Math 31A - Review of Precalculus

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1 Math 31A - Review of Precalculus 1 Introduction One frequent complaint voiced by reluctant students of mathematics is: Why do I have to study this? When will I need math in real life? While it is my personal belief that mathematics is a beautiful subject that should be studied (worshipped?) for aesthetic reasons alone, a simpler answer to the question is that mathematics is the language through which we express our understanding of the physical world. (This contentious claim may raise some interesting philosophical questions, but they are beyond the scope of this course.) In most numerate disciplines, our knowledge is encapsulated in mathematical models, which more often than not involve calculus. Calculus is one of the most widely applied areas of mathematics, because it allows us to examine the relationship between different variables. For instance, in kinetics we are interested in studying velocity - the rate of change of position with time - and acceleration - the rate of change of velocity with time. Calculus makes this possible. While some of the ideas of calculus have been around for thousands of years, the formal theory was developed (independently) by Newton and Leibniz in the 1600 s. This course provides an introduction to the subject, which will allow you to apply the concepts to problems you will encounter in other classes (and in the Real World). However, it is necessary to have a firm grasp of the prerequisites (usually covered in a precalculus course), and these notes are intended to act as a review of functions and graphs, which will appear throughout the quarter. 2 Functions Just as the operations of arithmetic (addition, subtraction, multiplication, division, etc) are applied to numbers, the operations of calculus (differentiation and integration) are applied to functions. Hence a firm understanding of functions is required before we can begin our study of calculus. 2.1 Definition of a function You can think of a function as a machine that accepts an input value and returns an output value. For a function f and an input x, we denote the output as f(x). It is important that for every input value, there is precisely one output value. Figure 1: A function as a machine While a function can be defined more generally, we will typically consider functions whose input and output values are real numbers. 2.2 Key Terminology The following are some definitions of terminology related to functions: The domain of a function is the set of input values for which the function is defined. For example, the domain of the function f(x) = 1 x is the set of non-zero real numbers; we cannot divide by 0, so 0 is not in 1

2 the domain of f. The range of a function is the set of possible output values of the function (when applied to its domain). For instance, the range of the function g(x) = x 2 is the set of non-negative numbers, as x 2 0 for any value x. The following table gives some examples of domains and ranges of functions. See if you can fill in the blanks. (R represents the set of all real numbers.) f(x) Domain Range x {x : x 0} {y : y 0} cos x R {y : 1 y 1} 1 2x 1 {x : x 1 2 } {y : y 0} tan x x2 2 The independent variable is the input of the function, usually denoted by x. If writing an equation in the form y = f(x), y is called the dependent variable because its value depends on the independent variable x. A zero (or root) of a function f is an input value x such that the corresponding output value is zero, or f(x) = Graphs of Functions While the algebraic form of the functions presented above is useful for computing exact values, it does not give us a clear picture of the overall behaviour of the function. An alternate way to present a function is its graph - this is a pictorial representation that illustrates global features very clearly. The graph of a function is obtained by plotting the points (x, f(x)) on a coordinate plane. The x-axis represents the independent variable (the input values), while the y-axis represents the dependent variable (the output values). The height of the function above/below the x-axis at a point x represents the value f(x) at that point. An example of the graph of the function f(x) = 2x + 10 is given below. Figure 2: Graph of the function f(x) = 2x + 10 Note that if we are presented with a curve in a plane, the Vertical Line Test allows us to tell whether it is the graph of a function or not. The test says that every vertical line (x = a) should intersect the curve 2

