Problem 34 - Precalculus: Mathematics for Calculus

Question

The given limit represents the derivative of a function \(f\) at a number \(a\). Find \(f\) and \(a\) $$\lim _{x \rightarrow 5} \frac{2^{x}-32}{x-5}$$

Step-by-Step Solution

Answer

The function is \( f(x) = 2^x \) and the point is \( a = 5 \).

Step 1: Recognize the Limit Definition of the Derivative

The problem presents a limit expression that reminds us of the definition of the derivative of a function at a point. The derivative of a function \( f \) at \( a \) is given by: \[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \] In the given problem, \( \lim_{x \to 5} \frac{2^x - 32}{x - 5} \), recognize that it fits this form, with \( a = 5 \).

Step 2: Identify the Function and the Point

In the limit \( \lim_{x \to 5} \frac{2^x - 32}{x - 5} \), compare the numerator \( 2^x - 32 \) with \( f(x) - f(a) \). Notice that \( 32 \) can be expressed as \( 2^5 \), which allows us to identify the function and the constant: Here, we see \( f(x) = 2^x \) and \( f(a) = 2^5 = 32 \). Thus, \( f \) is the exponential function \( 2^x \), and \( a = 5 \).

Step 3: Verify the Identification

Confirm that the identified function and point satisfy the derivative limit. If \( f(x) = 2^x \), then \( f(5) = 2^5 = 32 \). Thus, substituting these into the definition of the derivative, we have: \[ f'(5) = \lim_{x \to 5} \frac{2^x - 32}{x - 5} \] This matches exactly with the given limit expression, confirming that our identified \( f(x) \) and \( a \) are correct.

Key Concepts

Limit Definition of Derivative

The concept of the limit definition of the derivative is essential for understanding how derivatives work in calculus. The derivative of a function at a specific point provides us with the slope of the tangent line at that point. This is crucial for understanding rates of change. In general, the limit definition of the derivative is expressed as \( f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \). This notation might appear complicated at first, but it breaks down as follows:

• \( f'(a) \) is the derivative of the function \( f \) at the point \( a \).
• The expression \( \lim_{x \to a} \) means we are looking at the value that \( \frac{f(x) - f(a)}{x - a} \) approaches as \( x \) gets closer to \( a \).

By applying this limit definition, you can find the derivative for any function at any point, simply by identifying \( f(x) \) and \( f(a) \) from the problem description, such as in the equation \( \lim_{x \rightarrow 5} \frac{2^{x}-32}{x-5} \). Here, you recognize \( a = 5 \) and that the expression fits the form of \( \frac{f(x) - f(a)}{x-a} \). Understanding this will be vital as you continue to work through calculus problems.

Exponential Functions

Exponential functions are a cornerstone in mathematics, widely used in various natural phenomena such as population growth, radioactive decay, and financial modeling. The function \( f(x) = 2^x \) illustrates an exponential function where the base is a constant (in this case, 2), and the exponent is a variable, \( x \). What makes exponential functions unique is their rate of growth or decay:

• Exponential functions grow or decay rapidly. For instance, doubling at each increment in the case of \( f(x) = 2^x \).
• The base of the exponential function determines the rate of growth or decay. If the base is greater than one, the function grows. If it's between zero and one, it decays.

In calculus, dealing with exponential functions often involves finding derivatives to understand how the function's rate of change behaves, as seen in the exercise involving the limit \( \lim_{x \rightarrow 5} \frac{2^{x}-32}{x-5} \). This involves comparing the exponential function with its evaluated value at a certain point (like \( a = 5 \)), helping to pinpoint its behavior at that point.

Calculus Problem Solving

Solving calculus problems effectively requires a structured approach and understanding of core principles. Whether you're dealing with derivatives, integrals, or limits, a few steps can facilitate the problem-solving process:

• Thoroughly read the problem to understand what's being asked.
• Identify the appropriate formula or principle, like using the derivative's limit definition.
• Substitute known values and simplify the expression. Look for patterns that match known forms.

For example, when solving \( \lim_{x \to 5} \frac{2^x - 32}{x - 5} \), begin by recognizing that it fits the form of the derivative's limit definition. Identify \( f(x) = 2^x \) and \( f(5) = 32 \), then confirm your findings by substituting back into the definition to validate the solution. Practicing this structured method not only solves problems more easily but also aids in understanding the underlying calculus concepts thoroughly.