Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(f^{\prime}(3)\) if \(f(t)=t^{2}-t\)

Step-by-Step Solution

Answer

The derivative of the function at \(t = 3\) is 5.

Step 1: Understanding the Problem

We need to find the derivative of the function \(f(t) = t^2 - t\) and evaluate it at \(t = 3\). This means we're looking for \(f^{ ext{'}(3)}\).

Step 2: Find the Derivative

The given function is \(f(t) = t^2 - t\). To find its derivative, we use the power rule. Differentiating \(t^2\) gives \(2t\), and differentiating \(-t\) gives \(-1\). So, \(f'(t) = 2t - 1\).

Step 3: Evaluate the Derivative at the Given Point

Now, substitute \(t = 3\) into the derivative \(f'(t) = 2t - 1\) to find \(f'(3)\). Calculate it as follows: \[f'(3) = 2(3) - 1 = 6 - 1 = 5\].

Step 4: Verify the Derivative Calculation

Double-check the calculation: substitute \(t = 3\) into \(f'(t) = 2t - 1\). This gives \(2(3) = 6\), and subtract 1 to get 5. The value is correctly calculated as 5.

Key Concepts

Power Rule

The power rule is an essential tool in calculus for finding derivatives of functions. It's a simple formula that allows you to differentiate polynomial functions easily. The rule states that if you have a function of the form \(t^n\), where \(n\) is any real number, the derivative is \(nt^{n-1}\). For example, when differentiating \(t^2\), you multiply the exponent 2 by the coefficient (which is 1 here, often omitted) and then reduce the exponent by 1, resulting in \(2t^{2-1} = 2t\).

• Remember: Each term of a polynomial can be differentiated separately.
• Apply the rule: Multiply by the exponent, then decrease the exponent by one.

Using the power rule simplifies problems that might seem complicated at first. For instance, the derivative of \(t^2 - t\) becomes \(2t - 1\), as shown in the solution.

Evaluation of Functions

Evaluating functions is the process of finding the output of a function given a specific input. This involves substituting the input value into the function and performing the necessary calculations. In the context of derivatives, once you've determined the derivative function, you substitute the given point into this derivative. For instance, with the derivative \(f'(t) = 2t - 1\), to evaluate it at \(t = 3\), you replace \(t\) with 3, resulting in \(f'(3) = 2(3) - 1\). The calculation then simplifies to \(6 - 1\), giving the final result of 5.

• Steps for evaluation:
• Identify the function or derivative to evaluate.
• Substitute the given value into the function.
• Simplify the expression to find the result.

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the derivative of a function. The derivative represents the rate of change of a function with respect to its variable. Differentiation enables you to determine how a function behaves at any given point and find its slope at that point.

• Key Points of Differentiation:
• The derivative of a constant is zero.
• The derivative of \(x^n\) using the power rule is \(nx^{n-1}\).

Through differentiation, you can not only solve for slopes and tangent lines but also identify turning points and overall behavior of functions. Mastery in differentiation is crucial for understanding more advanced concepts in calculus.

Calculus Problem Solving

Calculus problem solving involves applying various techniques and rules to analyze and solve problems related to changes. This process often includes finding derivatives, evaluating functions at specific points, and interpreting results to understand physical phenomena or abstract mathematical concepts.

• Typical Steps in Calculus Problem Solving:
• Understand the problem and identify what is being asked.
• Choose the appropriate mathematical method or rule to apply.
• Perform calculations carefully and verify your results.

The given exercise encapsulates calculus problem-solving perfectly. We first find the derivative using differentiation rules, then substitute the needed value, which is a key part of evaluation. Finally, verifying each step ensures accuracy, demonstrating the efficacy and completeness in solving calculus problems.