Problem 2 - University Calculus: Early Transcendentals

Question

Using the definition, calculate the derivatives of the functions in Exercises \(1-6 .\) Then find the values of the derivatives as specified. $$F(x)=(x-1)^{2}+1 ; \quad F^{\prime}(-1), F^{\prime}(0), F^{\prime}(2)$$

Step-by-Step Solution

Answer

\(F'(-1) = -4\), \(F'(0) = -2\), \(F'(2) = 2\).

Step 1: Write Down the Function

We are given the function \( F(x) = (x - 1)^2 + 1 \). We need to find its derivative using the definition of a derivative and calculate \( F'(x) \) at specific points.

Step 2: Set Up the Derivative Definition

The derivative of a function \( F(x) \) is defined as \( F'(x) = \lim_{h \to 0} \frac{F(x + h) - F(x)}{h} \). First, we need to compute \( F(x + h) \).

Step 3: Compute \( F(x + h) \)

Substitute \( x + h \) into the function: \[ F(x + h) = ((x + h) - 1)^2 + 1 = (x + h - 1)^2 + 1 \].

Step 4: Expand \( (x + h - 1)^2 \)

Expand \( (x + h - 1)^2 \) to get \( (x - 1)^2 + 2(x - 1)h + h^2 \). Thus, \( F(x + h) = (x^2 - 2x + 1) + 2(x - 1)h + h^2 + 1 \).

Step 5: Simplify \( F(x + h) - F(x) \)

Compute \( F(x + h) - F(x) = ((x^2 - 2x + 1) + 2(x - 1)h + h^2 + 1) - ((x-1)^2 + 1) = 2(x-1)h + h^2 \).

Step 6: Divide by \( h \) and Take the Limit

Now divide by \( h \): \( \frac{2(x-1)h + h^2}{h} = 2(x-1) + h \). Taking the limit as \( h \to 0 \), we get \( F'(x) = 2(x-1) \).

Step 7: Evaluate \( F'(x) \) at Specified Points

Substitute the specified points into the derivative:\( F'(-1) = 2((-1)-1) = 2(-2) = -4 \),\( F'(0) = 2(0-1) = 2(-1) = -2 \),\( F'(2) = 2(2-1) = 2(1) = 2 \).

Key Concepts

Derivative Definition

The derivative of a function represents how the function's output value changes relative to its input value. In mathematical terms, the derivative of a function at a given point provides the slope of the tangent line to the function's graph at that point. For the function \( F(x) = (x-1)^2 + 1 \), we aim to find its derivative using this fundamental concept. The derivative \( F'(x) \) is calculated via the limit definition of the derivative, which rigorously defines the slope as the limit of average rates of change as the intervals between points shrink towards zero.

Limit Process

The limit process is a cornerstone of calculus, fundamental to understanding derivatives. It involves computing the derivative of \( F(x) = (x-1)^2 + 1 \) using the foundational formula:

• \( F'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} \).

In this formula, \( h \) is a tiny increment added to \( x \). We first compute \( F(x+h) \), then find the difference \( F(x+h) - F(x) \), and finally divide by \( h \). Taking the limit as \( h \) approaches zero gives us the instantaneous rate of change at \( x \), which corresponds to the derivative \( F'(x) \). The execution of this limit reveals the specific algebra necessary to handle expressions in terms of \( x \) and \( h \).

Function Evaluation

Function evaluation involves plugging specific values into the derivative we derived. Once we have found \( F'(x) = 2(x-1) \), we calculate the derivative at particular points:

• For \( x = -1 \), \( F'(-1) = 2(-1-1) = -4 \).
• For \( x = 0 \), \( F'(0) = 2(0-1) = -2 \).
• For \( x = 2 \), \( F'(2) = 2(2-1) = 2 \).

These evaluations show how \( F \) behaves at those points: the slope is negative at \( x = -1 \), zero at \( x = 1 \), and positive at \( x = 2 \), revealing how the function's graph changes across different regions.

Algebraic Manipulation

Algebraic manipulation is key in deriving the function's derivative. After computing \( F(x+h) = (x+h-1)^2 + 1 \), we expand it to obtain:

• \( (x+h-1)^2 = x^2 - 2x + 1 + 2(x-1)h + h^2 \).

By subtracting \( F(x) = (x-1)^2 + 1 \) from \( F(x+h) \), we simplify to find \( 2(x-1)h + h^2 \). Dividing each term by \( h \) results in \( 2(x-1) + h \). As \( h \) approaches zero, it vanishes, leaving us with \( 2(x-1) \). This algebraic groundwork simplifies the broader calculus process, confirming the derivative calculation's accuracy.