Problem 21 - 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. \(f(t)=5 t-9 t^{2}\)

Step-by-Step Solution

Answer

The derivative of the function is \(f'(t) = 5 - 18t\). Both the function and its derivative have the domain \(\mathbb{R}\).

Step 1: Understand the Function

The function given is a polynomial function, specifically a quadratic, given by \( f(t) = 5t - 9t^2 \). Our task is to find its derivative using the definition of the derivative which is the limit of the difference quotient.

Step 2: Apply the Definition of Derivative

The derivative of \( f(t) \), denoted as \( f'(t) \), using the definition is\[ f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h}. \]

Step 3: Calculate \( f(t+h) \)

Calculate \( f(t+h) \), where \( f(t) = 5t - 9t^2 \).\[f(t+h) = 5(t+h) - 9(t+h)^2 = 5t + 5h - 9(t^2 + 2th + h^2). \]

Step 4: Simplify \( f(t+h) - f(t) \)

Plug \( f(t+h) \) into the difference quotient:\[ f(t+h) - f(t) = (5t + 5h - 9t^2 - 18th - 9h^2) - (5t - 9t^2). \]This simplifies to:\[ 5h - 18th - 9h^2.\]

Step 5: Compute the Difference Quotient

Substitute \( f(t+h) - f(t) \) into the difference quotient:\[ \frac{5h - 18th - 9h^2}{h} = 5 - 18t - 9h.\]

Step 6: Take the Limit as \( h \to 0 \)

To find \( f'(t) \), take the limit of the difference quotient as \( h \to 0 \):\[f'(t) = \lim_{h \to 0} (5 - 18t - 9h) = 5 - 18t.\]

Step 7: Determine the Domains

The domain of the original function \( f(t) \) is all real numbers \( \mathbb{R} \) because it is a polynomial. The domain of its derivative \( f'(t) = 5 - 18t \) is likewise all real numbers \( \mathbb{R} \), as polynomials are defined everywhere on the real number line.

Key Concepts

Definition of Derivative

The concept of a derivative is fundamental in calculus. It represents the rate at which a function is changing at any given point. In simple terms, it shows how a small change in the input, like moving slightly along the x-axis, affects the output of the function.
Using calculus terms, we define the derivative of a function as the limit of the difference quotient as the change in the input, denoted as \( h \), approaches zero. This limit can be mathematically expressed as:

• \( f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h} \)

The definition emphasizes understanding how a function behaves at very small scales, giving insight into its rates of change.

Difference Quotient

The difference quotient is a formula used for approximating the derivative of a function. It involves calculating the slope of the secant line between two points on the function. These points are close together.
The difference quotient is given by:

• \( \frac{f(t+h) - f(t)}{h} \)

Here, \( f(t+h) \) represents the function's value at a small distance \( h \) from \( t \), and \( f(t) \) is the function's value at \( t \). By taking the limit of the difference quotient as \( h \) approaches zero, we move from the average rate of change to the instantaneous rate of change. This is essentially the definition of the derivative in action.

Polynomial Function

Polynomial functions are expressions involving variables raised to whole number powers, and they can include terms like \( t^2 \), \( t \), and constants such as 5 or -9.
The function in our exercise, \( f(t) = 5t - 9t^2 \), is a quadratic polynomial with a degree of 2. This means its graph is a parabola. Polynomial functions are very smooth and continuous, meaning there are no gaps, jumps, or breaks. They are defined everywhere over the real numbers.
This simplicity makes polynomial functions a favorite in calculus, as their derivatives can be calculated easily using the rules of differentiation.

Domain of a Function

The domain of a function defines the set of all possible input values (often \( t \) for time or \( x \) for a horizontal coordinate). It tells us where the function is valid and defined. For instance, the polynomial function \( f(t) = 5t - 9t^2 \) has its domain as all real numbers

• \( \mathbb{R} \)

This is because polynomial functions, like this one, don't have restrictions like divisions by zero or square roots of negative numbers.
For derivatives, the domain follows closely from the original function's domain. As such, the domain for the derivative \( f'(t) = 5 - 18t \) is also all real numbers \( \mathbb{R} \), maintaining the properties of being defined continuously over the same set.