Problem Sets for MATH1110 Fall 2017

Transcription

1 Problem Sets for MATH1110 Fall 2017 Matt Hin November 27, 2017 Contents A Lectures L1-1 A.1 Lecture 1-22 August L1-1 A.2 Lecture 2-24 August L2-1 A.3 Lecture 3-29 August L3-1 A.4 Lecture 4-31 August L4-1 A.5 Lecture 5-5 September L5-1 A.6 Lecture 6-7 September L6-1 A.7 Lecture 7-12 September L7-1 A.8 Lecture 8-14 September L8-1 A.9 Lecture 9-19 September L9-1 A.10 Lecture September L10-1 A.11 Lecture September L11-1 A.12 Lecture September L12-1 A.13 Lecture 13-3 October L13-1 A.14 Lecture 14-5 October L14-1 A.15 Lecture October L15-1 A.16 Lecture October L16-1 A.17 Lecture October L17-1 A.18 Lecture October L18-1 A.19 Lecture October L19-1 A.20 Lecture October L20-1 A.21 Lecture 21-2 November L21-1 A.22 Lecture 22-7 November L22-1 A.23 Lecture 23-9 November L23-1 A.24 Lecture November L24-1 A.25 Lecture November L25-1 A.26 Lecture November L26-1 I

2 MATH 1110 Calculus I - Fall August 2017 Lecturer: Matt Hin Lecture 1 1.1: Functions and Their Graphs Mathematics is a language and Calculus is the dialect of functions. In order to starting talking, we need a few words under our belt. This session will be heavy on the terminology, but light on theory. Functions are special rules or relationships between one set of objects to another set of objects. For example this function f relates or maps these numbers to these colors: f(1) = red, f(1) = green, f(1) = blue (1) Specifically, f is a function that maps a set D to a set Y if for every x in D, there exists a unique y in Y. We write functions as y = f(x), read as y is equal to f of x. Here, f is the function, x is the independent variable, and y is the dependent variable. Note that a function has to have a domain D and a codomain Y defined. We call the set of objects that f acts on the domain while the set of objects that get mapped to by f as the range. Note that the range may or may not encompass all of the codomain Y. Technically, whenever a function is addressed, you want to specify its domain. However when no domain is specified, we assume the natural domain which is the largest set of things that the function can operate on. Of course, the set of all the corresponding images makes up the natural range. We say a function is real-valued, if the natural domains and ranges are subsets of the real line. In fact for this semester, we ll be focusing on these real-valued functions. With most of mathematics, whenever there s an algebraic or analytic way of looking at things, there is often a geometric interpretation as well. We can write down the rule for a function, or we can look at its graph. Mathematically, a graph is defined as the set of points: {(x, f(x)), x D}, read as The set of points (x, f(x)) where x is in the domain D of f. We can always plot a graph by picking and choosing x values and computing the corresponding f(x) values. Graphing computers like Desmos ( perform a similar computation. We can tell a graph represents a function by using the vertical line test. This test requires one to scan the graph of a function with a vertical line, checking if the graph intersects each vertical line at most once. Hence, this graphically verifies that for every x there is a unique y. So what kinds of functions are there, and what are their (natural) domains and (natural) ranges? Piecewise Functions ( and ) (Monotonic) Increasing/Decreasing Even and Odd Functions Linear(y = x, y = kx) Power functions (y = k/x, y = n x) Polynomials (coefs, degree) Rational Algebraic Trigonometric (periodic) Exponential Logarithmic Transcendental L1-1

3 1.2: Combining Functions; Shifting and Scaling Graphs Functions can be added, subtracted, multiplied, and divided just like real numbers. Functions can also be composed with one another. (f ± g)(x) = f(x) ± g(x), (f g)(x) = f(x) g(x), (fg)(x) = f(x)g(x), (f g)(x) = f(g(x)). Example. Let f(x) = x 2 and g(x) = x + 1, determine (a) (f + g)(x), (b) (fg)(x), (c) (f g)(x), and (d) (g f)(x). (a) (f + g)(x) = x 2 + x + 1 (b) (fg)(x) = x 3 + x 2 (c) (f g)(x) = (x + 1) 2 (d) (g f)(x) = x We can also shift the graph of a function vertically and horizontally. y = f(x) + k Vertical Shift: Shifts graph up by k units if k > 0 and down by k units if k < 0, y = f(x + h) Horizontal Shift: Shifts graph the to the left by h units if h > 0 and to the right by h units if h < 0. We can also stretch and reflect the graph of a function through scaling. y = cf(x) Vertically stretches the graph by c > 1, y = 1 c f(x) Vertically compresses the graph by c > 1, y = f(cx) Horizontally compresses the graph by c > 1, y = f(x/c) Horizontally stretches the graph by c > 1, y = f(x) Reflection across x-axis, y = f( x) Reflection across y-axis. (2) (3) Example. For f(x) = x, determine an expression for the desired function and plot its graph: (a) The function f(x) but shifted right 1 unit and horizontally stretched by a factor of 3. L1-2

