Problem 22 - Precalculus Mathematics for Calculus

Question

Find \(f^{\prime}(a),\) where \(a\) is in the domain of \(f .\) $$f(x)=-\frac{1}{x^{2}}$$

Step-by-Step Solution

Answer

The derivative at the point \(a\) is \(f'(a) = \frac{2}{a^3}\).

Step 1: Understand the Problem

We are given a function \(f(x) = -\frac{1}{x^2}\) and need to find the derivative of the function, \(f'(x)\), and then evaluate this derivative at a specific point \(x = a\).

Step 2: Find the Derivative

To find \(f'(x)\), we need to differentiate the function \(f(x) = -\frac{1}{x^2}\). We will use the power rule for differentiation, which states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).First, rewrite \(f(x) = -x^{-2}\). Now, apply the power rule: \[f'(x) = -(-2)x^{-3} = 2x^{-3}\] or equivalently, \(f'(x) = \frac{2}{x^3}\).

Step 3: Evaluate the Derivative at \(a\)

Now that we have \(f'(x) = \frac{2}{x^3}\), substitute \(x = a\) to find \(f'(a)\): \[f'(a) = \frac{2}{a^3}\].

Key Concepts

Understanding the Power Rule

The power rule is a fundamental concept in calculus that simplifies the process of differentiation. It is particularly handy when dealing with polynomial functions or functions that can be rewritten to involve powers of a variable. The general form of the power rule states that if you have a function of the form \(f(x) = x^n\), the derivative of this function, \(f'(x)\), will be \(nx^{n-1}\). This means you multiply the power by the coefficient (if there is one), and reduce the power by one.

Let's consider our function \(f(x) = -\frac{1}{x^2}\). We first rewrite the function using the rule that \(x^{-n} = \frac{1}{x^n}\), giving us \(f(x) = -x^{-2}\). Now, applying the power rule: multiply \(-2\) by \(-1\) (from the negative sign) to get \(2\), and then reduce the power from \(-2\) to \(-3\). This gives us the derivative \(f'(x) = 2x^{-3}\).

• The power rule simplifies calculation of derivatives.
• It's crucial for polynomials and can be adapted for negative powers.
• Remember to adjust negative exponents correctly.

Differentiation Demystified

Differentiation is the process of finding a derivative, which represents the rate of change of a function with respect to a variable. It answers questions like how fast is something changing at any given point. In mathematical terms, this is like finding the slope of a function at any specific point.

In our example, \(f(x) = -\frac{1}{x^2}\), differentiating involves using the power rule, as previously discussed. By taking the derivative, \(f'(x)\), you obtain \(2x^{-3}\), which describes how the function \(f(x)\) changes as \(x\) changes. The derivatives help in understanding the graph's behavior, such as increasing or decreasing trends, and identifying maximum or minimum points.

• Derivatives indicate the rate of change.
• Use power rule to differentiate functions like polynomials.
• Understanding trends of the function through its derivative.

Evaluation at a Point

Once we've found the derivative, we often need to evaluate it at a specific point, which helps in understanding the function's behavior precisely at that point. This means substituting a specific value into our derivative function to get a numerical answer.

In the problem we are solving, after finding \(f'(x) = \frac{2}{x^3}\), we need to evaluate this at a point \(x = a\). By plugging \(a\) into the derivative, we find \(f'(a) = \frac{2}{a^3}\). This value represents the rate of change of \(f(x)\) when \(x\) is exactly equal to \(a\). Essentially, this gives you the slope of the tangent line to the function at \(x = a\).

• Evaluation at a point gives precise function behavior at specific value.
• Helps in practical applications like velocity in physics.
• Finds the slope of function curves at particular points.