Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

The given limit is a derivative, but of what function and at what point? (See Example 6.) $$ \lim _{h \rightarrow 0} \frac{2(5+h)^{3}-2(5)^{3}}{h} $$

Step-by-Step Solution

Answer

The limit represents the derivative of \( f(x) = 2x^3 \) at \( x = 5 \).

Step 1: Recall the Definition of a Derivative

The derivative of a function \( f(x) \) at a point \( x = a \) is given by the limit \[ \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]. This formula tells us how the limit given in the problem is related to the derivative of some function at some point.

Step 2: Identify the Function and Point X=a

Compare the given limit \( \lim_{h \to 0} \frac{2(5+h)^3 - 2(5)^3}{h} \) with the derivative definition. Notice that the limit has the structure of \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \), suggesting that \( f(x) = 2x^3 \) and \( a = 5 \).

Step 3: Conclude the Solution

Since the expression matches the structure of the derivative, we can conclude that the provided limit represents the derivative of the function \( f(x) = 2x^3 \) at the point \( x = 5 \). No further transformation or calculation is needed.

Key Concepts

Limit Definition of Derivative

The limit definition of the derivative is a foundational concept in calculus, used to find the derivative of a function at a specific point. This definition uses limits to explore how a function behaves when its input is changed by an infinitely small amount. Imagine that you have a function, say \( f(x) \), and you are interested in how this function changes as \( x \) changes. You do this by examining the change in function value, \( f(a+h) - f(a) \), divided by the change in \( x \), represented as \( h \), as \( h \) approaches zero.

Mathematically, this is expressed as:

\[\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\]

This calculation gives us the slope of the tangent line to the graph of the function at the point \( x = a \).

• Helps determine the instantaneous rate of change of the function.
• Important for understanding motion, slopes, and many more real-world phenomena.

In the given problem, identifying the function and the point helps understand exactly where this limit originates from. Once you match the structure of the given problem with the limit definition, as done with \( f(x) = 2x^3 \) and \( a = 5 \), it becomes clear which function and point were used.

Differentiation

Differentiation is a key concept in calculus that involves finding the derivative of a function. The process of differentiation finds how a function changes as its input changes, offering valuable information like slopes of curves, rates of change, and more.

In the problem mentioned, differentiation was already done through its limit form, helping to clarify that \( f(x) = 2x^3 \) at \( x = 5 \). Differentiation can be done using several methods:

• Power Rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
• Product Rule: \( (uv)' = u'v + uv' \).
• Quotient Rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).
• Chain Rule: \( (f(g(x)))' = f'(g(x))g'(x) \).

With the function \( 2x^3 \), applying the power rule, you'd get \( 6x^2 \), which aligns with recognizing \( a \) as a specific point of interest in the limit form of the derivative.

Calculus Problems

Calculus problems often test understanding of concepts like limits, derivatives, integrals, and their applications. Solving calculus problems brings together these ideas to analyze and interpret various mathematical scenarios, be it motion, growth, or decay over time.

This exercise, focusing on identifying which function corresponds to a given derivative limit, tests the skill of recognizing the application of the limit definition of derivatives. It's about matching an abstract form to a specific function and point.

To effectively solve such calculus problems, keep the following in mind:

• Understand the problem’s structure and what concept it relates to.
• Look for patterns that match known definitions and rules.
• Ensure precision in computation and a clear understanding of what each component represents.

These are crucial for mastering calculus and solving diverse equations efficiently, leading to stronger analytical skills and a deeper understanding of mathematical principles.