Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Using the limit de nition of the derivative, nd \(f^{\prime}(x)\) if \(f(x)=(x-1)^{2}\).

Step-by-Step Solution

Answer

The derivative \(f^{\prime}(x)\) of the function \(f(x)=(x-1)^{2}\) is \(2x\).

Step 1: Express \(f(x + \Delta x)\)

To find the derivative, we first need to find the expression for \(f(x + \Delta x)\). For our function \((x-1)^{2}\), we will substitute \(x + \Delta x\) in the place of \(x\). Hence, \(f(x + \Delta x) = ((x + \Delta x)-1)^{2} = (x + \Delta x - 1)^{2}\).

Step 2: Compute the difference quotient

Now we formulate the difference quotient which is \((f(x + \Delta x) - f(x))/\Delta x\). Replacing \(f(x + \Delta x)\) and \(f(x)\) with their actual expressions, it becomes: \(( (x + \Delta x - 1)^{2} - (x-1)^{2} )/\Delta x\)

Step 3: Simplify the difference of squares

The numerator of the difference quotient is a difference of squares, which can be simplified using the formula \(a^2-b^2 = (a-b)(a+b)\). When we apply this formula here, we get : \( [(x + \Delta x - 1)-(x-1)][(x + \Delta x - 1)+(x-1)]/\Delta x\). This simplifies to \(\Delta x * 2x/\Delta x\). Dividing by \(\Delta x\), we now get \(2x\).

Step 4: Find the derivative

The derivative \(f'(x)\) is the limit of the difference quotient as \(\Delta x\) approaches 0. As \(\Delta x\) doesn't appear in our simplified expression \(2x\), the limit as \(\Delta x\) approaches 0 is also \(2x\). Hence, \(f'(x) = 2x\)

Key Concepts

Difference Quotient

When learning calculus, the concept of the difference quotient is like observing the world through a zoom lens to understand local details, specifically how a function changes at a single point. It provides a way to calculate the slope of the secant line that passes through two points on the graph of a function. This slope is a snapshot of the function's behavior at those points.

The difference quotient is defined as:\[\begin{equation}\frac{f(x + \(\Delta x\)) - f(x)}{\(\Delta x\)}\end{equation}\]The numerator represents the change in the function's value as the input changes from \(x\) to \(x + \(\Delta x\)\), while the denominator is the change in the input itself. When simplified, this form of the function can reveal patterns in the function's behavior as \(\(\Delta x\)\) gets infinitely small, leading us toward the concept of the derivative. Simplification often involves expanding polynomials, factorization, and canceling out terms, aiming at a form that resolves any division by zero issues as \(\(\Delta x\)\) approaches zero.

Derivative of a Function

To comprehend the essence of change, you can liken the derivative of a function to capturing the exact moment the shutter of a camera closes—highlighting the instantaneous rate of change of the function concerning its input. Mathematically, it is defined as the limit of the difference quotient as \(\(\Delta x\)\) approaches zero.

Formally, the derivative \(f'(x)\) is given by:\[\begin{equation}\lim_{{\(\Delta x\) \to 0}} \frac{f(x + \(\Delta x\)) - f(x)}{\(\Delta x\)}\end{equation}\]When we take this limit, we find the slope of the tangent line to the curve at point \(x\). The derivative has profound implications in various fields, such as physics for velocity and acceleration, in biology for rates of reaction, and economics for marginal cost and revenue.

Simplifying Algebraic Expressions

Mathematics, in some ways, is an art form that relies on simplification to reveal the core elements of an algebraic statement. Simplifying algebraic expressions is a foundational skill in calculus and is imperative when dealing with the difference quotient and derivatives. Simplification can turn a complex, overwhelming problem into a more manageable one.

Simplifying involves processes like:

• Expanding products of polynomials,
• Combining like terms,
• Factoring,
• Canceling common factors in numerators and denominators.

These steps reduce the expression to its simplest form, making it easier to apply further operations such as taking limits. A common example is the difference of squares, the technique used in the solution provided to simplify \((x + \(\Delta x\) - 1)^2 - (x - 1)^2\) using the identity \(a^2 - b^2 = (a - b)(a + b)\).

Limits and Continuity

The concept of limits is the bridge that allows us to traverse the gap between the finite and the infinite, revealing the behavior of functions as inputs approach a certain value but never actually reach it. Limits are essential in defining the derivative and understanding continuity—a function's property that ensures its graph can be drawn without picking up the pen.

The limit of a function as \(x\) approaches a certain value \(a\) is the value that \(f(x)\) gets closer to as \(x\) comes nearer and nearer to \(a\). Continuity at a point means there's no abrupt change in the function's value, which is a necessary condition for a function to be differentiable at that point.

• If a function \(f(x)\) is continuous at \(x = a\) and the limit of the difference quotient exists, then \(f\) is differentiable at \(a\).
• However, a function being continuous doesn't always guarantee its differentiability. The function must also have a defined slope at that point, which means it must not have any corners or cusps.

Understanding these concepts reveals how calculus can describe dynamic changes in a static framework of numbers and equations.