4 Figure 1: Graphs to Solutions. Note the domain and range of each transformed function. (b) The function f(x) but shifted down 4 units and vertically compressed by a factor of 2. (c) The function f(x) but shifted up 3 units and to the left 2 unit and then reflected across the origin. (d) The function (g f)(x), where g(x) = x 2. (a) h a (x) = 3(x 1) (b) h b (x) = x 4 2 ( (x ) (c) h c (x) = + 2) + 3 (d) h d (x) = x for x 0. L1-3

5 MATH 1110 Calculus I - Fall August 2017 Lecturer: Matt Hin Lecture 2 1.3: Trigonometric Functions Trigonometric functions are those that relate angles to ratios. There are two ways to measure angles: degrees and radians. A 360 degrees sweeps out a full circle, while 2π radians do the same. Radians are typically used over degrees since there is a simple relationship between the angle θ in radians and the length of the arc s swept out by a sector of a circle of radius r: s = rθ. Hence to convert between these measurement units we have: 1 radian = 180 π degrees or 1 degree = radians. (4) π 180 We impose a direction on how we measure angles as well: if the angle is swept out in a counterclockwise fashion from the position x-axis, the angle is positive, else negative. These angles can also extend beyond 2π as the terminal ray continues to sweep around the origin. There are 6 trigonometric functions that relate an angle with a ratio of side lengths of a right triangle. sin θ = y r, cos θ = x r, tan θ = y x, csc θ = r y, sec θ = r x, cot θ = x y. (5) Example. Recreate this table of exact values. Degrees θ (Rads) π 3π/4 π/2 π/4 0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π 3π/2 2π sin θ cos θ tan θ A function is periodic, if there is a positive number p such that f(x + p) = f(x) for every value of x. The smallest of such numbers p is called the period. Example. What are the periods of the 6 basic trigonometric functions? The sine, cosine, secant, and cosecant functions are all periodic with period 2π while the tangent and cotangent functions are periodic with period π. Due to their highly geometric nature, trigonometric functions have many properties. For angles α, β and L2-1

6 side lengths (a, b, c) of a triangle ABC with angle θ opposite c, we have cos 2 θ + sin 2 θ = 1 cos(α ± β) = cos α cos β sin α sin β sin(α ± β) = sin α cos β ± cos α sin β cos(2θ) = cos 2 θ sin 2 θ sin(2θ) = 2 sin θ cos θ cos cos 2θ θ = 2 sin 2 1 cos 2θ θ = 2 c 2 = a 2 + b 2 2ab cos θ Pythagorean Identity Angle Sum/Difference Formula Double Angle Formula Half-Angle Formula Law of Cosines (6) Since trigonometric functions like the sine function are frequently used in signal processing, there s a special structure to transforming sine functions, called the general sine function: ( ) 2π f(x) = A sin (x C) + D (7) B Here, A is the amplitude, B is the period, C is the horizontal shift, and D is the vertical shift. 1.5: Exponential Functions Exponential behavior occurs when a quantity or value increases by a constant factor. For example, the exponential function with base a is given b f(x) = a x. Note that this is one of the fastest growing functions we know of since with a unit increase in x, we multiply the value by the base a. Here are some properties that are important about exponential functions: a p/q = q a p, a x a y = a x+y, a x a y = ax y, (a x ) y = (a y ) x = a xy, a x b x = (ab) x, ( a b ) x = a x b x. (8) A special type of exponential function is the natural exponential where the base is the transcendental number e In fact, we can express any exponential function as a scaled version of this natural exponential since y = a x = exp(x ln a). Natural exponentials are used typically in models of growth and decay such as economics and ecology. There we typically encounter the function in the form: f(x) = e kx. If k is negative, then we say f(x) is a model of exponential decay, else exponential growth. Example. A 5-year US Treasury Note in August 2017 has an annual return rate of 1.8%. If we use an exponential function to model the growth of $1000 invested into a Treasury Note, what would be the value of my investment in 2022 when it matures? Let t denote the number of years since we purchased a Note. Since the rate k = and our initial investment P 0 = 1000, we have the model: Then for the value of the investment in 2022 corresponds to f(t) = P 0 e kt = 1000e 0.018t. (9) f(5) = 1000e L2-2

