Problem 28 - Precalculus: Mathematics for Calculus

Question

Find the following for the given function \(f:\) (a) \(f^{\prime}(a),\) where \(a\) is in the domain of \(f,\) and (b) \(f^{\prime}(3)\) and \(f^{\prime}(4)\) $$f(x)=-\frac{1}{x^{2}}$$

Step-by-Step Solution

Answer

(a) \( f'(a) = \frac{2}{a^3} \); (b) \( f'(3) = \frac{2}{27} \) and \( f'(4) = \frac{1}{32} \).

Step 1: Understand the Function

The given function is \( f(x) = -\frac{1}{x^2} \), which is a rational function where the denominator is \( x^2 \). This tells us that \( x \) cannot be 0 because it would make the denominator zero, so the domain of \( f \) is \( x eq 0 \).

Step 2: Differentiate the Function

To find \( f'(x) \), use the power rule for differentiation. Rewrite \( f(x) \) as \( f(x) = -x^{-2} \). The power rule states that \( \frac{d}{dx}[x^n] = nx^{n-1} \). Apply this to \( f(x) = -x^{-2} \):\[ f'(x) = (-1) \cdot (-2)x^{-3} = 2x^{-3} \]This can be rewritten as \( f'(x) = \frac{2}{x^3} \).

Step 3: Evaluate \( f'(a) \)

For part (a), where \( a \) is any point in the domain of \( f \), the derivative \( f'(a) \) can be expressed by replacing \( x \) with \( a \) in \( f'(x) \):\[ f'(a) = \frac{2}{a^3} \].

Step 4: Calculate \( f'(3) \)

Substitute \( x = 3 \) into the derivative \( f'(x) = \frac{2}{x^3} \) to find \( f'(3) \):\[ f'(3) = \frac{2}{3^3} = \frac{2}{27} \].

Step 5: Calculate \( f'(4) \)

Substitute \( x = 4 \) into the derivative \( f'(x) = \frac{2}{x^3} \) to find \( f'(4) \):\[ f'(4) = \frac{2}{4^3} = \frac{2}{64} = \frac{1}{32} \].

Key Concepts

Rational Functions

A rational function is a type of function that is expressed as the ratio of two polynomials. In mathematical terms, it's written as \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) eq 0 \). This condition on \( Q(x) \) ensures the function is defined for all values where the denominator is not zero.

For example, if we consider the function \( f(x) = -\frac{1}{x^2} \), it is a rational function because it can be expressed as \( \frac{-1}{x^2} \). The numerator is a constant polynomial (\(-1\)), and the denominator is \( x^2 \), a polynomial with a degree of 2.

Key aspects of rational functions to remember include:

• They can have vertical asymptotes, which occur where the denominator is zero.
• The domain of a rational function consists of all real numbers except where the denominator is zero.
• They can also have horizontal or oblique asymptotes, giving insight into the function's behavior as \( x \) approaches positive or negative infinity.

Understanding these properties helps in solving, analyzing, and graphing rational functions.

Power Rule

The power rule is a basic principle in differentiation used to find the derivative of functions of the form \( x^n \), where \( n \) is any real number. According to the power rule, the derivative of \( x^n \) is \( nx^{n-1} \). This rule allows for quick differentiation without the need for more complex techniques.

To apply the power rule to the function \( f(x) = -x^{-2} \), we follow these steps:

• Identify the exponent: here, \( n = -2 \).
• Differentiate using the power rule: the derivative is \( (-2)x^{-2-1} = -2x^{-3} \).
• Simplify the derivative: \( -2x^{-3} \) can be rewritten as \( \frac{-2}{x^3} \).

This rule is versatile and applies to many types of functions, making it an essential tool in calculus. Whenever you see a power of \( x \), think "power rule" for a quick way to find the derivative.

Domain of a Function

The domain of a function is a set of all input values (typically \( x \)-values) for which the function is defined. In simpler terms, it's all the numbers you can plug into a function without causing mathematical errors, such as division by zero or taking the square root of a negative number.

For the function \( f(x) = -\frac{1}{x^2} \), determining the domain involves finding all the values of \( x \) that do not make the denominator zero. Because the denominator \( x^2 \) equals zero when \( x = 0 \), we have to exclude this value.

• The domain of \( f(x) \) is \( x eq 0 \), which can also be expressed as all real numbers except zero.

Understanding the domain is crucial because it tells you where the function exists and can be evaluated. It connects directly to the concept of continuity, ensuring that every point within the domain is defined and smooth, as seen in rational and polynomial functions.