Problem 58 - Precalculus

Question

Refer to the following: In calculus, we find the derivative, \(f^{\prime}(x),\) of a function \(f(x)\) by allowing \(h\) to approach 0 in the difference quotient \(\frac{f(x+h)-f(x)}{h}\) of functions involving exponential functions. Find the difference quotient of the exponential decay model \(f(x)=P e^{-k x},\) where \(P\) and \(k\) are positive constants.

Step-by-Step Solution

Answer

The difference quotient simplifies to \(-kP e^{-kx}\). This represents the derivative of the exponential decay model.

Step 1: Write the Difference Quotient Formula

Start with the difference quotient formula: \(\frac{f(x+h)-f(x)}{h}\). This formula is used to find the derivative of a function as \(h\) approaches 0.

Step 2: Substitute the Exponential Function

Substitute the given function \(f(x) = Pe^{-kx}\) into the difference quotient formula: \[\frac{Pe^{-k(x+h)} - Pe^{-kx}}{h}\].

Step 3: Simplify the Expression

Factor \(Pe^{-kx}\) out of the numerator: \[\frac{Pe^{-kx}(e^{-kh} - 1)}{h}\].

Step 4: Evaluate as h Approaches 0

To find the derivative \(f'(x)\), evaluate the limit as \(h\) approaches 0: \[\lim_{h \to 0} \frac{Pe^{-kx} (e^{-kh} - 1)}{h}\].

Step 5: Recognize the Limit of Exponential Function

The key limit to recognize here is \(\lim_{h \to 0} \frac{e^{-kh} - 1}{h} = -k\). This is a standard limit in calculus.

Step 6: Simplify the Limit

Substitute the recognized limit into the equation: \[Pe^{-kx} \cdot (-k)\]. Therefore, this simplifies to: \(-kP e^{-kx}\).

Key Concepts

Derivative

In calculus, the concept of a derivative holds pivotal importance. The derivative represents how a function changes as its input changes. This rate of change is effectively captured through the difference quotient, which forms the bedrock of differential calculus.

To understand derivatives thoroughly, think of them as the slope of a tangent line to the graph of the function at a particular point. As we refine this slope by making intervals smaller, we get the derivative.

The difference quotient is computed by the formula \(\frac{f(x+h) - f(x)}{h}\), where we let \(h\) approach zero.

• The expression \(f(x+h)\) evaluates the function at a slightly increased value.
• The subtraction \(f(x)\) gives the change in function value.
• Dividing by \(h\) gives us the average rate of change over the interval \([x, x+h]\).

This expression, in its limit form, helps us find the precise instantaneous rate of change, i.e., the derivative. Such computations are powerful in determining behavior of functions including exponential ones.

For the exponential function given, the derivative helps understand how rapidly or slowly the function value changes with respect to the input.

Exponential Functions

Exponential functions exhibit some intriguing characteristics that differentiate them from polynomial functions. An exponential function takes the form \(f(x) = a \cdot e^{bx}\), where \(a\) and \(b\) are constants, and \(e\) is the base of natural logarithms, approximately equal to 2.71828.

Key features of exponential functions include:

• An ever-increasing or decreasing rate, driven by the exponent's sign (positive for growth, negative for decay).
• The function value never reaches zero, aligning with its asymptotic nature towards the x-axis.
• Natural exponential functions are widely used in multiple fields such as finance, biology, and physics due to their modeling capacity for continuous growth or decay.

In the given problem, the function \(f(x) = P e^{-kx}\) indicates an exponential decay. Here, \(P\) is the initial value or starting point of the function, and \(k\) is the decay rate. Such exponential decay functions often model real-world phenomena like radioactive decay or cooling processes.

Calculus

Calculus is a vast field of mathematics centered around the ideas of change and motion. It is divided mainly into two subfields: differential calculus and integral calculus. In the exercise provided, differential calculus is used to find the derivative of an exponential function.

Differential calculus, for its part, focuses heavily on rates of change. By honing in on small-scale changes using limits, it provides a framework for modeling dynamic systems.

Key concepts in calculus that have been employed include:

• Limits, which allow us to examine the behavior of functions as inputs approach a specific point, essential in derivative calculations.
• Derivatives, pinpointing the "rate of change" or "slope" of a function at any particular input.

Learning calculus does not just equip students with mathematical tools; it opens up understanding of the world in motion and change. It gives insight into how seemingly simple concepts can have profound applications in science and engineering, offering crucial understanding in interpreting gradients and rates, as shown in the problem solution provided.