7 1.6: Inverse Functions and Logarithms A function is one-to-one on a domain D if f(x 1 ) f(x 2 ) whenever x 1 x 2. We simply say that a function is one-to-one if this defining property holds for its natural domain. Just like we can check if a graph is one for a function using the vertical line test, we can check if a function is one-to-one if it passes the horizontal line test. Hence: a function is one-to-one if and only if its graph intersects each horizontal line at most once. If a function f is one-to-one on a domain D with range R, then it has an inverse function f 1 defined by The domain of f 1 is R and the range of f 1 is D. f 1 (b) = a if f(a) = b. Example. Which of these functions are one-to-one? domain be restricted to obtain this property? (a) f(x) = 3x + 1 (b) f(x) = sin x (c) f(x) = x 2 (d) f(x) = e x For those that are not one-to-one, how could the (e) f(x) = 1 x (f) f(x) = { x 4, 0 x π 2 tan x, π 2 < x 0 The functions for (a), (d), (e), and (f) are all one-to-one. The function f(x) = x 2 is one-to-one when restricted to the set of nonnegative real numbers. Similarly, the function f(x) = sin x is one-to-one when restricted to the interval π 2 x π 2. Example. Let f(x) = x 2, x 0. Determine the graph of its inverse. In order to obtain the graph of a functions inverse, we can utilize the graph of the original function itself and reflect across the line: y = x. Hence, we have Figure A.2. Example. Let f(x) = x 2, x 0. Determine the inverse as a function of x. To find an algebraic form for the inverse, we first solve for x as a function of y and then interchange our variables: y = x 2 = y = x 2, (10) y = x, (11) y = x = y = x. (12) For a > 0, a 1, a logarithmic function of base a (y = log a x) is defined as the inverse of an exponential function of base a (y = a x ). A special logarithmic function is the natural logarithm: y = ln x which is the inverse of the natural exponential function. Here are some important properties of natural logarithms: ln bx = ln b + ln x, ln b x = ln b ln x, ln xr = r ln x, log a x = ln x ln a. Example. Radioactive materials have a property called its half-life which is defined as the amount of time it takes for half of the material to decay. Cesium-137 is a radioactive isotope that has a half-life of about 30 years, what would be the rate constant we would use for an exponential decay model for this isotope? L2-3

8 Figure 2: Graphs for Example A.2. An appropriate exponential decay model for this isotope would take the form: f(t) = y 0 e kt, (13) where y 0 is the initial quantity of the isotope and t is measured in years. In order to find the rate constant k, we impose the condition that f(30) = y0 2. Thus: y 0 2 = f(30), (14) = y 0 e 30k, (15) e 30k = 1 2, (16) 30k = ln 1 = ln 2, (17) 2 k = ln (18) Since trigonometric functions are periodic, they are not one-to-one and thus does not have inverses. However, if we restrict the domain, then we can define an inverse over that domain. We typically restrict the L2-4

9 trigonometric functions and thus define their inverses in the following fashion: [ f(x) = sin x, x π 2, π ] f 1 (x) = arcsin(x), x [ 1, 1], 2 f(x) = cos x, x [0, π] f 1 (x) = arccos(x), x [ 1, 1], ( f(x) = tan x, x π 2, π ) f 1 (x) = arctan(x), x (, ), 2 f(x) = csc x, x [ π ) ( 2, 0 0, π ] f 1 (x) = arccsc(x), x [, 1] [1, ], [ 2 f(x) = sec x, x 0, π ) ( π ] 2 2, π f 1 (x) = arcsec(x), x [, 1] [1, ], f(x) = cot x, x (0, π) f 1 (x) = arccot(x), x (, ). The best way to work with these functions is through right triangles. A value for arcsin x corresponds to the angle (or arc of a circle) in a right triangle with hypotenuse 1 and opposing side x. Example. Prove the following identities: (a) sin(arccos x) = 1 x 2 (b) tan(arcsin( x)) = Proof: x 1 x 2 (a) Using triangles, we note that θ = arccos x is the angle between the unit hypotenuse and side of length x of a right triangle. By the Pythagorean Theorem, the opposite side of the triangle has length 1 x 2. Hence the sine of this angle is the ratio of the lengths of the opposite side and the hypotenuse: sin(arccos x) = 1 x 2 1 = 1 x 2. (b) Using triangles, we note that θ = arcsin( x) is the angle of a right triangle in the fourth quadrant with a hypotenuse of 1 and altitude of x. By the Pythagorean Theorem, the remaining side as length 1 x2. Hence the tangent of this angle is the ratio of the signed lengths of the opposite side and the adjacent side: x tan(arcsin( x)) =. 1 x 2 L2-5

10 MATH 1110 Calculus I - Fall August 2017 Lecturer: Matt Hin Lecture 3 2.1: Rates of Change and Tangent Lines to Curves One of the earliest notions of limits came from a discussion about speed. In the late 16th century Galileo found that if you drop an object off of a tall location, the distance fallen in feet after t seconds is y = 16t 2 feet. If we re interested in the speed of this object, we can compute its average speed by dividing the distance traveled by how much time the object takes to traverse that distance. In the parlance of functions, if y = f(t) measures distance traveled at time t, the average speed between [t 0, t 1 ] is given by: y t = f(t 1) f(t 0 ). t 1 t 0 But what about instantaneous speed at a time t, when t is really close to zero? Example. Say we drop a rock, what is the instantaneous speed at time t = 1? t = 2? We can try to compute this by checking average speeds over time intervals that get smaller and smaller. Let t = t 1 t 0 = h. Then, our definition for average speed becomes: Using this formula, we find the following progression: y t = f(t 0 + h) f(t 0 ) = 16(t 0 + h) 2 16t 2 0 h h h Average Speed t 0 = It starts to look like as h 0, the average speeds at t 0 = 1 approach a limiting value of 32. Lets try to compute this algebraically: 16(t 0 + h) 2 16t 2 0 h = 16(1 + h)2 16 h = 32h + 16h2 h = h 32. Another way of looking at this is through the graph of the function y = 16x 2. These average rates of change is simply the slope of secant lines drawn on the graph. When we fix t = t 0 and reduce the size of h, we are effectively looking at secant lines that get closer and closer to the line tangent to the graph at t = t 0. L3-1

