Midterm I Review:

Midterm I Review: 1.1 -.1 Monday, October 17, 016 1 1.1-1. Functions Definition 1.0.1. Functions A function is like a machine. There is an input (x) and an output (f(x)), where the output is designated by some sort of rule. Each input has EXACTLY one output. A function can be thought of like a machine. You put x into a machine, which is programmed to do something with it, and then spits out f(x). This Figure 1: Function Machine machine is only a function if each input has exactly one output. Table 1: A Function Input x 1 4 Output f(x) 5 6 9 This table shows an example of a function. Each input has exactly one output. Table : NOT A Function Input x 1 Output f(x) 5 6 9 1

This table shows an example of NOT a function. The input value x = corresponds to both f(x) = 5 and f(x) = 6. This also corresponds to what we call the Vertical Line Test. Given a graph, if we draw a vertical line that intersects a graph more than once, then the graph does NOT define a function. Figure : Here, (a) and (b) both define functions. We can draw a vertical line anywhere on them, and that line will intersect the graph only once. The last graph (c), however, is not a function. A vertical line drawn here will intersect the graph twice.

Function Domains and Ranges Definition 1.0.. Domain The domain of a function is the input values, or in our examples, x. It is what we put into the machine. In math, this is said, Each element x of the set X is called the domain D. (a) Some plotted points (b) A Parabola (c) The square root function Figure : Example: Find the domain of the graphs of these functions. The domain of a function is often determined by where the function actually exists. There are certain functions (like the square root function) where we can only input positive values of x. Example. The following functions are not defined at certain values of x. Where are they not defined? f(x) = 1 x ; f(x) = 1 (x 1)(x ) ; f(x) = x 1 Definition 1.0.. Range The range of a function is the possible output values, or in our examples, f(x). It is what comes out of the machine. Equations of Lines and Periodic Functions Definition 1.0.4. Linear Function A linear function is a function which takes the form f(x) = y = mx + b where m is the slope of the line and b is the y-intercept. Definition 1.0.5. Slope The slope of a line tells us that, if we increase x by 1 unit, f(x) or y will increase by m units. The equation which defines the slope is m = y y 1 x x 1 where (x 1, y 1 ) and (x, y ) are two points on the line. This is colloquially referred to as rise over run. Definition 1.0.6. Y-Intercept The y-intercept is the value of f(x) or y when x = 0. It always exists on the y-axis and is the value at which f(x) crosses the y-axis.

Example. We will do a few different types of examples on how you would find the equation of a line. 1. Given Slope and a Point Slope, passing through (1, ). Given Two Points Passing through ( 1, ) and (0, 1). Given a Graph Figure 4: Find the equation representation y = mx + b Definition 1.0.7. Periodic Function A periodic function is a function which is exactly repeated over and over again, or, mathematically, for which f(x) = f(x + T ) for all x and for some T > 0. functions. The best example of this is any of the trig Trig Functions There are six trig functions. You should know them, plus the value of them at specific angles. Table : Important Trig Function Values 0 π 6 sin(x) 0 1 π 4 π π π π π 1 0-1 0 cos(x) 1 1 0-1 0 1 tan(x) 0 1 undef 0 undef 0 sec(x) 1 undef -1 undef 1 csc(x) undef 1 undef -1 undef cot(x) undef 1 0 undef 0 undef 4

Figure 5: Trig Functions and their Graphs Linear and Periodic Functions Example. Can you find the equation for a line given two points? Find the line between (1, 1) and (4, 6). m = y y 1 = 6 1 x x 1 4 1 = 6 = : Find the slope y = x + b: Now we have to find the intercept b 6 = 4 + b: We plug in either one of the points 14 = b Final equation: y = + 14 Reminder that this relates to a rate of change. For a linear function, the slope represents a rate of change, and it generally has the units of output per input, 5

like miles per hour. Example. Jane wants to know how far her school is from her house. When she drives on the 5 mph roads, it takes her 10 minutes. How far is her school from her house? Solution. This problem deals mainly with units. We know from the phrasing of the problem that our function is d(t) = 5t, where t is measured in hours. We must convert 10 minutes into hours. 10 minutes is one-sixth of an hour, so t = 1/6. d(1/6) = 5(1/6) = 5/6 This means that the school is roughly 4 miles away. Example. Can you determine amplitude and period from a graph? (See Figure 6.) Proportionality Example. When two things are proportional, it means they are equal given some constant, i.e., y x y = Ax; h y h = By By the way, this means that not only is h proportional to y, but it is also proportional to x because of transitivity, i.e., Example. h y x = h x y x y = Ax If I tell you that y is proportional to x, and that y = 15 when x =, you can solve for A: y = Ax: Plug in y = 15, x = 15 = A : Solve for A A = 5 Example. The circumference C of a circle is proportional to its diameter d. The circumference of a circle of diameter.5 cm is 11 cm. What is the circumference of a circle of diameter 4. cm? Solution. C = Ad 11 = A(.5) A = 11/.5 = /7 C = 7 4. = 1.cm 6

