{1}{(2+h)^2+1} - \frac{1}{5}}{h}\) allowed us to find the derivative at \(x = 2\) and the specific function involved. By practicing these calculations, you develop a robust understanding of how limits lead to derivatives, making complex changes in functions easier to comprehend.