Problem 20 - Single Variable Calculus

Question

Find the instantaneous rate of change of the given function when \(x=a .\) \(f(x)=\frac{1}{x-2} ; \quad a=1\)

Step-by-Step Solution

Answer

The instantaneous rate of change of the function \(f(x) = \frac{1}{x - 2}\) when \(x = a = 1\) is: \(f'(1) = -1\).

Step 1: Find the derivative of the function

To find the derivative of the function \(f(x) = \frac{1}{x - 2}\), we can rewrite the function as \(f(x) = (x - 2)^{-1}\), and now we use the power rule to find the derivative of the function with respect to x. The power rule states that \(\frac{d}{dx} (x^n) = n * x^{(n-1)}\), where n is a constant. Applying this rule to our function, we get: \(f'(x) = -1 * (x - 2)^{-2} * \frac{d}{dx}(x - 2)\) Since the derivative of the inner function (\(x - 2\)) with respect to x is 1, we have \(f'(x) = -1 * (x - 2)^{-2} * 1\) So, the derivative of the function is: \(f'(x) = -\frac{1}{(x - 2)^2}\)

Step 2: Evaluate the derivative at \(x = a\)

Now that we have the derivative of the function, we need to evaluate it at \(x = a = 1\). Substitute \(x = 1\) into the derivative: \(f'(1) = -\frac{1}{(1 - 2)^2}\) This simplifies to: \(f'(1) = -\frac{1}{(-1)^2}\) \(f'(1) = -\frac{1}{1}\) So, the instantaneous rate of change of the function when \(x = a = 1\) is: \(f'(1) = -1\)

Key Concepts

Derivative

A derivative is a fundamental concept in calculus representing how a function changes at a specific point. It gives us the instantaneous rate of change of the function's value with respect to changes in its input. For a function \( f(x) \), the derivative, denoted as \( f'(x) \) or \( \frac{df}{dx} \), provides the slope of the tangent line to the graph of the function at any point \( x \).

• The derivative indicates how the output value of a function changes as its input value changes.
• In practical terms, it can represent speed, growth rate, or other changing quantities.

Understanding derivatives allows us to predict and analyze the behavior of dynamic systems. It's a crucial tool for solving real-world problems where change is a constant factor.

Power Rule

The power rule is one of the simplest and most used rules for finding derivatives. It states that if you have a function in the form of \( x^n \), where \( n \) is any constant, its derivative is \( n \times x^{n-1} \). This rule makes it easy to differentiate functions that are powers of \( x \).
For example, if we have \( f(x) = x^3 \), the derivative using the power rule would be \( f'(x) = 3x^2 \).

• The power rule greatly simplifies finding derivatives, especially for polynomial functions.
• It’s fundamental for more complex differentiation techniques as well.

In our exercise, we use the power rule to find the derivative of \( f(x) = (x - 2)^{-1} \), resulting in \( f'(x) = -\frac{1}{(x - 2)^2} \). The power rule is essential for breaking down more complex terms into manageable parts.

Function Evaluation

Function evaluation involves substituting specific input values into a function to find the corresponding output. When we evaluate a function's derivative at a particular point, we determine the instantaneous rate of change at that point.
In the exercise, after finding the derivative \( f'(x) = -\frac{1}{(x - 2)^2} \), we evaluate it at \( x = 1 \) to determine the rate of change when \( x = a = 1 \).

• Plug \( x = 1 \) into the derivative: \( f'(1) = -\frac{1}{(1 - 2)^2} \).
• Simplify the expression to find \( f'(1) = -1 \).

This value tells us how quickly the function's output is changing at that particular \( x \) value.

Calculus

Calculus is a branch of mathematics focused on studying change. It comprises two main topics: differentiation and integration.
Differentiation, as seen in this exercise, deals with finding derivatives. It helps us understand how quantities change instantaneously.

• Calculus provides tools to solve problems involving motion, growth, and area.
• It's applicable in diverse fields such as physics, engineering, economics, and biology.

Integration, on the other hand, concerns summing parts to find whole quantities, like areas under curves. Understanding calculus is crucial for analyzing dynamic systems and modeling real-world phenomena efficiently.