Question

Simplify the difference quotient\(\frac{f(x)-f(a)}{x-a}\) for the following functions. $$f(x)=\frac{1}{x}-x^{2}$$

Short answer

The simplified difference quotient for the given function is $$\frac{f(x)-f(a)}{x-a} = \frac{a(1 - x^2) - x(1 - a^2)}{x(x - a)}$$.

Step-by-step solution

Write down the given function and difference quotient formula

The function given is: $$f(x) = \frac{1}{x} - x^2$$ The difference quotient formula is: $$\frac{f(x) - f(a)}{x - a}$$

Substitute the function into the difference quotient formula

We will now substitute the expression for f(x) into the difference quotient formula: $$\frac{(\frac{1}{x} - x^2) - (\frac{1}{a} - a^2)}{x - a}$$

Find a common denominator

In order to simplify the expression further, we need to find a common denominator for the fractions in the numerator: $$\frac{\frac{1 - ax^2 - x^3a}{ax} - \frac{1 - a^3x - a^2}{a}}{x - a}$$

Combine the fractions in the numerator and simplify

Combine the two fractions in the numerator: $$\frac{\frac{1 - ax^2 - x^3a - a^3x + a^4x^2}{ax}}{x - a}$$ Now, we can simplify the fraction further by factoring out an "a" in the numerator: $$\frac{\frac{a(1 - x^2) - x(1 - a^2)}{x}}{x - a}$$

Simplify the entire expression

Now we'll multiply the numerator and denominator by the reciprocal of the denominator to simplify the entire expression: $$\frac{a(1 - x^2) - x(1 - a^2)}{x(x - a)}$$ So the simplified difference quotient for the given function is: $$\frac{f(x)-f(a)}{x-a} = \frac{a(1 - x^2) - x(1 - a^2)}{x(x - a)}$$

Key concepts

Calculus

Calculus is an advanced branch of mathematics that deals with continuous change. It's divided mainly into two areas: differential calculus, which concerns the rate of change of functions, and integral calculus, which focuses on the accumulation of quantities. The concept of the difference quotient falls under differential calculus and provides a means of calculating the derivative—an essential tool expressing how a function changes as its input changes.

When looking at the difference quotient, \(\frac{f(x)-f(a)}{x-a}\), we're evaluating the ratio of the change in the function values to the change in the input values as the separation becomes infinitesimally small. This process leads to the derivative, which at its core represents an instantaneous rate of change and is deeply connected to real-world concepts like velocity and acceleration in physics.

Simplifying Expressions

In mathematics, simplifying an expression is the process of reducing it to its simplest form. This involves combining like terms, reducing fractions, and factoring, among other operations. When dealing with complex formulas, such as the difference quotient, simplification is crucial for better understanding and working with the expression.

For instance, in our problem, we simplify the difference quotient by finding a common denominator, combining fractions, and factoring. This process makes the expression manageable and prepares it for further operations, such as taking limits. Simplification is not just a mechanical process but a strategy to reveal the underlying structure of the expression, allowing us to work towards our goal, whether that's computing a derivative or evaluating a limit.

Limits

The concept of a limit is foundational in calculus. It involves assessing the behavior of a function as the input approaches a certain point. When a function approaches a distinct value as the input nears some point, we say the function has a limit at that point. Limits allow us to deal with situations where direct substitution isn't possible due to indeterminacy or infinity.

In the context of the difference quotient, limits play a pivotal role. To find the derivative of a function at a point, we take the limit as \( x \) approaches \( a \) of the difference quotient. This tells us how the function behaves instantaneously around \( a \). Although the textbook solution doesn't show the limit process, understanding that the simplification is a stepping stone to taking the limit helps clarify the ultimate goal of finding a derivative through the difference quotient.