Module 1 Notes (Filled In Class) PDF

# MATH 1113 | PRECALCULUS ## Module 1: Functions ### Warm Up: Factor the following quadratic polynomials. **(a)** p(x) = x² + x - 30 ac = -30 b = 1 p(x) = (x + 6)(x - 5) FOIL to check. **(b)** q(z) = 5z² + 8z + 3 can use grouping: q(z) = 5z² + 5z + 3z + 3 ac = 5 *3 = 15 b = 8 = 5 + 3 = 5z [z...

# MATH 1113 | PRECALCULUS ## Module 1: Functions ### Warm Up: Factor the following quadratic polynomials. **(a)** p(x) = x² + x - 30 ac = -30 b = 1 p(x) = (x + 6)(x - 5) FOIL to check. **(b)** q(z) = 5z² + 8z + 3 can use grouping: q(z) = 5z² + 5z + 3z + 3 ac = 5 *3 = 15 b = 8 = 5 + 3 = 5z [z + 1] + 3 [z + 1] = [z + 1] (5z + 3) = (z + 1) (5z + 3) ## Functions - A set of ordered pairs (x, y) is called a **relation** in x and y. - The set of x-values in the ordered pairs is called the **domain** of the relation. - The set of y-values in the ordered pairs is called the **range** of the relation. - Given a relation in x and y, we say that y is a **function** of x if for each value of x in the domain, there is exactly one value of y in the range. This means that the relation passes a **vertical line test (VLT)**. ## Representation of Functions - **Verbal:** Two more than twice a number. - **Algebraic:** Y = 2x + 2 - **Numeric:** (maybe a table of values) | x | y | |:---:|:---:| |-1 | 0 | | 0 | 2| | 1 | 4| | 2 | 6| - **Graphic:** A graph is drawn with a line passing through the points (-1, 0), (0, 2), (1, 4) and (2,6). The y-axis is labeled 'y' and the x-axis is labeled 'x'. # Example 1.1: Use the graph of the function y = g(x) below (on the left) to answer the questions. **(a)** Is g(5) negative? g(5) = -3, yes **(b)** For which value(s) of x is g(x) = 0? x = -4 and x = 4 **(c)** For which value(s) of x is g(x) < 0? [-5, -4) U (4, 5] The graph shows a curve going up, then down, and then up again with a minimum point at x=0 and y=-4. **Example 1.2:** The entire graph of the function h is shown in the figure above (on the right). Write the domain of h as intervals or unions of intervals. x-vals D: [-5, -2) U [2, 5] The graph shows a curved line going down and then up with a minimum at x=0 and a y-value of -5. ## Algebraically Finding Domains When finding the domain of functions algebraically, there are currently two restrictions to which we must pay particular attention. **1. Division by zero** is undefined. Therefore, for rational functions (a function that can be written as a ratio (fraction) of two algebraic expressions), we need to ensure that the **denominator** does not equal zero. That is, for $f(x) = \frac{p(x)}{q(x)}, q(x) \neq 0$. **2. For radical functions of even roots** (functions that can be written with a symbol, the **radicand** (expression found inside/under the radical symbol) must be **nonnegative** (greater than or equal to zero). That is, for $f(x) = \sqrt[n]{p(x)}$, where n is even, $p(x) \geq 0$. **Example 1.3:** Find the domains of the functions f and g defined as follows. **(a)** $f(x) = \frac{x + 5}{x² - 10x + 25}$ Figure out where denom=0 and don't include those x-vals: in the domain x² - 10x + 25 = 0 (x - 5)² = 0 x = 5 **(b)** $g(x) = \frac{x - 4}{x² - 16}$ x² - 16 = 0 (x - 4)(x + 4) = 0 x = ±4 **Example 1.4:** Find the domains of the functions f and g, and h. Write your answers using