(a) Example 1: Amplitude =, Period = 4 (b) Example : Amplitude = 5, Period = 4π (c) Example : Amplitude = 1, Period = π Figure 6: Amplitude and Period Examples 7

1.4-1.6 Exponential Growth Definition.0.1. Exponential Function An exponential function is a function of the form f(x) = y = a x where a 1, the parameter, is a constant and x is a variable. If a > 1, the function is increasing for all x; if a = 1, the function is a constant function, aka, a horizontal line; if 0 < a < 1, the function is decreasing for all x. Example. Exponential Growth It s very common to use the formula A(t) = Ce kt when describing population growth. In this formula, C is the original population, k is growth rate, t is the time in some units (days, months, years, etc.), and A(t) is the population at time t. Let s try the following problem: A biologist is researching a newly-discovered species of bacteria. At time t = 0 hours, he puts one hundred bacteria into what he has determined to be a favorable growth medium. Six hours later, he measures 450 bacteria. Assuming exponential growth, what is the growth constant k for the bacteria? (Round to two decimal places.) Solution. We note here that t is mentioned in hours, so we will assume that the units are hours. What is the original population? The problem tells us: 100! We also note that at t = 6, we have a new population of 450, so we want to solve the equation: A = Ce kt 450 = 100e 6k 4.5 = e 6k ln(4.5) = 6k k = 0.5 Let s try another one! A certain type of bacteria, given a favorable growth medium, doubles in population every 6.5 hours. Given that there were approximately 100 bacteria to start with, how many bacteria will there be in a day and a half? Solution. We again note that t is in hours. In order to solve this problem, we must complete two steps: (1) find k; and () find A after 1.5 days. Let s start with finding k. We are told that the population doubles in 6.5 hours, i.e., at t = 6.5, A = 00 (since the original population is 100). So we follow the same 8

steps as the above problem: 00 = 100e 6.5k = e 6.5t ln() = 6.5k k = ln() 6.5 Now that we have k, we can use it to solve for the population! But we have to convert 1.5 days into hours. If a day has 4 hours, that means 1.5 days is 6 hours. Thus we have, at t = 6: ln() 6 A(6) = 100e 6.5 = 4647.75 The same formula is used for Exponential Decay, except we see that the population will decline, which correlates to a negative growth rate, i.e., k < 0. Example. Compound Interest We end up having two formulas for a future value of an original investment. The first is the same as for exponential growth, except replacing k with r: A = P e rt : Continuously Compounded Interest The second formula is A = P (1 + r n )nt This formula correlates to an original investment P which is compounded n times per year at the rate r, where the units on t are years. Say you invest $10,000 at time 0 into two separate accounts, both of which have an annual interest rate of 5%, but one which is compounded continuously and one which is compounded semiannually. How much will you have at the end of the year? Solution. We will start with the account that is compounded semiannually. This means that n =, r = 0.05, and t = 1: A = 10, 000(1 + 0.05 )1 = 10, 506.5 Conversely, we will see what happens when it is compounded continuously: A = 10, 000e 0.05 1 = 10, 51.71 This doesn t seem like a big difference, but if we look at a time period of 10 years instead of 1, the number changes drastically! Continuously: 16,487.1. Semiannually: 16,86.16. The difference continues to grow! 9