3 at most once. That is to say, if any vertical line intersects the curve twice or more, then the curve does not represent a function. This is because the function can only have one output value for every input value. 2.4 Properties of Functions As mentioned above, the graph of a function allows us to identify certain properties of the function very quickly. We list some such properties below: A function is increasing if larger input values result in larger output values. Formally, this expressed as: if x 2 x 1, then f(x 2 ) f(x 1 ). The function is strictly increasing if a slightly stronger condition holds: if x 2 > x 1, then f(x 2 ) > f(x 1 ). Graphically, this means the graph goes up as it goes to the right. Similarly, a function is decreasing if larger input values result in smaller output values. More formally, if x 2 x 1, then f(x 2 ) f(x 1 ). As you may suspect, a function is said to be strictly decreasing if whenever x 2 > x 1, then f(x 2 ) < f(x 1 ). Graphically, the graph goes down as it goes to the right. A function is even if f( x) = f(x). This means the graph is symmetric with respect to the y-axis. A function is odd if f( x) = f(x). This means the graph is symmetric with respect to the origin (i.e. the line joining (x, f(x)) and ( x, f( x)) passes through the origin). Thus the first two properties determine the direction of the graph, while the latter two determine the symmetry of the graph. These properties are useful if trying to sketch the graph of a given function. 3 Polynomials There are many, many functions (infinitely many, and a very big type of infinity), and some functions are nicer than others. Perhaps the nicest class of (interesting) functions is the polynomials. These functions are sums of multiples of non-negative integer powers of x. For example, f(x) = 3.54x 19 +2x x+21 is a polynomial (the constant term is valid as x 0 = 1), while f(x) = x 3 2 is not a polynomial, since the power is not a non-negative integer. The degree of a polynomial is the highest power of x that appears (with a non-zero coefficient). Thus the degree of the polynomial in the example is 19. There are many reasons why polynomials are nice functions (perhaps even infinitely many), which include the fact that they are not only continuous but can be differentiated infinitely many times, and that any continuous function can be approximated by polynomials (with an error as small as you want). Most of these reasons will have to wait until a course on analysis, but in this course we will be able to appreciate some of the nice calculus-related properties they have. 3.1 Linear Polynomials The simplest polynomials are the linear polynomials, which are polynomials of degree at most one. The name comes from the fact that linear polynomials represent straight lines Equations As these are polynomials of degree at most one, they can be represented in the standard form f(x) = ax + b where a and b are real numbers. This is sometimes called the slope-intercept form of the equation of the line. This is because the quantity a represents the slope of the line y = f(x) (the increase in y per unit increase in x), while the quantity b is the y-intercept (the value where the line crosses the y-axis). 3

4 There are a couple of alternative forms of the equation of a line. If we know that the slope of the line is m and the line passes through the point (x 0, y 0 ), then we can express the line as This is the point-slope form of the equation. y y 0 = m(x x 0 ) Alternatively, if we know the line passes through the points (x 1, y 1 ) and (x 2, y 2 ), then the two-point equation for the line is y y 1 = y 2 y 1 x 2 x 1 (x x 1 ) One can easily switch between the different forms with a little bit of algebraic manipulation Graphs As stated above, linear polynomials represent straight lines. In particular, every non-vertical straight line in the plane is the graph of a linear polynomial (vertical lines, of course, fail the vertical line test, and hence are not functions). Below we show the graphs of some sample linear polynomials: The function y = 3 The function y = 1.3x 4 The function y = 3x + 2 If a = 0, then we get a horizontal line. If a 0, then we get a non-horizontal line; if a > 0, it is increasing, while if a < 0, it is decreasing. In particular, this line must have a unique root. We can solve for it algebraically: if f(x) = ax + b = 0, then x = b a. To plot the graph of a linear polynomial, just find two points on the line and plot the unique straight line between them. These two points can be obtained by substituting any two input values x. Letting x = 0 gives the y-intercept, and can often be a useful point for plotting. 3.2 Quadratic Polynomials While the simplicity of linear polynomials may make them a favourite of students, unfortunately it also means that they are not complex enough to be used in many situations. Thus we need also consider polynomials of degree two, or quadratic functions Equations of Quadratics As a polynomial of degree two, quadratics have the standard form f(x) = ax 2 + bx + c where a 0 (otherwise it would be linear). As before, some algebra can give us different forms of the equation. Another useful form of the quadratic is the vertex form, where we write f(x) = A(x x 0 ) 2 + y 0 4

