Problem 55 - Calculus Volume 1

Question

Use the definition of a derivative to find \(f^{\prime}(x)\). $$ f(x)=2-3 x $$

Step-by-Step Solution

Answer

The derivative \( f'(x) = -3 \).

Step 1: Recall the Derivative Definition

The derivative of a function at a point is defined by the limit: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] We will use this definition to find the derivative of the function \( f(x) = 2 - 3x \).

Step 2: Calculate \( f(x + h) \)

We need \( f(x + h) \) for the definition of the derivative. If \( f(x) = 2 - 3x \), then substituting \( x + h \) for \( x \) gives: \[ f(x + h) = 2 - 3(x + h) = 2 - 3x - 3h \]

Step 3: Substitute into the Derivative Formula

Substitute \( f(x + h) \) and \( f(x) \) into the derivative formula: \[ f'(x) = \lim_{h \to 0} \frac{(2 - 3x - 3h) - (2 - 3x)}{h} \]

Step 4: Simplify the Expression

Simplify the expression inside the limit: \[ f'(x) = \lim_{h \to 0} \frac{2 - 3x - 3h - 2 + 3x}{h} \] This simplifies to: \[ f'(x) = \lim_{h \to 0} \frac{-3h}{h} \]

Step 5: Cancel Terms and Evaluate the Limit

Cancel \( h \) in the numerator and denominator: \[ f'(x) = \lim_{h \to 0} -3 \] Since the expression is now constant, the limit is simply \[ f'(x) = -3 \]

Key Concepts

Limit Definition in Calculus

The limit definition of a derivative is fundamental in calculus. It allows us to understand how functions behave as inputs approach a certain value. The derivative measures how a function changes; essentially, it's the function's slope at any given point.

For any function \( f(x) \), this definition is given by the formula: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]This formula represents the slope of the tangent line to the curve at point \( x \). As \( h \) approaches zero, we get the most accurate slope. This concept is foundational to many applications in mathematics and science.

To apply it, consider substituting \( x+h \) into the function. Then, calculate the difference \( f(x+h) - f(x) \). Finally, divide by \( h \) and take the limit as \( h \) approaches zero. With practice, using the limit definition becomes intuitive and straightforward.

Understanding Linear Functions

Linear functions are among the simplest and most important functions in algebra. A linear function can be expressed in the form: \[ f(x) = mx + b \]where \( m \) is the slope and \( b \) is the y-intercept.

For the function \( f(x) = 2 - 3x \), it is a linear function with a slope \( m = -3 \) and a y-intercept \( b = 2 \). The slope \( -3 \) means that for every unit increase in \( x \), the value of \( f(x) \) decreases by 3.

Linear functions have graphs that are straight lines. Their simplicity makes them useful for modeling directly proportional relationships. Understanding how the slope and y-intercept affect the line can help you quickly graph the function and solve related problems. In calculus, linear functions have constant derivatives, which simplifies their differentiation significantly.

Steps in Differentiation

Differentiating a function involves several systematic steps. Here, we'll outline these steps using the function \( f(x) = 2 - 3x \) as an example.

**1. Recall the formula:** The derivative of a function, using the limit definition, is described by
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{ h} \]

**2. Calculate \( f(x+h) \):** Substitute \( x+h \) into the function: \[ f(x + h) = 2 - 3(x + h) = 2 - 3x - 3h \]

**3. Plug into the formula:** Substitute into the derivative formula: \[ f'(x) = \lim_{h \to 0} \frac{2 - 3x - 3h - (2 - 3x)}{h} \]

**4. Simplify and reduce:** Simplify the expression: \[ f'(x) = \lim_{h \to 0} \frac{-3h}{h} \]

**5. Evaluate the limit:** Cancel \( h \) in the numerator and denominator, giving: \[ f'(x) = -3 \]

This process yields the derivative, \( f'(x) = -3 \), indicating the rate of change is constant for this linear function. Once these steps are understood, applying them to other functions becomes simpler.