Problem 25 - Calculus: Early Transcendentals

Question

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative. \( f(x) = x^2 - 2x^3 \)

Step-by-Step Solution

Answer

Derivative: \(f'(x) = -4x^2\). Domain of both \(f(x)\) and \(f'(x)\) is \(\mathbb{R}\).

Step 1: Definition of Derivative

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

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

Substitute \( x + h \) into \( f(x) \) to find \( f(x + h) \):\[f(x + h) = (x + h)^2 - 2(x + h)^3\]Simplify it to:\[(x^2 + 2xh + h^2) - 2(x^3 + 3x^2h + 3xh^2 + h^3)\]

Step 3: Simplify \( f(x + h) - f(x) \)

Subtract \( f(x) \) from \( f(x + h) \):\[f(x + h) - f(x) = (x^2 + 2xh + h^2 - 2(x^3 + 3x^2h + 3xh^2 + h^3)) - (x^2 - 2x^3)\]Distribute and simplify:\[= 2xh + h^2 - 2(3x^2h + 3xh^2 + h^3) \]\[= 2xh + h^2 - 6x^2h - 6xh^2 - 2h^3\]\[= -4x^2h - 4xh^2 - 2h^3\]

Step 4: Formulate the Difference Quotient

The difference quotient is:\[\frac{f(x+h) - f(x)}{h} = \frac{-4x^2h - 4xh^2 - 2h^3}{h}\]Simplify by cancelling an \( h \) from each term:\[= -4x^2 - 4xh - 2h^2\]

Step 5: Take the Limit as \( h \) Approaches 0

Calculate the limit as \( h \to 0 \):\[f'(x) = \lim_{{h \to 0}} (-4x^2 - 4xh - 2h^2) = -4x^2\]Therefore, the derivative of the function is:\[f'(x) = -4x^2\]

Step 6: Identify the Domain of the Function and Its Derivative

The given function \( f(x) = x^2 - 2x^3 \) is a polynomial, which is defined for all real numbers, hence its domain is \( \mathbb{R} \). Similarly, the derivative \( f'(x) = -4x^2 \) is also a polynomial and is defined for all real numbers, so its domain is \( \mathbb{R} \) too.

Key Concepts

Understanding Polynomial Functions

Polynomial functions are essential in mathematics. These functions are composed of variables with whole number exponents and coefficients. In our exercise, we are dealing with the polynomial function \( f(x) = x^2 - 2x^3 \).
A polynomial can be as simple as \( x^2 \) or more complex like \( x^2 - 2x^3 \). Each term in a polynomial is a combination of a coefficient and a variable raised to an exponent.

• The number in front of the variable is called the coefficient. For example, in \( -2x^3 \), \( -2 \) is the coefficient.
• The letter \( x \) is the variable.
• The number to which the variable is raised is the exponent. In \( -2x^3 \), \( 3 \) is the exponent.

Polynomial functions like ours are smooth and continuous. This smoothness makes them differentiable, allowing us to find their derivatives.

Exploring the Difference Quotient

The difference quotient is a crucial step when finding the derivative of a function. It is essentially the formula: \[ \frac{f(x+h) - f(x)}{h} \] This formula helps in measuring how much \( f(x) \) changes when \( x \) changes slightly by \( h \). For our function \( f(x) = x^2 - 2x^3 \),we compute \( f(x+h) \), subtract \( f(x) \), and then divide by \( h \). Each part helps us approach the instantaneous rate of change as \( h \) approaches zero.
The difference quotient is essential because it leads us to the definition of the derivative, allowing us to precisely quantify how a function's output changes with its input.

Delving into Limits in Calculus

Limits are foundational to calculus. They help us understand behavior at points that standard algebra cannot solve. When finding a derivative, we often need to calculate the limit as \( h \) approaches 0. This process was employed in our exercise to finalize the derivative: \[ \lim_{{h \to 0}} (-4x^2 - 4xh - 2h^2) = -4x^2 \] Limits in calculus allow us to evaluate functions at specific points, and identify behavior of functions as inputs change very slightly. Without limits, deriving the instantaneous rate of change would not be possible. In this problem, the limit takes us from a difference quotient to a clean equation: the derivative.

Determining the Domain of a Function

The domain of a function refers to all the input values for which the function is defined. It's the set of all possible \( x \)-values you can use without encountering undefined operations, such as division by zero. Our function \( f(x) = x^2 - 2x^3 \) is polynomial.
Polynomials are well-behaved functions:

• They have domains that include all real numbers.
• There are no restrictions like divisions or square roots involving negative numbers

Therefore, both the function and its derivative, \( f'(x) = -4x^2 \), are defined everywhere on the real number line. So, the domain for each is \( \mathbb{R} \), the entire set of real numbers.