5 The point (x 0, y 0 ) is called the vertex, and as we shall see when we plot quadratics, this is the turning point of the graph. To see the relationship between the coefficients of the standard form and the vertex form, we expand the vertex form to get f(x) = A(x x 0 ) 2 + y 0 = A(x 2 2x 0 x + x 2 0) + y 0 = Ax 2 2Ax 0 x + (Ax y 0 ) Comparing this to the standard form f(x) = ax 2 +bx+c, we find that a = A, b = 2Ax 0 and c = (Ax 2 0 +y 0 ). To get the vertex form from the standard form, we solve for A, x 0 and y 0 in terms of a, b and c to get A = a, x 0 = b 2a and y 0 = c b2 4a. The process of converting the standard form into the vertex form is known as completing the square. Finally, when the quadratic has real roots (more on that in the next section), we can also write the equation in factored form, namely f(x) = A(x x 1 )(x x 2 ) Here A is the coefficient of the x 2 term (so A = a), and x 1 and x 2 are the roots of the quadratic (so f(x 1 ) = f(x 2 ) = 0) Roots of Quadratics Not every quadratic has a (real) root (and thus not every quadratic can be written in factored form). For instance, if f(x) = x 2 + 1, then since x 2 0 for every x, we have f(x) 1 for every x, and hence f(x) is never zero. The quadratic formula tells us exactly when a given quadratic has roots, and what they are (if they exist). According to the formula, the roots of the quadratic f(x) = ax 2 + bx + c are b ± b 2 4ac 2a Note that this expression is only defined when the term inside the square root, b 2 4ac, is non-negative. This term has a special name; it is called the discriminant, and is usually denoted by = b 2 4ac. In summary, given a quadratic in standard form, we wish to determine whether it has roots, and if so; what they are. We first compute the discriminant, and the following table shows the possible cases: Discriminant No. of distinct roots Roots < 0 0 none = 0 1 b 2a > 0 2 b± 2a If the quadratic is not given in standard form, then determination of the roots is even easier. In vertex form (f(x) = A(x x 0 ) 2 + y 0 ), we have y 0 No. of distinct roots Roots Ay 0 < 0 2 x 0 ± y 0 A y 0 = 0 1 x 0 Ay 0 > 0 0 none If the quadratic is given in factored form (f(x) = A(x x 1 )(x x 2 )), then the roots are given to us explicitly; they are x = x 1 and x = x 2. 5

6 3.2.3 Graphs of Quadratics The shape of the graph of a quadratic function should be a familiar shape - it is a parabola! It can come in one of two varieties: Happy parabola : a > 0 Sad parabola : a < 0 Note that the turning point of the parabola (its local minimum or maximum) is located at the vertex as defined in the vertex form of the equation. Also, if the parabola has its vertex at (x 0, y 0 ), then it is symmetric through the line x = x 0. Thus, as described in the preceding sections, from the equation of the parabola we can extract the following information: Whether it is upwards or downwards (a > 0 or a < 0) Where its vertex is How many roots it has, and where they are This is enough for us to be able to sketch a fairly accurate graph of the equation. Below are some more examples of quadratics and their graphs: f(x) = 2x 2 + x 1 f(x) = x 2 + 3x 2 An upward parabola with two distinct roots A downward parabola with two distinct roots f(x) = 4x 2 4x + 1 f(x) = 2x 2 4x + 3 An upward parabola with one root An upward parabola with no roots 6

