Question

Simplify the difference quotients \(\frac{f(x+h)-f(x)}{h}\) and \(\frac{f(x)-f(a)}{x-a}\) by rationalizing the numerator. $$f(x)=-\frac{3}{\sqrt{x}}$$

Short answer

Based on the given function \(f(x) = -\frac{3}{\sqrt{x}}\), we are asked to simplify two difference quotients. By rationalizing the numerator, we have found that: 1. \(\frac{f(x+h)-f(x)}{h} = \frac{x}{h\sqrt{x(x+h)}(\sqrt{x} + \sqrt{x+h})}\) 2. \(\frac{f(x)-f(a)}{x-a} = \frac{ax}{(x-a)\sqrt{ax}(\sqrt{a} + \sqrt{x})}\)

Step-by-step solution

1. Simplify the difference quotient \(\frac{f(x+h)-f(x)}{h}\) by rationalizing the numerator

First, let's substitute the function, \(f(x) = -\frac{3}{\sqrt{x}}\), into our difference quotient: $$\frac{f(x+h)-f(x)}{h} = \frac{-\frac{3}{\sqrt{x+h}} + \frac{3}{\sqrt{x}}}{h}$$ Now, we want to combine the fractions in the numerator of our difference quotient. To do this, we need a common denominator, which would be \(\sqrt{x(x+h)}\). So we have: $$\frac{f(x+h)-f(x)}{h} = \frac{-3\sqrt{x} + 3\sqrt{x+h}}{h\sqrt{x(x+h)}}$$ To rationalize the numerator, we'll multiply the top and bottom of the fraction by the conjugate of the numerator, which is \((3\sqrt{x} + 3\sqrt{x+h})\): $$\frac{f(x+h)-f(x)}{h} = \frac{-3\sqrt{x} + 3\sqrt{x+h}}{h\sqrt{x(x+h)}}\cdot\frac{3\sqrt{x} + 3\sqrt{x+h}}{3\sqrt{x} + 3\sqrt{x+h}}$$ When we multiply the numerators together, the result is: $$\left(-3\sqrt{x}\right)\left(3\sqrt{x}\right) - \left(-3\sqrt{x}\right)\left(3\sqrt{x+h}\right) + \left(3\sqrt{x+h}\right)\left(3\sqrt{x}\right) - \left(3\sqrt{x+h}\right)^2$$ Simplifying the numerator and multiplying the denominators together, we have: $$\frac{9x - 9x - 9h + 9x + 9h}{h\sqrt{x(x+h)}(3\sqrt{x} + 3\sqrt{x+h})}$$ Now we can simplify: $$\frac{9x}{h\sqrt{x(x+h)}(3\sqrt{x} + 3\sqrt{x+h})}$$ Now, we can divide the numerator and denominator by 9, which gives us the simplified difference quotient: $$\frac{x}{h\sqrt{x(x+h)}(\sqrt{x} + \sqrt{x+h})}$$

2. Simplify the difference quotient \(\frac{f(x)-f(a)}{x-a}\) by rationalizing the numerator

Now, we'll take our second difference quotient and substitute the function, \(f(x) = -\frac{3}{\sqrt{x}}\). We get: $$\frac{f(x)-f(a)}{x-a} = \frac{-\frac{3}{\sqrt{x}} + \frac{3}{\sqrt{a}}}{x-a}$$ Next, we'll get a common denominator for the fractions in the numerator: $$\frac{f(x)-f(a)}{x-a} = \frac{-3\sqrt{a} + 3\sqrt{x}}{\sqrt{ax}(x-a)}$$ To rationalize the numerator, we'll multiply the top and bottom of the fraction by the conjugate of the numerator, \((3\sqrt{a} + 3\sqrt{x})\): $$\frac{f(x)-f(a)}{x-a} = \frac{-3\sqrt{a} + 3\sqrt{x}}{\sqrt{ax}(x-a)}\cdot\frac{3\sqrt{a} + 3\sqrt{x}}{3\sqrt{a} + 3\sqrt{x}}$$ When we multiply the numerators together, the result is: $$\left(-3\sqrt{a}\right)\left(3\sqrt{a}\right) - \left(-3\sqrt{a}\right)\left(3\sqrt{x}\right) + \left(3\sqrt{x}\right)\left(3\sqrt{a}\right) - \left(3\sqrt{x}\right)^2$$ Simplifying the numerator and multiplying the denominators together, we have: $$\frac{9a - 9a + 9ax + 9x}{\sqrt{ax}(x-a)(3\sqrt{a} + 3\sqrt{x})}$$ Now we can simplify: $$\frac{9ax}{(x-a)\sqrt{ax}(3\sqrt{a} + 3\sqrt{x})}$$ Lastly, divide the numerator and denominator by 9 to obtain the simplified difference quotient: $$\frac{ax}{(x-a)\sqrt{ax}(\sqrt{a} + \sqrt{x})}$$

Key concepts

Difference Quotient

The difference quotient is an expression representing the average rate of change of a function over an interval. It is often written as \( \frac{f(x+h) - f(x)}{h} \) or \( \frac{f(x) - f(a)}{x-a} \). This formula is a foundational concept in calculus as it forms the basis for differentiation.

• When you calculate the difference quotient for \( f(x) = -\frac{3}{\sqrt{x}} \), you are effectively measuring how the function changes as \( x \) changes by a small amount \( h \) or \( x-a \).
• This expression simplifies by operations like addition and subtraction, but to precisely handle it, fractions must be combined by finding common denominators.
• The goal is to simplify the quotient such that it becomes easier to take a limit.

By simplifying the difference quotient, one finds an expression that helps in understanding the derivative of a function at a point.

Rationalizing the Numerator

Rationalizing the numerator involves transforming an expression to make the square roots or other irrational numbers vanish from the numerator. This technique is particularly useful in calculus when simplifying difference quotients.

• You often multiply both numerator and denominator by the conjugate of the numerator to achieve rationalization.
• The conjugate of a binomial \(a + b\sqrt{c}\) is \(a - b\sqrt{c}\), which helps to cancel out square roots in the numerator.
• After multiplying with the conjugate, differences of squares come into play simplifying the square roots.

This process helps in illuminating the overall structure of the expression and allows further steps, such as taking limits, to be carried out more smoothly.

Limits

The concept of a limit is fundamental in calculus and serves as the foundation for defining derivatives and integrals. It represents the value that a function approaches as the input approaches a certain point.

• Calculating the limit of the difference quotient as \( h \) or \( x - a \) approaches zero reveals the derivative of the function.
• Limits allow us to transition from average rates of change to instantaneous rates of change.
• In practice, simplifying the expression is vital for effectively finding the limit.

The process of evaluating limits transforms the difference quotient from a mere approximation to an exact rate of change, key to fully understanding the behavior of functions.

Derivative

A derivative represents an instantaneous rate of change of a function with respect to one of its variables, commonly denoted as \( f'(x) \). It is a core concept in calculus, developed from the limit of a difference quotient.

• The derivative tells us how a function is changing at any point and can be visualized as the slope of the tangent line to the function's graph at that point.
• This concept is widely used in many fields, such as physics for describing motion and in economics for cost and revenue changes.
• For the function \( f(x) = -\frac{3}{\sqrt{x}} \), once we simplify the difference quotient and take the limit, we extract the precise derivative function.

Understanding derivatives through the lens of difference quotients and limits bridges our understanding from intuitive to mathematical precision.