11 2.2: Limit of a Function and Limit Laws What is the difference between f(x) = x2 x and g(x) = x? Note that despite this hole in the graph of f, the function itself behaves pretty much the same as g near x = 0. Namely, that as x 0, f should approach zero like g does. This gives us an intuition for what it means to take the limit of a function. Informally, we say that a function f approaches the limit L as x approaches c if f(x) is arbitrarily close to L for all x sufficiently close to c. We write this as lim x c f(x) = L. Note that the limit of a function doesn t care about what the value of the function is at the point. Example. Let f(x) = x2 x. What is the limit of f(x) as x approaches 1? 0? Does the value of f(c) matter to the limit? For any c R, we find x 2 lim f(x) = lim x c x c x = lim x c x = c. Note that while lim x 1 f(x) = 1 = f(1), this is not the case with x = 0. The value of f(c) doesn t influence the value of the limit. Example. The following functions don t have limits as x 0: (a) (b) (c) U(x) = { 0, x < 0 1, x 0 The function here jumps from zero to one at x = 0. Note that as you approach the origin from the left, we have U(x) = 0 whereas from the right we find U(x) = 1. This disagreement tells us that there is no single limit for U(x) at x = 0. g(x) = { 1 x, x 0 0, x = 0 The function here grows too large as we approach x = 0 despite the function itself being finite at x = 0. As we approach from the left, the functions blows up to while from the right blows up to. f(x) = sin 1 x The function oscillates too much near x = 0. As x approaches zero from either direction, the sine function causes the value of f(x) to oscillate faster between 1 and 1. Properties. Let f(x) and g(x) be real-valued functions with limits L = lim x c f(x) and K = lim x c g(x). Then, the following properties hold: (i) lim x c (kf(x)) = k lim x c f(x), (ii) lim x c (f(x) ± g(x)) = lim x c f(x) ± lim x c g(x), f(x) (iii) lim x c g(x) = lim x c f(x) lim x c g(x) for K 0, (iv) lim x c (f(x)g(x)) = lim x c f(x) lim x c g(x), ( ) k, (v) lim (f(x)) k = lim f(x) x c x c Theorem 1 (Sandwich Theorem). Suppose that g(x) f(x) h(x) for all x in an open interval containing c, except possibly at x = c. Suppose also that Then lim x c f(x) = L. lim g(x) = lim h(x) = L x c x c L3-2

12 MATH 1110 Calculus I - Fall August 2017 Lecturer: Matt Hin Lecture 4 More With 2.2: Limit of a Function and Limit Laws Example. For nice enough functions f, the limit always exists for any x = c in the natural domain of the function. For example, consider the limit of the function near x = c: (a) f(x) = x 3 + 2x 2 4x + 5, c = 1 (b) g(x) = e x cos x, c = 0 (a) lim f(x) = lim x3 + 2x 2 4x + 5, x 1 x 1 = lim x 1 x 3 + lim x 1 2x 2 lim x 1 4x + lim x 1 5, = , = 2. (b) lim x 0 g(x) = lim x 0 (e x cos x) = (lim x 0 e x ) (lim x 0 cos x) = 1 0 = 0. Example. The following functions don t have limits as x 0: (a) U(x) = { 0, x < 0 1, x 0 The function here jumps from zero to one at x = 0. Note that as you approach the origin from the left, we have U(x) = 0 whereas from the right we find U(x) = 1. This disagreement tells us that there is no single limit for U(x) at x = 0. (b) g(x) = { 1 x, x 0 0, x = 0 The function here grows too large as we approach x = 0 despite the function itself being finite at x = 0. As we approach from the left, the functions blows up to while from the right blows up to. (c) f(x) = sin 1 x The function oscillates too much near x = 0. As x approaches zero from either direction, the sine function causes the value of f(x) to oscillate faster between 1 and 1. L4-1

