Question

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Find the difference quotient for \(f(x)=-\frac{1}{x^{2}} .\) Express the answer as a single fraction.

Short answer

\( \frac{2x + h}{x^{2}(x + h)^{2}} \)

Step-by-step solution

- Write the Difference Quotient Formula

Recall that the difference quotient is given by \( \frac{f(x + h) - f(x)}{h} \).

- Find \( f(x + h) \)

Substitute \( x + h \) into the function: \( f(x + h) = -\frac{1}{(x + h)^{2}} \).

- Substitute the Values into the Difference Quotient Formula

Substitute \( f(x) = -\frac{1}{x^{2}} \) and \( f(x + h) = -\frac{1}{(x + h)^{2}} \) into the difference quotient formula: \[ \frac{-\frac{1}{(x + h)^{2}} - \left( -\frac{1}{x^{2}} \right)}{h} \].

- Simplify the Numerator

Combine the terms in the numerator: \[ \frac{-\frac{1}{(x + h)^{2}} + \frac{1}{x^{2}}}{h} \].

- Combine the Fractions in the Numerator

Find a common denominator to combine the fractions: \[ \frac{-\frac{x^{2}}{x^{2}(x + h)^{2}} + \frac{(x + h)^{2}}{x^{2}(x + h)^{2}}}{h} \].

- Simplify the Combined Fraction in the Numerator

Combine the terms in the numerator: \[ \frac{-x^{2} + (x + h)^{2}}{x^{2}(x + h)^{2}h} \].Simplify the expression: \[ \frac{-x^{2} + x^{2} + 2xh + h^{2}}{x^{2}(x + h)^{2}h} \].Combine like terms: \[ \frac{2xh + h^{2}}{x^{2}(x + h)^{2}h} \].

- Factor and Simplify

Factor out \( h \) in the numerator: \[ \frac{h(2x + h)}{x^{2}(x + h)^{2}h} \].Cancel the \( h \) terms: \[ \frac{2x + h}{x^{2}(x + h)^{2}} \].

Key concepts

function evaluation

To understand the difference quotient, we first need to evaluate a function. Function evaluation is about finding the value of a function at a particular point. For instance, if we have a function, say, \( f(x) = -\frac{1}{x^2} \), and we need to find \( f(x+h) \), we substitute \( x+h \) in place of \( x \). This gives us \( f(x + h) = -\frac{1}{(x + h)^{2}} \). This step is crucial for the difference quotient because we need to evaluate the function at \( x \) and \( x+h \).

fraction simplification

Simplifying fractions is an essential skill, especially when dealing with algebraic expressions. Fractions can become complex, but with the right approach, they can be simplified easily. When combining fractions, like in the numerator of our difference quotient, finding a common denominator is the first step. For example, from \( \frac{-\frac{1}{(x + h)^{2}} + \frac{1}{x^{2}}}{h} \), we get a common denominator: \( x^{2}(x + h)^2 \). The next step is to rewrite each fraction with this common denominator, yielding \( \frac{-\frac{x^{2}}{x^{2}(x + h)^{2}} + \frac{(x + h)^{2}}{x^{2}(x + h)^{2}}}{h} \). Combining terms in the numerator, we can then simplify further until unnecessary terms, like \( h \), can be canceled out.

algebraic manipulation

Algebraic manipulation involves re-arranging expressions and equations to simplify or solve them. This often includes combining like terms, factoring, and reducing fractions. Let's look at how this works with our difference quotient: Starting with \( \frac{-x^{2} + x^{2} + 2xh + h^{2}}{x^{2}(x + h)^{2}h} \), we combine like terms to simplify the numerator: \( 2xh + h^{2} \). Then, we factor out common terms, in this case, \( h \): \( \frac{h(2x + h)}{x^{2}(x + h)^{2}h} \). By canceling \( h \) from the numerator and denominator, we simplify the expression to \( \frac{2x + h}{x^{2}(x + h)^{2}} \). Techniques like these are pivotal in solving complex expressions and need regular practice to master.