7 3.3 Cubic Polynomials We end this section by moving one degree higher and looking at polynomials of degree three, or cubic polynomials Equations and Roots of Cubics As polynomials of degree three, cubic functions have the standard form where a 0. f(x) = ax 3 + bx 2 + cx + d Unlike quadratics, which could possibly have no real roots, cubic equations must always have at least one real root. (In fact, it is generally true that any odd-degree polynomial must have at least one real root, while even-degree polynomials need not have any roots. The maximum number of roots a polynomial of degree n can have is n.) Unfortunately, there is no counterpart of the quadratic formula for cubics; there is no nice formula for the roots of a cubic polynomial. If you are asked to find the roots of a cubic polynomial, you are generally in one of two situations: You are allowed to use a computer/graphing calculator, which have solvers that can find a root for you. You are expected to do this by hand, and there is an easy root - usually a small integer (e.g. 0, 1, -1, 2, -2) that you can find by trial and error. Suppose you have now found a root, x = x 1, so that f(x 1 ) = 0. Then you can use polynomial long division to factor the cubic into a linear term times a quadratic term (there will be no remainder) f(x) = ax 3 + bx 2 + cx + d = (x x 1 )(Ax 2 + Bx + C) If you would like a review of how polynomial long division works, try checking out We can then apply our discriminant analysis and quadratic formula to determine whether the quadratic factor has roots. If so, then these roots are also roots of the cubic polynomial. For example, consider the polynomial f(x) = x 3 + 3x 2 7x + 3. By trial and error, we find f(1) = = 0, and so 1 is a root of the cubic. This means that x 1 is a factor of the cubic polynomial. Indeed, if we carry out polynomial division, we find f(x) = x 3 + 3x 2 7x + 3 = (x 1)(x 2 + 4x 3) Evaluating the quadratic factor x 2 +4x 3, we see the discriminant is = ( 3) = = 28 > 0. Hence the quadratic has two distinct roots, which by the quadratic formula are 4± 28 2 = 2 ± 7. Thus the roots of the cubic polynomial are 2 7, 2 + 7, Graphs of Cubics In a parabola, we saw that the behaviour of the function is the same as x goes to positive or negative infinity. For instance, in the case of an upward parabola, if x is very large and positive, then f(x) is very large and positive. If x is very large and negative, then f(x) is still very large and positive. With cubics, the behaviour at the two ends of the real axis is always different. When the leading coefficient is positive, a > 0, then f(x) is very large and positive for large positive x, while f(x) is very large and negative for large negative x. If a < 0, then the relationship is inverted - f(x) is large and negative for large positive x, and f(x) is large and positive for large negative x. The two examples below illustrate this property. 7

8 f(x) = x 3 + 3x 2 4x + 2 f(x) = x 3 + 3x 2 4x + 2 So whenever you have to sketch the graph of a cubic, always start with the leading coefficient - use it to determine the direction of the cubic. We also have to look out for roots of the cubic. The previous section explained how to find the roots of the cubic - try to find one by trial and error, divide to get a quadratic, and then use the discriminant and quadratic formula to find the remaining roots. Depending on the form of the quadratic factor, you can have 1, 2 or 3 different roots (but you must always have at least 1). The pictures below show the different possibilities. Upward Cubic with One Root Upward Cubic with Two Roots Upward Cubic with Three Roots f(x) = x 3 3x 2 + 3x 1 f(x) = x 3 + x 2 x 1 f(x) = x 3 + 2x 2 x 2 Downward Cubic with One Root Downward Cubic with Two Roots Downward Cubic with Three Roots f(x) = x 3 + 3x 2 3x + 1 f(x) = x 3 x 2 + x + 1 f(x) = x 3 2x 2 + x + 2 Note that when there are multiple roots, there should be a turning point between every pair of consecutive roots. This should help guide your sketch. 8