13 Example (Polynomials and Rational Functions). As we ve seen with the sum, difference, product and quotient rules, we can also find the limits of some rational functions: x 2 + x 2 lim x 2 x 2 = lim ( x 2 x 2 + x 2 ) x lim x 2 (x 2 = 4 x) 2 = 2. But what happens when we try to take the limit near when the denominator goes to zero like x = 1? We cannot simply substitute x = 1 into the expression and simplify. Sometimes, the rational function can be reduced: x 2 + x 2 (x 1)(x + 2) x + 2 lim x 1 x 2 = lim = lim = 3 x x 1 x(x 1) x 1 x 1 = 3. Other times, the rational function can t be reduced as in the case near x = 0. In such a case, we get that the limit doesn t converge, doesn t exist and instead diverges to positive/negative infinity. More on that when we talk about asymptotes of functions. Sometimes we can t compute the limit of a function by decomposing it in terms of limit properties. Another tool we have to evaluate these limits is the Sandwich Theorem. Example. Prove that for any function f, lim x 0 f(x) = 0 implies that lim x 0 f(x) = 0. Note that lim x 0 f(x) = 0 also implies that lim x 0 f(x) = 0. Since f(x) f(x) f(x), by the sandwich theorem, we prove the implication. 2.4: One-Sided Limits Thus far we have considered two-sided limits and as we ve seen this type of limit exists when the values of f(x) approach the same quantity from either side of x = c. We can relax this condition and consider one-sided limits: and so we have the left-hand limit and the right-hand limit. Notationally, we represent these quantities by: f(x) = L and lim f(x) = M. + lim x c Example. Consider the step function: U(x) = x c { 0, x < 0, 1, x 0 The left-hand limit as x 0 is lim U(x) = 0 while the right-hand limit is lim x 0 earlier this disagreement means that the two-sided limit does not exist. Example. Consider the function: f(x) = 1 x 2. U(x) = 1. As we saw x 0 + The two-sided limit exists in the interval ( 1, 1). When x = 1, only the right-hand limit exists: 0. Similarly, only the left-hand limit exists when x = 1: lim f(x) = 0. x 1 lim f(x) = x 1 + With these examples, we have the following theorem regarding the relationship between two-sided and one-sided limits: Theorem 2. Suppose a function f is defined on an open interval containing c, except perhaps at c itself. Then the function f(x) has a limit as x approaches c if and only if it has left-hand and right-hand limits there and that these two limits are equal: lim x c f(x) = L lim x c x c f(x) = lim f(x) = L. + L4-2

14 Example. Prove that sin x lim x 0 x = 1. We will prove this statement by using the sandwich theorem to prove that the right-hand limit converges to 1 and the evenness of the function to conclude that the left-hand limit and thus the two-sided limit also converges to 1. To use the sandwich theorem, we need two functions to sandwich sin x x near x = 0. We can find one by returning to the unit circle and drawing two right triangles. From this diagram it is apparent that the area of the smaller triangle is less than the area of the sector which itself is less than the area of the larger triangle. From our homework we know that the area of the small triangle is sin θ θ, the area of the sector is θ tan θ 2, and the area of the large triangle is 2. Thus, We compute the right-hand limit as: sin θ < θ < tan θ = cos θ < sin θ θ < 1. sin θ sin θ lim cos θ < lim < lim 1 = lim = 1. x 0 + x 0 + θ x 0 + x 0 + θ By the even symmetry of the function, we find that the same holds for the left-hand limit. Since both these one-sided limits exist and are equal, we find that the two-sided limit exist and equals 1. L4-3

15 MATH 1110 Calculus I - Fall September 2017 Lecturer: Matt Hin Lecture 5 2.5: Continuity In high-school algebra, you may have heard that a function is continuous if you can draw the graph of said function with an unbroken curve. Today we ll characterize this idea using a rigorous definition. First and foremost, the idea of continuity is based around pointwise continuity or continuity at a point. This idea is defined in the follow way: Definition 1. Let c be a real number in the domain of a function f. Then the function f is continuous at c if lim x c f(x) = f(c). left-continuous at c if lim f(x) = f(c). x c right-continuous at c if lim x c + f(x) = f(c). If the function f is continuous at all points x [a, b], we say that f is continuous over [a, b]. If f is not continuous at x = c, then we say that f is discontinuous at c. Example. Determine the where f is continuous, left-continuous, and right-continuous for the following functions: { 0, x < 0, (a) U(x) = 1, x 0 (b) f(x) = 1 x 2 (a) U(x) is continuous and left-continuous over all the reals except at x = 0: {x R : x 0}. The function is right continuous over all the reals. (b) f is continuous over the open interval ( 1, 1), is left-continuous over the interval ( 1, 1], and rightcontinuous over [ 1, 1). The process here is to check three conditions if they all hold, then f is continuous at x = c: (i) f(c) exists (or c is in the domain of f) (ii) lim x c f(x) exists (iii) lim x c f(x) = f(c) L5-1

16 Remark 1. There are several types of discontinuities: removable, jump, infinite (asymptotic), and oscillating. Only the removable kind can have a continuous extension, where the hole is filled in using a piecewise function. For example, we can continuously extend the function f(x) = sin x x by the new function: { sin x F (x) = x, x 0 1, x = 0. Rational functions are common functions that have removable discontinuities and thus a continuous extension. A function is said to be continuous if the function is continuous at every point in its domain, else it is called a discontinuous function. In addition, the algebraic combinations of continuous functions are still continuous (sum, difference, scaling, products, quotients, powers, roots). In addition, the composition of two continuous functions is still continuous. Example. Prove that every polynomial: P (x) = a n x n a 0 is continuous. Note that from our discussion regarding limits, we saw that lim P (x) = P (c), x c for any polynomial P and real number c. Since the domain of P is the real line, we have that P (c) exists for any real number c, the limit exists for any real number c, and that these limits are equal to the value of P (c) for any real number c. Example. lim x 0 ( ) ( ) x + 1e tan x = lim x + 1 exp lim tan x, x 0 x 0 = 1 e 0, = 1. Theorem 3 (The Intermediate Value Theorem for Continuous Functions). If f is a continuous function on a closed interval [a, b] and if y 0 is any value between f(a) and f(b), then there exists some c, such that a c b and f(c) = y 0. Remark 2. The idea behind this theorem, is that if f is continuous then in order to go between f(a) and f(b), the function f needs to return every value between as x ranges from a to b. Remark 3. The IVT implies that the graph is connected that the graph is a single unbroken curve. Remark 4. The IVT also implies that there is an easy way to find a root or zero or solution to the equation f(x) = 0 by finding an interval where f(a) and f(b) have different signs. There must be some c [a, b] where f(c) = 0. Example. Use the IVT to prove that the equation 2x + 5 = 4 x 2 has a solution. Let s rewrite the equation as: f(x) = 2x x 2 4 = 0. Since x is defined for all real numbers, our domain is restricted by the function 2x + 5: [ 5/2, ). Note that f(0) = 5 4 < 0 while f(2) = 3 > 0. The function f is continuous over its full domain and thus continuous over the subinterval [0, 2]. Since y 0 = 0 is between f(0) and f(2), we have that there exists some value of x between [0, 2] that solves the equation. L5-2

