Problem 23 - Essential Calculus Early Transcendentals

Question

\(23-36=\) Find the critical numbers of the function. $$f(x)=4+\frac{1}{3} x-\frac{1}{2} x^{2}$$

Step-by-Step Solution

Answer

The critical number is \(x = \frac{1}{3}\).

Step 1: Find the Derivative

To determine the critical numbers of the function, we first need to find its derivative. The function is given by \(f(x) = 4 + \frac{1}{3}x - \frac{1}{2}x^2\). The derivative is found by applying the power and constant multiple rules.The derivative of \(f(x)\) is:\[f'(x) = \frac{d}{dx}\left(4 + \frac{1}{3}x - \frac{1}{2}x^2\right) = 0 + \frac{1}{3} - x = \frac{1}{3} - x\].

Step 2: Set the Derivative to Zero

To find the critical numbers, set the derivative equal to zero and solve for \(x\):\[\frac{1}{3} - x = 0\].Rearrange the equation to solve for \(x\):\[x = \frac{1}{3}\].

Step 3: Conclusion on Critical Numbers

The critical number is the value of \(x\) where the derivative is zero or undefined. Since our derivative is defined for all \(x\), the critical number is the solution found in Step 2. Therefore, the critical number for the function is \(x = \frac{1}{3}\).

Key Concepts

Understanding Derivatives

A derivative is a fundamental concept in calculus. It represents the rate at which a function is changing at any given point. Think of it as the slope of the function's graph at a specific point. When we compute the derivative of a function, we are essentially finding a new function that gives us this rate of change. In the context of the original exercise, we found the derivative of the function \(f(x) = 4 + \frac{1}{3}x - \frac{1}{2}x^2\). This required knowing the subtle art of manipulating expressions to get the rate of change function: \(f'(x) = \frac{1}{3} - x\). Finding derivatives is crucial for identifying critical numbers, which help us understand where a function's graph might peak or dip, among other things.

To sum it up:

• Derivatives measure how a function changes at any point.
• They are used to find slopes of tangent lines and critical points on graphs.
• They are essential for analyzing trends and making predictions based on mathematical models.

The Power Rule

The power rule is a simple but powerful tool used to find derivatives of polynomial functions. If you have a term in the form of \(ax^n\), the power rule states that the derivative of this term is \(anx^{n-1}\). This rule helps to quickly evaluate the derivative of polynomial terms by reducing the power of \(x\) by one and multiplying it by the original coefficient.

For the function \(f(x) = 4 + \frac{1}{3}x - \frac{1}{2}x^2\), applying the power rule to the term \(-\frac{1}{2}x^2\) yields \(-x\), after reducing the power of \(x\) by one. Simple applications like these save time and simplify the derivative calculation.

Key points to remember:

• The power rule applies to terms where a variable is raised to a power.
• It simplifies the process of taking derivatives of polynomials.
• It's crucial in finding critical points efficiently by analyzing slopes.

Applying the Constant Multiple Rule

The constant multiple rule is another basic rule in calculus, and it complements the power rule nicely. This rule states that the derivative of a constant times a function is simply the constant times the derivative of that function. Mathematically, if \(c\) is a constant and \(f(x)\) is a function, then \(\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)\).

In our exercise, the constant multiple rule was used to compute the derivative of \(\frac{1}{3}x\) and \(\frac{1}{2}x^2\). For \(\frac{1}{3}x\), the derivative is simply \(\frac{1}{3}\) because the derivative of \(x\) is 1, and we just multiply by the coefficient \(\frac{1}{3}\). This rule makes the process straightforward, removing the need for more complex operations when handling constants.

Consider these takeaways:

• The constant multiple rule simplifies dealing with constant coefficients.
• It is essential when terms involve constants multiplying functions.
• It helps in streamlining calculations in calculus, particularly for polynomials.