9 4 Composition of Functions Most functions are very complicated, and direct analysis would be very messy. One way to make them easier to study is to build them up through composition of simpler functions. 4.1 Definition of composed functions Recall that we can think of functions as machines - we input one number and get another number as the output. Composition of functions is stringing these machines together (in series), so that the output of one machine becomes the input of another. This is analogous to a manufacturing line; a series of machines work together, with each completing a simple task before passing the result onto the next machine. In this way, complex products can be built one simple step at a time. Figure 3: Composition is Linking the Function Machines Together Suppose we have two functions, f(x) and g(x). The composition f g (read f composed with g ) is the function that results when we apply g to x, and then feed the output g(x) as input to f. Thus (f g)(x) = f(g(x)). Note that the order the functions are applied is inside-out, or right-to-left in the composition notation. 4.2 Evaluation of compositions Given f and g, we get an expression for f g by simply replacing x in the equation for f with g(x). Then we can substitute in the equation for g(x) (remember to use parenthesis - constants should factor through). For instance, suppose f(x) = x 2 + 2x + 1 and g(x) = x 3. Then we have (f g)(x) = f(g(x)) = (g(x)) 2 + 2(g(x)) + 1 = (x 3) 2 + 2(x 3) + 1 If necessary, we could expand further to get (f g)(x) = x 2 4x + 4. This is a good place to note that the order of the composition matters! If we were to instead evaluate g f, then we get (g f)(x) = g(f(x)) = g(x 2 + 2x + 1) = (x 2 + 2x + 1) 3 = x 2 + 2x 2 (f g)(x) We can repeat the process for any number of functions. If f(x) = x 2, g(x) = x + 1 and h(x) = x, then (f g h)(x) = f(g(h(x))) = f(g( x)) = f(( x) + 1) = (( x) + 1) 2 = x + 2 x + 1 Note, however, that if we are only interesting in particular values of the composition, then we can avoid the algebraic manipulation. In the previous example, if we wanted to find (f g h)(4), then we note that h(4) = 4 = 2, g(h(4)) = g(2) = = 3, and so (f g h)(4) = f(g(h(4))) = f(3) = 3 2 = Decomposing Functions It is often useful to be able to decompose functions as well - given a complicated function, we want to write it as the composition of simpler functions. This is useful because it can separate the complexity of the original function into separate parts, which we can then deal with later (you will see this in action when we learn the chain rule). 9

10 This is perhaps easiest to see by example. Suppose we have the function f(x) = 1 cos 3 x. There are two complicated expressions here - the square root and the cosine cubed. If we parenthesise the cosine cubed (f(x) = 1 (cos 3 x)), this suggests letting the inner function be h(x) = cos 3 x, and then the outer function is g(x) = 1 x. We then have f(x) = (g h)(x), reducing the function to a composition of simpler functions. Note that this is not the only way to decompose the function. We could also have used G(x) = x and H(x) = 1 cos 3 x, and then f(x) = (G H)(x). Moreover, we could have decomposed it into three functions: f 1 (x) = x, f 2 (x) = 1 x 3, and f 3 (x) = cos x. Then f(x) = (f 1 f 2 f 3 )(x). These three functions are even simpler, but the trade-off is that we have more functions. 4.4 Domains and Ranges of Compositions Recall that the domain of a function is the set of possible inputs, while the range is the set of possible outputs. Determining the domain and range of a composed function can be a little tricky, since the domains and ranges of the two functions interact a little. In my experience, it is easier to try and find the domain first. For example, let f(x) = x and g(x) = 4 x 2. We wish to find the domain of (f g)(x) = f(g(x)). If x is in the domain, then f(g(x)) must make sense, so g(x) must be in the domain of f. f(x) = x, which is defined for x 0. Hence for f(g(x)) to be defined, we must have g(x) = 4 x 2 0. Solving this, we find x 2 4, so 2 x 2. Hence the domain of (f g) is [ 2, 2]. Now what about the range? We know from our above calculations that when x is in [ 2, 2], g(x) 0. However, g(x) = 4 x 2, and x 2 is always non-negative, so g(x) 4. Hence we are feeding the interval [0, 4] into f. If 0 y 4, then 0 y 2, and so 0 f(g(x)) 2 for x in the domain [ 2, 2]. So we see the range is [0, 2]. To review, we first list the domain D of the outer function f. Then we have to find which values of x can go into the inner function g such that the result g(x) lies in D. This gives the domain for the composed function. To get the range of the composed function, we have to see what g does to the domain, and then see what f does to that set. If we have more than two functions being composed, we can just continue this process. Note that F = f g h = f (g h), so if we let G = g h, then F = f G. We can use the method to find the range and domain of G, and then use that information to find the range and domain of F. 5 Conclusion I realise these notes may be a little rushed, but they are intended to hopefully awaken your memories of the material rather than teach it. Also, you may have been taught different methods for approaching these kind of problems. Provided your method works and you understand why it works, that is perfectly fine. If you have any questions about any of the above, please do me. I wish you all the very best for this course. Have a great quarter! 10