17 MATH 1110 Calculus I - Fall September 2017 Lecturer: Matt Hin Lecture 6 2.6: Limits Involving Infinity; Asymptotes of Graphs There are two ways for limits to involve infinite quantities. The first is when the argument of the limit goes off to infinity: Definition 2. The function f(x) has the limit L as x approaches (plus/minus) infinity lim f(x) = L x ± if for every number ɛ > 0, there exists a corresponding number M such that for all x in the domain of f: Example. f(x) L < ɛ whenever x > / < M. 1 1 lim = 0 and lim x x x x = 0 Remark 5. Note that all of the limit laws and properties we ve talked about prior hold for these limits where the argument heads off to ±. The second is when the limit itself is an infinite quantity. In this case, we say, that the limit diverges to positive or negative infinity. Example. 1 1 lim = and lim x 0 + x x 0 x = These limits also lend themselves to notion of asymptotes, which are classified by their orientation. Vertical Asymptotes Definition 3. A line x = a is a vertical asymptote of the graph of a function y = f(x) if either lim x a x a f(x) = ± or lim f(x) = ±. + Example. Consider the function f(x) = x2 +1 x 2 1. Then we find that this function has a vertical asymptote at x = ±1. Horizontal Asymptotes Definition 4. A line y = b is a horizontal asymptote of the graph of a function y = f(x) if either lim f(x) = b or lim f(x) = b. x x Example. Consider the function f(x) = x2 +1 x 2 1. Then we find that this function has a horizontal asymptote at y = 1. L6-1

18 Example. ( lim x ) x = lim x x x + x , = lim x = x, x 2 x x = 0. Oblique Asymptotes Example. [ f(x) = x2 3 (x ) ( )] 1 2x 4 = x 4 As x ±, the function f(x) approaches g(x) = 1 2 x + 1. Dominant Terms Example. Let f(x) = ( ( ) x 2 + 1) 1 + 2x 4. We say that x dominates or is the dominant term when x 1 approaches ±. Similarly, 2x 4 dominates when x approaches 0. These dominant terms helps us predict the behavior of these functions near these values of x. We can show these analytically by consider the limit of the ratio: and f(x) lim x 0 1 2x 4 lim x [ ( x ) ] = lim (2x 4) x = 1 f(x) x = lim x 0 (2x 4)(x/2 + 1) = 1. L6-2

19 MATH 1110 Calculus I - Fall September 2017 Lecturer: Matt Hin Lecture 7 3.1: Tangent Lines and the Derivative at a Point Definition 5. The derivative of a function f at a point x 0, denoted by f (x 0 ) is provided that this limit exists. f f(x 0 + h) f(x 0 ) (x 0 ) = lim, h 0 h Definition 6. The slope of a curve y = f(x) at the point P = (x 0, f(x 0 )) is the quantity: f(x 0 + h) f(x 0 ) lim h 0 h The tangent line to the curve at P is the line through P with this slope. Example. Let f(x) = 1/x. (a) Find the slope of the curve at a point x = a. What is the slope at x = 1? (b) Where is the derivative equal to 1/4? (c) What happens to the derivative when a changes? (a) Hence the slope is f ( 1) = 1. f f(a + h) f(a) (a) = lim = lim h 0 h h 0 1 a+h 1 a h 1 = lim h 0 a(a + h) = 1 a 2. (b) With our expression for the slope of the curve, or derivative, we can solve for the a for which the derivative is equal to 1 4 : 1 4 = 1 a 2 = a 2 = 4 = a = ±2. (c) Note that as a 0, the derivative decays to zero, while a 0 the derivative blows up. Definition 7. The derivative of a function f(x) with respect to the variable x is the function f whose value at x is f f(x + h) f(x) (x) = lim, h 0 h provided that the limit exists. We say that f is differentiable at x if the limit f (x) exists. We can extend this to the full function and say f is differentiable if f is differentiable at every point in its domain. Example. We ve seen what the derivative is for the reciprocal function: ( ) d 1 = 1 dx x x 2, x 0. L7-1