Function Building (aka, Translations) Definition.0.. Vertical Shift Given a function f(x), we can define a new function g(x) called a vertical shift of f(x) where g(x) = f(x) + k. All values of f(x) are increased by k. If k is positive, it s a shift up; if k is negative, it s a shift down. Vertical shifts affect the output/outside of the function. Definition.0.. Horizontal Shift Given a function f(x), we can define a new function g(x) called a horizontal shift of f(x) where g(x) = f(x + k). If k is positive, it s a shift left. If k is negative, it s a shift right. A horizontal shift will change the input of the function. Figure 7: Vertical and Horizontal Transformations Definition.0.4. Reflections A vertical reflection of a function f(x) occurs when we define the new function g(x) = f(x). This is also called a reflection about the x-axis. A horizontal reflection of a function f(x) occurs when we define the new function g(x) = f( x). This is also called a reflection about the y-axis. Definition.0.5. Stretch and Compression Given a function f(x), we can define a new function g(x) = kf(x). We call this function a vertical stretch or compression. If k > 0, the graph is stretched; if 0 < k < 1, the graph is compressed; if k < 0, it s a mix of stretch or compression and vertical reflection. Given a function f(x), we can define a new function g(x) = f(kx). We call this function a horizontal stretch or compression. If k > 1, the graph is stretched by 1 k. If 0 < k < 1, the graph is stretched by 1 k. If k < 0, it s a mix of stretch or compression and horizontal reflection. When combined, the stretch/compression must happen first, then followed by the shifts. 10

Definition.0.6. Composition of Functions We can write the output of one function as the input of another; this is called a composition of functions. We write it as h(x) = f(g(x)) or (f g)(x) and read it as f of g of x or f composed with g at x. (a) What is y = g(h(1))? (b) What is y = h(f( ))? Figure 8: Two ways to evaluate composition functions: Graph and Table Example. We can evaluate compositions of functions using formulas as well as graphs and tables. Let s say we have the two functions f(x) = x + x 1 and g(x) = x +, and let h(x) = f(g(x)). What is h(1)? h(1) = f(g(1)); g(1) = 1 + = 4; f(4) = 4 + 4 1 = 19; h(1) = 19 Example. Given the functions q(x) = 1 x and m(x) = x 4, state the domains of the following functions using interval notation. 1. Domain of p(x) m(x). Solution. We note that p(x) m(x) = 1 1 x x 4 = 1 x(x )(x + ) We know that x 0 because of x. We also know that x, x, and x 0 because then we would divide by zero. All of this together combines to make the interval (0, ) (, ) 11

. Domain of p(m(x)). Solution. We note that p(m(x)) = p(x 4) = 1 x 4 We know that x 4 > 0, which makes the number line below: In interval notation, this becomes (, ) (, ). Domain of m(p(x)). Solution. We note that m(p(x)) = m( 1 x ) = ( 1 x ) 4 = 1 x 4 This is a more difficult problem that it first seems. We think, just upon observation, that the only value we cannot input is x = 0. This is wrong, however. When composing functions, the domain restrictions of the input function must follow through the whole composition. That means we cannot put in any negative values because of the square root function. Thus, our actual interval is (0, ) Inverse Functions and Logarithms Definition.0.7. One-to-One A function is said to be one-to-one if each input corresponds to an output, and each output corresponds to an input. This concept corresponds to what we call the Horizontal Line Test. Much like the Vertical Line Test, if we draw a horizontal line on a graph and it intersects that graph more than once, then the graph does NOT define a one-to-one function. We will refer back to the graphs in Figure. Obviously, (c) is not a one-to-one function since it s not even a function to begin with. What about (a) and (b), though? Using the Horizontal Line Test, we see that (a) is not one-to-one, but (b) is! It s true that all non-horizontal and non-vertical lines are always one-to-one. The following figure also displays this idea in another form: 1

Figure 9: Difference Between One-to-One and Not Definition.0.8. Inverse For a function f(x) = y, its inverse g(y) is the function for which g(y) = x. We write this as f 1 (x). The inverse of a function swaps the domain and range, i.e., the domain of the function is the range of its inverse. Only one-to-one functions can have inverses, however. So, for example, the function f(x) = x does not have an inverse on its whole domain; it s one-to-one on [0, ), though, and thus has an inverse here. The graph of an inverse is always a reflection of the original function about the line y = x. Figure 10: Some Inverse Graphs Example. Finding an Inverse Find the inverse of f(x) = x x +. We must write the equation as y = x x +. 1

Then we swap the x and the y values and then solve for y: x = y y + x(y + ) = y xy y = x y(x 1) = x y = x x 1 Definition.0.9. Logarithm We call the formula y = log a x a y = x the logarithm of x with base a, assuming that a > 0 and a 1. Properties of Logarithms 1. Addition: log a x + log a y = log a xy. Subtractive: log a x log a y = log a x y. Multiplication: y log a x = log a (x y ) 4. Change of base: log a x = log b x log b a 5. Cancellation: log a (a x ) = x = a log a x 1.7-.1: To be added... 14