Problem 59 - Essential Calculus

Question

$$ f^{\prime}(x)=2 e^{-x / 4}, \quad(0,1) $$

Step-by-Step Solution

Answer

The derivative of the function at (0,1) is 2.

Step 1: Identify the Function and the Point

The function is \(f'(x) = 2e^{-x/4}\) and the point is (0,1). The task is to evaluate the function at x=0.

Step 2: Substitute x with 0

Substitute x with 0 in the function: \(f'(0) = 2e^{-0/4}\).

Step 3: Evaluate the Expression

Simplify the expression, remembering that any number to the power of 0 equals 1, so \(e^{0} = 1\). So, \(f'(0) = 2e^0 = 2 \cdot 1 = 2\)

Key Concepts

Derivative Calculation

Understanding how to compute derivatives is a foundational skill in calculus. The derivative of a function at a specific point gives us the rate at which the function's value is changing at that point. It can be visualized as the slope of the tangent line to the function's graph at that given point. In our exercise, we deal with the derivative of an exponential function, \( f'(x) = 2e^{-x/4} \).

When you're asked to find \( f'(x) \) for a particular value of \( x \) like \( x=0 \), this means you need to plug \( 0 \) into the derivative formula to find the instantaneous rate of change of the function at \( x=0 \). The calculation is usually straightforward: substitute the value into all instances of \( x \) and simplify. Remember that the derivative itself is not the original function, but it tells us how rapidly the original function is changing at every point.

Exponential Functions

Exponential functions are powerful tools in mathematics, embodying growth and decay processes in real-world scenarios, like populations, interest rates, and natural phenomena. An exponential function typically has the form \( f(x) = a\cdot e^{bx} \), where \( e \) is Euler's number (approximately 2.71828), and \( a \) and \( b \) are constants determining the scale and growth rate, respectively.

In the context of our exercise, the function \( f'(x) = 2e^{-x/4} \) involves the exponential term \( e^{-x/4} \) which represents exponential decay. As \( x \) increases, the function's value decreases at a rate proportionate to its current value, reflecting a decay process. This characteristic makes exponential functions like this suitable to describe processes that decrease rapidly at first and then more slowly over time.

Function Evaluation at a Point

Function evaluation at a point simply means finding the value of a function for a particular input. It is fundamental to calculus as it lets you understand the behavior of functions in a precise and local perspective. To perform a function evaluation, like we see in our exercise, the specific value of \( x \) is substituted into the function, and then the expression is simplified.

It is important to note that even though in our situation the function represents a derivative, the process of evaluation remains the same. When we substitute \( 0 \) for \( x \) in \( f'(x) = 2e^{-x/4} \) and simplify, we get \( f'(0) = 2 \), which tells us the rate of change of the original function at \( x=0 \) is 2. This means, at the exact point \( x=0 \) on the graph of the original function \( f(x) \)—not given in our exercise—the slope of the tangent line, or the rate at which the function is increasing or decreasing, is 2.