Problem 33 - Precalculus: Mathematics for Calculus

Question

The given limit represents the derivative of a function \(f\) at a number \(a\). Find \(f\) and \(a\) $$\lim _{h \rightarrow 0} \frac{(1+h)^{10}-1}{h}$$

Step-by-Step Solution

Answer

Function is \(f(x) = x^{10}\); \(a = 1\).

Step 1: Interpret the Limit Expression

The given limit expression \( \lim _{h \rightarrow 0} \frac{(1+h)^{10}-1}{h} \) suggests that it represents the derivative of a function of the form \( f(x) = x^{10} \) evaluated at some point \( a \). The goal is to identify \( f(x) \) and \( a \).

Step 2: Recognize Structure

Recognize that \( \frac{(1+h)^{10}-1}{h} \) is in the form \( \frac{f(a+h) - f(a)}{h} \). Comparing it with the expression for the derivative, \( f(x) = x^{10}, \) and \( a = 1 \), since \( (1+h)^{10} \) corresponds to \( (1+h) \) raised to a power.

Step 3: Identify Function \( f(x) \)

From the expression \((1+h)^{10}\), it is clear that \( f(x) = x^{10} \). This is because the form \((1+h)^{10}\) represents \(f\) evaluated at \(1 + h\).

Step 4: Determine \( a \)

Since the original expression is \((1+h)^{10} - 1\), the function is evaluated at \(a = 1\), which means \(a = 1\) is the point where the derivative is evaluated.

Key Concepts

Derivative

The concept of a derivative is foundational in calculus, allowing us to understand how a function changes at any given point. Simply put, the derivative measures the rate at which a function's value changes as its input changes. Imagine you’re deciphering the speed of a car at a particular moment. That speed is like the derivative of the car’s position over time. In mathematical terms, the derivative of a function \( f(x) \) at a point \( a \) is given by:
\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\]
This formula expresses how \( f(x) \) changes around the specific point \( a \). It captures the instantaneous rate of change, or the slope of the tangent line to the curve at that point.
Understanding derivatives involves recognizing patterns and applying rules like the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \). By interpreting limit expressions like \( \lim _{h \rightarrow 0} \frac{(1+h)^{10}-1}{h} \), we are essentially finding the derivative of the function.

Limit

A limit in calculus is a value that a function "approaches" as the input "approaches" some value. Limits help us understand behaviors of functions at points that might not be explicitly defined. The notation \( \lim_{h \to 0} \) tells us what a function's behavior is tending toward as \( h \) gets arbitrarily close to zero.
Why do limits matter? They provide the foundation upon which calculus is built, especially in the definition of derivatives and integrals. Limits help in finding precise values for expressions that may otherwise look undefined, like dividing by zero scenarios.

• Limits analyze how functions behave near specific points.
• They are crucial for defining and understanding derivatives.

In the given exercise, the expression \( \lim _{h \rightarrow 0} \frac{(1+h)^{10}-1}{h} \) explores what happens to the function \( (1+h)^{10} \) as \( h \) tends to zero, giving insights into the derivative at that point.

Function Evaluation

Function evaluation refers to the process of finding the output of a function given an input value. In simplest terms, if \( f(x) \) represents a function, then evaluating this function at \( x = a \) means computing \( f(a) \). It's like putting something into a machine to see what comes out on the other side.
In calculus problems, especially when dealing with derivatives and limits, evaluating the function at certain points becomes essential to derive accurate results. For example, if \( f(x) = x^{10} \), evaluating \( f(x) \) at \( x = 1 + h \) gives us \((1+h)^{10}\). The expression \((1+h)^{10}-1\) in our problem involves function evaluation at two points: \( 1+h \) and \( 1 \). The difference \( f(1+h) - f(1) \), obtained by evaluating the function at these points, assists in understanding changes between these values.

Precalculus

Precalculus lays the groundwork for calculus by consolidating essential mathematical concepts such as algebraic operations, understanding functions, and graphing. It equips students with the skills to manipulate and transform functions, which is crucial when dealing with calculus problems like derivatives.

• Familiarity with powers and polynomials.
• An understanding of algebraic expressions.

These foundational skills from precalculus help to simplify complex calculus problems. Consider \( (1+h)^{10} \), which expands using binomial theorem knowledge often covered in precalculus. This helps simplify problems and find derivatives efficiently. For instance, knowing how to expand \( (1+h)^{10} \) can clarify why \( f(x) = x^{10} \) in the exercise. Being adept at these precalculus skills eases the transition to more advanced calculus concepts.