Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the definition $$f^{\prime}(c)=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$$ to find the indicated derivative. $$ f^{\prime}(3) \text { if } f(t)=t^{2}-t $$

Step-by-Step Solution

Answer

The derivative \(f'(3)\) is 5.

Step 1: Understand the Definition

The derivative of a function \(f(t)\) at \(t = c\) is defined as \(f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}\). This formula calculates the rate at which \(f(t)\) changes at the point \(t = c\).

Step 2: Identify the Function and Point

The given function is \(f(t) = t^2 - t\), and we need to find the derivative at \(t = 3\). Thus, \(c = 3\) in the definition.

Step 3: Substitute into the Difference Quotient

Calculate \(f(3+h) - f(3)\) first. We have \(f(t) = t^2 - t\), so \(f(3+h) = (3+h)^2 - (3+h)\) and \(f(3) = 3^2 - 3\). Substitute these into the difference quotient: \(\frac{f(3+h) - f(3)}{h}\).

Step 4: Simplify \(f(3+h)\) and \(f(3)\)

Calculate \(f(3+h) = (3+h)^2 - (3+h) = 9 + 6h + h^2 - 3 - h = h^2 + 5h + 6\). We also know \(f(3) = 9 - 3 = 6\). Thus, \(f(3+h) - f(3) = h^2 + 5h + 6 - 6 = h^2 + 5h\).

Step 5: Form and Simplify the Difference Quotient

The difference quotient becomes \(\frac{h^2 + 5h}{h}\). We can simplify this to \(h + 5\) by canceling \(h\) in the numerator and denominator (provided \(h eq 0\)).

Step 6: Compute the Limit

Apply the limit as \(h \to 0\) to \(h + 5\). Since it's a simple linear expression, the limit is straightforward: \(\lim_{h \to 0} (h + 5) = 5\).

Step 7: State the Derivative

Since the limit of the difference quotient as \(h \to 0\) is \(5\), we find that the derivative of \(f(t) = t^2 - t\) at \(t = 3\), \(f'(3)\), is \(5\).

Key Concepts

Difference Quotient

The difference quotient is a foundational concept in calculus that is used to understand the rate of change of a function. In simple terms, it provides a way to measure how a function's output values change in response to varying its input values by small amounts. The formula for the difference quotient is given by:

• \( \frac{f(c+h) - f(c)}{h} \)

Here, \( f(c+h) \) and \( f(c) \) represent the function values at \( c+h \) and \( c \) respectively, and \( h \) is the change in the input value.
The key idea is to measure the average rate of change of the function over an interval \( h \). As \( h \) becomes smaller, this average rate more accurately reflects the function's behavior at \( c \). The difference quotient essentially forms the backbone of the derivative, highlighting changes in the function over extremely small intervals.

Limit Process

The limit process is crucial in calculus as it allows us to make approximations about a function's behavior more precise as our interval of observation narrows. When you take the limit of a difference quotient, you make \( h \) approach zero:

• \( \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} \)

This limit operation assesses what happens to the difference quotient as the changes in \( h \) practically disappear.
In practical terms, by shrinking the interval \( h \) towards zero, the limit helps us identify the instantaneous rate of change of the function at a particular point \( c \). For example, in the solved exercise, shrinking \( h \) to zero leads us to correctly conclude that \( f'(3) = 5 \). This process helps to convert an average rate into an exact, instantaneous rate of change.

Derivative Definition

The derivative of a function is one of the central concepts of calculus. It describes how a function changes at any given point, providing the rate of change or slope of the function's graph at that point. The formal definition of a derivative at a point \( c \) is expressed as:

• \( f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} \)

Understanding this definition involves recognizing that the derivative function, \( f'(t) \), is itself another function derived from the original. It encapsulates the original's changing behavior at each point across its domain.
This understanding allows us to directly compute the derivative of various functions by following a structured approach as seen in our solution steps. It’s also why derivatives are fundamental in diverse fields like physics and engineering where understanding rates of change is crucial.