interval notation. **(a)** h(x) = √9 - 9x Figure out where denom=0 and don't include those x-vals: in the domain 9 - 9x ≤ 0 -9x ≤ -9 x ≤ 1 **(b)** f(x) = √4x + 3 This will be your domain 4x + 3 ≥ 0 4x ≥ -3 x ≥ -3/4 **(c)** g(x) = √x + 9 x + 9 ≥ 0 x ≥ -9 **Example 1.5:** Find the domain of the function $f(x) = \frac{√7 + x}{-10 + 2x}$ Write your answer using interval notation. even root: radicand ≥ 0 7 + x ≥ 0 x ≥ -7 AND denom ≠ 0 -10 + 2x ≠ 0 2x ≠ 10 x ≠ 5 # Warm-Up: Write each expression as a single quotient. Simplify answers as much as possible. **(a)** $\frac{2}{x+3} + \frac{x}{x²-9}$ = $\frac{2}{x+3} + \frac{x}{(x - 3)(x + 3)}$ = $\frac{2(x-3)}{(x+3)(x-3)} + \frac{x}{(x - 3)(x + 3)}$ = $\frac{2(x - 3) + x}{(x - 3)(x + 3)}$ = $\frac{2x - 6 + x}{(x - 3)(x + 3)}$ = $\frac{3x - 6}{(x - 3)(x + 3)}$ **(b)** $\frac{2}{x} * \frac{x²-9}{x+3}$ = $\frac{2}{x} * \frac{(x - 3)(x + 3)}{(x + 3)}$ = $\frac{2(x - 3)}{x}$ **Example 1.6:** The functions f and g are defined as follows $f(x) = x² + 9x - 4, g(x) = \frac{7}{3 - x}$ Find f($\frac{x}{2}$) and g(x - 1). Write your answers without parentheses and simplify them as much as possible. $f(\frac{x}{2}) = f(\sqrt{x}) = (\sqrt{x})² + 9\sqrt{x} - 4$ = x + 9$\sqrt{x}$ - 4 $g(x - 1) = \frac{7}{3 - (x-1)}$ = $\frac{7}{3 - x + 1}$ = $\frac{7}{4 - x}$ **Example 1.7** The functions f and g are defined as follows $f(x) = \frac{6x+7}{5x+3}, g(x) = \sqrt{x² - 8x}$ Find f($\frac{5}{x}$) and g(x - 3). Write your answers without parentheses and simplify them as much as possible. $f(\frac{5}{x}) = \frac{6(\frac{5}{x}) + 7}{5(\frac{5}{x}) + 3}$ = $\frac{30}{x} + 7 * \frac{x}{x}$ =$\frac{30+7x}{x}$ =$ \frac{30+7x}{25 + 3x}$ $g(x-3) = \sqrt{(x - 3)² - 8(x - 3)}$ = $\sqrt{x² - 6x + 9 - 8x + 24}$ = $\sqrt{x² - 14x + 33}$ = $\sqrt{(x - 11)(x - 3)}$ **Difference Quotient** Recall that the average rate of change of a function f(x) from x = a to x = b is the slope of the secant line through the points (a, f(a)) and (b, f(b)). **AROC** = $\frac{f(b) - f(a)}{b-a}$ **Example 1.8:** Find the average rate of change of f(x) = 3x² - 4x from x = 2 to x = 7. **AROC** = $\frac{f(7) - f(2)}{7 - 2}$ f(7) = 3(7)² - 4(7) = 119 f(2) = 3(2)² - 4(2) = 4 **AROC** = $\frac{119 - 4}{5}$ = $\frac{115}{5}$ = 23 Similarly, the difference quotient is the slope of the secant line through the points (x, f(x)) and (x + h, f(x + h)), where h > 0. **Difference Quotient** = $\frac{f(x + h) - f(x)}{h}$ **Example 1.9:** Find the difference quotient $\frac{f(x + h) - f(x)}{h}$, where h ≠ 0 for the function f(x) = 4x² + 6x. Simplify your answer as much as possible. $\frac{f(x + h) - f(x)}{h}$ = $\frac{4(x + h)² + 6(x + h) - (4x² + 6x)}{h}$ = $\frac{4(x + h)(x + h) + 6(x + h) - 4x² - 6x}{h}$ = $\frac{4(x² + 2xh + h²) + 6x + 6h - 4x² - 6x}{h}$ = $\frac{4x² + 8xh + 4h² + 6x + 6h - 4x² - 6x}{h}$ = $\frac{8xh + 4h² + 6h}{h}$ = $\frac{h(8x + 4h + 6)}{h}$ = 8x + 4h + 6 **Example 1.10:** Find the difference quotient $\frac{f(x + h) - f(x)}{h}$, where h ≠ 0 for the function $f(x) = \frac{7}{x + 1}$. Simplify your answer as much as possible. $\frac{f(x + h) - f(x)}{h}$ = $\frac{\frac{7}{x + h + 1} - \frac{7}{x + 1}}{h}$ = $\frac{7(x + 1) - 7(x + h + 1)}{(x + h + 1)(x + 1)}$ * $\frac{1}{h}$ = $\frac{-7h}{(x + h + 1)(x + 1)h}$ = $\frac{-7}{(x + h + 1)(x + 1)}$