20 MATH 1110 Calculus I - Fall September 2017 Lecturer: Matt Hin Lecture 8 3.2: Derivatives as a Function Recall that the derivative of a function is defined as f f(x + h) f(x) (x) = lim, h 0 h wherever this limit exists. Example. Differentiate Using the definition, we have: f(x) = x x 1. f f(x + h) f(x) (x) = lim, h 0 h = lim h 0 1 = lim h 0 h x+h x+h 1 x x 1 h, (x + h)(x 1) x(x + h 1), (x h 1)(x 1) 1 = lim h 0 h h (x + h 1)(x 1), 1 = (x 1) 2. Since derivatives are defined as limits, we also have the notion of one-sided derivatives for functions that are defined on a closed interval. As such the derivatives of these functions are defined using the corresponding one-sided derivative on the boundary. Example. Show that the absolute value function f(x) = x is differentiable over all the reals except for x = 0. Note that the absolute value function can be defined as the following piecewise function: { x, x 0, f(x) = x, x 0. For x (, 0), we find f (x) = 1 while for x (0, ) we have f (x) = 1. However, the one-sided derivatives are: 0 + h h 0 + h 0 L = lim = 1 and R = lim = 1. h 0 h h 0 h Since these one-sided limits don t agree, the derivative doesn t exist at x = 0. L8-1

21 Example. Find the domain of the derivative f(x) = x. The function f(x) has the domain [0, ). Let s compute the derivative in the interior of the domain: for a (0, ) z a f 1 (a) = lim = lim = 1 z a z a z a z + a 2 a. For x = 0, we require a right-hand derivative: 0 + h 0 lim h 0 + h 1 = lim h. h 0 This derivative diverges to infinity, so the derivative doesn t exist at x = 0. Therefore, the domain of the derivative is (0, ). Derivatives don t exist for functions with: (i) corners, where the two one-sided derivatives differ (ii) cusps, where the two one-sided derivatives diverge to ± (iii) vertical tangents, where the curve goes vertical and the derivative diverges to (iv) discontinuities (v) oscillations about a point, where the derivative continues to oscillate around the point Theorem 4 (Differentiability Implies Continuity). If f has a derivative at x = c, then f is also continuous at x = c. Note that the converse is not true: that if f is continuous at a point does not mean that f is differentiable at the point. L8-2

22 MATH 1110 Calculus I - Fall September 2017 Lecturer: Matt Hin Lecture 9 3.3: Differentiation Rules Properties (Power Rule). d dx xn = nx n 1. Properties (Derivative of a Natural Exponential). d dx ex = e x. Properties (Differentiation is Linear). For constants a, b R and differentiable functions u = u(x) and v = v(x): d (au(x) ± bv(x)) = adu dx dx ± b dv dx. Example. Differentiate the following function: f(x) = e π + e x + 2x 3 3x 2/3 + x x 4 + x 4/3 2 + x 2+π. f (x) = e x + 6x 2 + 2x 1/3 + 2x x x 7/ π xπ. 2 Properties (Product and Quotient Rule). If u = u(x) and v = v(x) are differentiable at x and v(x) 0, then d du (uv) = dx dx v + u dv dx and d dx Lo-de-Hi minus Hi-de-lo over lo squared Example. Differentiate the following functions: ( u ) = v du dx u dv dx v v 2 f(x) = 1 x ( x 2 + e x + 4 ) and g(t) = t2 1 t f (x) = ( 1 x 2 ) (x 2 + e x + 4 ) + 1 x (2x + ex ), = 1 + (x 1) ex x 2 4 x 2. L9-1

23 and g (t) = dg dt, = (t3 + 1)(2t) (t 2 1)(3t 2 ) (t 3 + 1) 2, = t4 + 3t 2 + 2t (t 3 + 1) 2. Properties (Higher-Order Derivatives). The second derivative of a function y = f(x) is written as: The nth derivative is thus: f (x) = d2 y dx 2 = d dx ( ) dy = dy dx dx = y = D 2 (f)(x) = Dxf(x). 2 f (n) (x) = y (n) = d dx y(n 1) = dn y dx n = Dn y. Example (Higher-order Derivatives of Polynomials). Find all derivatives of the polynomial: y = x 3 3x y = 3x 2 6x, y = 6x 6, y = 6, y (4) = 0. All higher order derivatives are zero as well. L9-2

24 MATH 1110 Calculus I - Fall September 2017 Lecturer: Matt Hin Lecture : The Derivative as a Rate of Change Definition 8. The instantaneous rate of change of f with respect to x at x 0 is the derivative: f f(x 0 + h) f(x 0 ) (x 0 ) = lim. x 0 h Example. How does the area of a circle change with respect to its radius when the diameter is 10 m? The area of a circle is given by the function: A(r) = πr 2. We are interested in Motion Along a Line A (5) = 2π(5) = 10π. Suppose we have an object (like a bead) is moving along a coordinate line, we can describe its position as a function of time: s = f(t). The displacement of an object over an interval [t, t + t] is s = f(t + t) f(t). Then, the (instantaneous) velocity is given as: v(t) = ds dt = lim f(t + t) f(t). t 0 t The speed is the absolute value of the velocity, disregarding information about direction. The acceleration and jerk is given by the second and third derivative of the position of an object: a(t) = d2 s dt 2 and j(t) = d3 s dt : The Derivative of Trigonometric Functions Properties (Derivative of Trig Functions). d (sin x) = cos x dx d (cos x) = sin x dx d dx (tan x) = sec2 x d (csc x) = csc x cot x dx d (sec x) = sec x tan x dx d dx (cot x) = csc2 x L10-1

