Problem 11 - Brief Applied Calculus

Question

For each function, find: a. \(f^{\prime \prime}(x)\) and b. \(f^{\prime \prime}(3)\). $$ f(x)=\frac{1}{6 x^{2}} $$

Step-by-Step Solution

Answer

\(f''(x) = \frac{1}{x^4}\), and \(f''(3) = \frac{1}{81}\).

Step 1: Identify the Function

The given function is \(f(x) = \frac{1}{6x^2}\). It is a rational function, and we can rewrite it as a power function: \(f(x) = \frac{1}{6}x^{-2}\).

Step 2: Find the First Derivative

To find the first derivative, \(f'(x)\), use the power rule of differentiation. The power rule states that \(\frac{d}{dx}x^n = nx^{n-1}\). Apply this to \(f(x) = \frac{1}{6}x^{-2}\):\[f'(x) = \frac{1}{6} \cdot (-2)x^{-3} = -\frac{1}{3}x^{-3}\].

Step 3: Find the Second Derivative

Now find the second derivative, \(f''(x)\), by differentiating \(f'(x)\) using the power rule once more:\(f'(x) = -\frac{1}{3}x^{-3}\).Apply the power rule:\[f''(x) = -\frac{1}{3} \cdot (-3)x^{-4} = x^{-4} = \frac{1}{x^4}\].

Step 4: Evaluate the Second Derivative at a Specific Point

To find \(f''(3)\), substitute \(x = 3\) into the second derivative function:\[f''(3) = \frac{1}{3^4} = \frac{1}{81}\].

Key Concepts

Derivative

A derivative represents the rate at which a function changes at any given point. It is a fundamental concept in calculus. When you calculate a derivative, you are essentially finding how much the output of a function changes if you make a small change to the input. This is crucial for understanding various physical phenomena, such as speed, growth rates, and trends.To calculate a derivative, there are several rules and methods. The most common of these is the power rule, which we'll discuss next. Derivatives can be used to find the slope of a tangent line to a curve at any point, making it a powerful tool in both mathematics and the sciences. In our example, we worked with the function \(f(x) = \frac{1}{6x^2}\) to explore this concept.

Power Rule

The power rule is a simple yet powerful tool in calculus used to take the derivative of power functions. Its formula is \(\frac{d}{dx}x^n = nx^{n-1}\). This means to calculate the derivative, you multiply by the exponent and subtract one from it.

• If \(f(x) = x^3\), then \(f'(x) = 3x^2\).
• If \(f(x) = x^{-2}\), then \(f'(x) = -2x^{-3}\).

In the given exercise, the original function was a rational function that we expressed as a power function. Specifically, \(f(x) = \frac{1}{6}x^{-2}\) was differentiated using the power rule. This resulted in the first derivative, \(f'(x) = -\frac{1}{3}x^{-3}\). In calculus, the power rule simplifies the process of differentiation, especially when dealing with polynomial expressions.

Rational Function

A rational function is any function that can be expressed as the quotient of two polynomials. In its simplest form, \(\frac{p(x)}{q(x)}\), where both \(p(x)\) and \(q(x)\) are polynomials, and \(q(x) eq 0\).In our exercise, the function \(f(x) = \frac{1}{6x^2}\) is a type of rational function. Initially, these functions may seem complex due to their fraction-based nature. However, the strategy of rewriting them as power functions, like \(f(x) = \frac{1}{6}x^{-2}\), can simplify the process of differentiation and further analysis.Rational functions have important applications in fields such as economics, engineering, and physics, where they can model real-world relationships with varying degrees of complexity. Understanding how to differentiate rational functions is integral to solving calculus problems involving rates of change.

Second Derivative

The second derivative is the derivative of the derivative of a function. It gives us information about the curvature or the concavity of the original function.

• If the second derivative is positive at a certain point, the function is concave up, resembling a "cup."
• If negative, the function is concave down, like a "cap."

In our problem, after finding the first derivative \(f'(x) = -\frac{1}{3}x^{-3}\), we differentiated again to find the second derivative. By applying the power rule, we found \(f''(x) = \frac{1}{x^4}\).Calculating the second derivative at a specific point, \(f''(3)\), involves substituting \(x = 3\) into \(f''(x)\), resulting in \(f''(3) = \frac{1}{81}\). This helps us understand how the function behaves at \(x=3\) and can offer insight into its nature, such as how steeply it "bends" at that point.