Lesson Objectives

- Learn how to Find the Difference Quotient

## How to Find the Difference Quotient

In this lesson, we will talk about the difference quotient. The difference quotient plays a crucial role in defining the derivative of a function in Calculus. The derivative provides a formula, in function form, for finding the slope of the tangent line to the graph of the function at a given point. Let's revisit our formula for the slope of a line. Let's suppose we have two points on a given line of P(x

Step 1) Replace x with x + h: $$f(x + h)=3(x + h)^2 - 4$$ $$=3(x^2 + 2xh + h^2) - 4$$ $$=3x^2 + 6xh + 3h^2 - 4$$ Step 2) We can now plug into our formula: $$\frac{f(x + h) - f(x)}{h}$$ $$=\frac{3x^2 + 6xh + 3h^2 - 4 - (3x^2 - 4)}{h}$$ $$=\frac{3x^2 + 6xh + 3h^2 - 4 - 3x^2 + 4}{h}$$ $$=\frac{6xh + 3h^2}{h}$$ $$=\require{cancel}\frac{\cancel{h}(6x + 3h)}{\cancel{h}}$$ $$=6x + 3h$$ Example #2: Find the Difference Quotient. $$f(x)=\sqrt{2 - x}$$ For this problem, there is an extra step that is required. We will want to cancel away the h in the denominator. The reason will become obvious when we get to Calculus. Notice that in the first example, this happened without any extra effort. Here, we will want to rationalize the numerator. To do that, recall the property from multiplying conjugates: $$(a + b)(a - b) = a^2 - b^2$$ Our goal is to find [f(x + h) - f(x)] / [h]. To do this, let's start by finding f(x + h).

Step 1) Replace x with x + h: $$f(x + h) = \sqrt{2 - (x + h)}$$ $$= \sqrt{2 - x - h}$$ Step 2) We can now plug into our formula: $$\frac{f(x + h) - f(x)}{h}$$ $$=\frac{\sqrt{2 - x - h} - \sqrt{2 - x}}{h}$$ Rationalize the numerator: $$=\frac{\sqrt{2 - x - h} - \sqrt{2 - x}}{h} \cdot \frac{\sqrt{2 - x - h} + \sqrt{2 - x}}{\sqrt{2 - x - h} + \sqrt{2 - x}}$$ $$=\frac{(2 - x - h) - (2 - x)}{h(\sqrt{2 - x - h} + \sqrt{2 - x})}$$ $$=\frac{2 - x - h - 2 + x}{h(\sqrt{2 - x - h} + \sqrt{2 - x})}$$ $$=\frac{-1 \cdot \cancel{h}}{\cancel{h}(\sqrt{2 - x - h} + \sqrt{2 - x})}$$ $$=\frac{-1}{\sqrt{2 - x - h} + \sqrt{2 - x}}$$ In this particular case, we would not rationalize the denominator.

Example #3: Find the Difference Quotient. $$f(x) = \frac{2x}{x - 1}$$ Our goal is to find [f(x + h) - f(x)] / [h]. To do this, let's start by finding f(x + h).

Step 1) Replace x with x + h: $$f(x + h) = \frac{2(x + h)}{x + h - 1}$$$$ = \frac{2x + 2h}{x + h - 1}$$ Step 2) We can now plug into our formula: $$\frac{f(x + h) - f(x)}{h}$$ $$=\frac{\large{\frac{2x + 2h}{x + h - 1} - \frac{2x}{x - 1}}}{h}$$ $$=\frac{\large{\frac{2x + 2h}{x + h - 1} - \frac{2x}{x - 1}}}{h} \cdot \frac{(x + h - 1)(x - 1)}{(x + h - 1)(x - 1)}$$ $$=\frac{(x - 1)(2x + 2h) - 2x(x + h - 1)}{h(x + h - 1)(x - 1)}$$ $$=\frac{2x^2 + 2xh - 2x - 2h - 2x^2 - 2xh + 2x}{h(x + h - 1)(x - 1)}$$ $$=\frac{-2 \cdot \cancel{h}}{\cancel{h}(x + h - 1)(x - 1)}$$ $$=\frac{-2}{(x + h - 1)(x - 1)}$$

_{1},y_{1}) and Q(x_{2},y_{2}): We can use our slope formula to calculate the slope of the line that passes through these given points: $$m=\frac{y_2 - y_1}{x_2 - x_1}, x_2 - x_1 ≠ 0$$ Now, let's change this notation up and use our function notation. So our leftmost point will now just be: $$P(x,f(x))$$ And let's say that the horizontal distance between the two points will be labeled as h. This means our rightmost point could be written as: $$Q((x + h), f(x + h))$$ Now using this new notation, we can show the slope as: $$m=\frac{f(x + h) - f(x)}{x + h - x}, h ≠ 0$$ We can further simplify this as: $$m=\frac{f(x + h) - f(x)}{h}, h ≠ 0$$ This is still the difference in y-values over the difference in x-values. We are just using function notation. This expression is known as the difference quotient.### The Difference Quotient:

$$\frac{f(x + h) - f(x)}{h}, h ≠ 0$$ Example #1: Find the Difference Quotient. $$f(x)=3x^2 - 4$$ Our goal is to find [f(x + h) - f(x)] / [h]. To do this, let's start by finding f(x + h).Step 1) Replace x with x + h: $$f(x + h)=3(x + h)^2 - 4$$ $$=3(x^2 + 2xh + h^2) - 4$$ $$=3x^2 + 6xh + 3h^2 - 4$$ Step 2) We can now plug into our formula: $$\frac{f(x + h) - f(x)}{h}$$ $$=\frac{3x^2 + 6xh + 3h^2 - 4 - (3x^2 - 4)}{h}$$ $$=\frac{3x^2 + 6xh + 3h^2 - 4 - 3x^2 + 4}{h}$$ $$=\frac{6xh + 3h^2}{h}$$ $$=\require{cancel}\frac{\cancel{h}(6x + 3h)}{\cancel{h}}$$ $$=6x + 3h$$ Example #2: Find the Difference Quotient. $$f(x)=\sqrt{2 - x}$$ For this problem, there is an extra step that is required. We will want to cancel away the h in the denominator. The reason will become obvious when we get to Calculus. Notice that in the first example, this happened without any extra effort. Here, we will want to rationalize the numerator. To do that, recall the property from multiplying conjugates: $$(a + b)(a - b) = a^2 - b^2$$ Our goal is to find [f(x + h) - f(x)] / [h]. To do this, let's start by finding f(x + h).

Step 1) Replace x with x + h: $$f(x + h) = \sqrt{2 - (x + h)}$$ $$= \sqrt{2 - x - h}$$ Step 2) We can now plug into our formula: $$\frac{f(x + h) - f(x)}{h}$$ $$=\frac{\sqrt{2 - x - h} - \sqrt{2 - x}}{h}$$ Rationalize the numerator: $$=\frac{\sqrt{2 - x - h} - \sqrt{2 - x}}{h} \cdot \frac{\sqrt{2 - x - h} + \sqrt{2 - x}}{\sqrt{2 - x - h} + \sqrt{2 - x}}$$ $$=\frac{(2 - x - h) - (2 - x)}{h(\sqrt{2 - x - h} + \sqrt{2 - x})}$$ $$=\frac{2 - x - h - 2 + x}{h(\sqrt{2 - x - h} + \sqrt{2 - x})}$$ $$=\frac{-1 \cdot \cancel{h}}{\cancel{h}(\sqrt{2 - x - h} + \sqrt{2 - x})}$$ $$=\frac{-1}{\sqrt{2 - x - h} + \sqrt{2 - x}}$$ In this particular case, we would not rationalize the denominator.

Example #3: Find the Difference Quotient. $$f(x) = \frac{2x}{x - 1}$$ Our goal is to find [f(x + h) - f(x)] / [h]. To do this, let's start by finding f(x + h).

Step 1) Replace x with x + h: $$f(x + h) = \frac{2(x + h)}{x + h - 1}$$$$ = \frac{2x + 2h}{x + h - 1}$$ Step 2) We can now plug into our formula: $$\frac{f(x + h) - f(x)}{h}$$ $$=\frac{\large{\frac{2x + 2h}{x + h - 1} - \frac{2x}{x - 1}}}{h}$$ $$=\frac{\large{\frac{2x + 2h}{x + h - 1} - \frac{2x}{x - 1}}}{h} \cdot \frac{(x + h - 1)(x - 1)}{(x + h - 1)(x - 1)}$$ $$=\frac{(x - 1)(2x + 2h) - 2x(x + h - 1)}{h(x + h - 1)(x - 1)}$$ $$=\frac{2x^2 + 2xh - 2x - 2h - 2x^2 - 2xh + 2x}{h(x + h - 1)(x - 1)}$$ $$=\frac{-2 \cdot \cancel{h}}{\cancel{h}(x + h - 1)(x - 1)}$$ $$=\frac{-2}{(x + h - 1)(x - 1)}$$

#### Skills Check:

Example #1

Find the difference quotient. $$f(x)=5x^2 - 4x + 3$$

Please choose the best answer.

A

$$3x + 7h - 3$$

B

$$10x + 5h - 4$$

C

$$8x + 9h + 7$$

D

$$-7x - 5h + 2$$

E

$$17x + 9h + 2$$

Example #2

Find the difference quotient. $$f(x)=9x - 7$$

Please choose the best answer.

A

$$7x - 3h$$

B

$$-3$$

C

$$9x - 4h$$

D

$$9$$

E

$$3$$

Example #3

Find the difference quotient. $$f(x)=\frac{1}{x - 19}$$

Please choose the best answer.

A

$$\sqrt{x - 19}+ 19h$$

B

$$x - 19xh - 3$$

C

$$\frac{15x}{(x + h - 19)(x - 19)}$$

D

$$-\frac{1}{(x + h - 19)(x - 19)}$$

E

$$-\frac{19}{(x - h - 19)(x - 19)}$$

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