25 MATH 1110 Calculus I - Fall September 2017 Lecturer: Matt Hin Lecture 11 Example Problems From Past Prelims Example (Spring 2016). Use a relevant theorem(s) from class to show that there exists a number a in [0, 1] such that a = cos a. This is an archetypal example where the Intermediate Value Theorem can be used. Let f(x) = x cos x. Note that f(0) < 0 while f(1) > 0. Since f(x) is continuous (since it is a difference of two continuous functions), there exists some a [0, 1] such that f(a) = 0. This in turn implies that a = cos a. Example (Fall 2016). True or False: The function f(x) = x 2 sin x x 3 is odd In order for f(x) to be odd, it must satisfy f( x) = f(x). Thus: f( x) = ( x) 2 sin( x) ( x) 3, = x 2 ( sin x) ( x 3), = ( x 2 sin x x 3), = f(x). Thus f is indeed odd, and so the statement is TRUE. Example (Spring 2017). Find the value(s) of c so that f(x) is continuous on all of R: { x 2, x 4, f(x) = c x, x > 4. Note that f(x) is continuous at all of the points away from x = 4. The only point of contention on the x-axis for the function is x = 4. In order for f(x) to be continuous at this point, we need the following condition: lim f(x) = f(4) = 16. x 4 This two-sided limit must be evaluated as individual one-sided limits on each branch. Hence, and lim f(x) = lim x2 = 16 x 4 x 4 c lim f(x) = lim x 4 + x 4 + x = c 4. Thus in order for the two-sided limit to exist, the right-hand limit must evaluate to the same finite value as the left-hand limit. Thus c = 64. L11-1

26 MATH 1110 Calculus I - Fall September 2017 Lecturer: Matt Hin Lecture : Derivatives of Trig Functions d (sin x) = cos x dx d (cos x) = sin x dx d dx (tan x) = sec2 x d (csc x) = csc x cot x dx d (sec x) = sec x tan x dx d dx (cot x) = csc2 x 3.6: The Chain Rule Theorem 5 (Chain Rule). If f(u) is differentiable at the point u = g(x) and g(x) is differentiable at x, then the composite function (f g)(x) = f(g(x)) is differentiable at x, and (f g) (x) = f (g(x))g (x). The intuition here is that a change in x induces a change throughout the composition. So a change in x induces a change in g which is multiplicatively compounded by the change in f from the change in g. Example. Differentiate the following function with respect to x: f(x) = cos(1 + x 2 + e x ) + e sin x + tan(e 1/x ) + (sin(xe x )) 2 df dx = (2x + ex ) sin(1 + x 2 + e x ) + e sin x cos x e1/x sec 2 (e 1/x ) x 2 + 2(e x + xe x ) sin(xe x ) cos(xe x ). L12-1

27 MATH 1110 Calculus I - Fall October 2017 Lecturer: Matt Hin Lecture : Implicit Differentiation Thus far we have considered relations between the variables x and y in terms of explicit equations, known as functions: y = f(x). For some more complex relationships or graphs, several functions are required to fully describe the dependency. Alternatively, the relationship can be described using an implicit equation like x 2 + y 2 = 1, which can represent the same graph that might have needed several explicit functions. With this implicit form, we can differentiate implicitly in order to compute the derivative of these relationships. Implicit differentiation is a two-step process: (1) differentiate the implicit equation with respect to the independent variable, taking care to use the chain rule whenever necessary, and (2) isolating the derivative quantity such as dy dx. Note that this process will yield the same derivative at each point done the typical manner. Example. Determine the derivative of the curves: (a) y 2 = x (b) x 2 + y 2 = 25 (c) y 2 = x 2 + sin xy (a) (b) (c) 2y dy dy = 1 = dx dx = 1 2y. 2x + 2y dy dy = 0 = dx dx = y x. dy 2x + y cos xy = dx 2y x cos xy Example. Determine the equation of the tangent line and normal line to the curve x 3 + y 3 9xy = 0 at P (2, 4). We find the derivative via implicit differentiation: dy 3y x2 = dx y 2 3x. The derivative at P is dy dx (P ) = 4 5. The tangent line has slope equal to this value and along with the point P, we find y = 4 5 x L13-1

28 The normal line has a slope that is the negative reciprocal of the slope of the tangent line. With the point P, we have: y = 5 4 x : Derivatives of Inverse Functions Theorem 6 (Derivative of an Inverse Function). If f has an interval I as a domain and f (x) exists and is never zero on I, then f 1 is differentiable at every point in its domain (the range of f). The value of (f 1 ) at a point b in the domain of f 1 is given by df 1 dx = x=b 1 df dx x=f 1 (b) Proof: We omit the proof of the first part of the theorem, and consider the second part. Since f 1 exists on the range of f over I, we can use implicit differentiation alongside the definition of the inverse function: Differentiating both sides with respect to x yields: f(f 1 (x)) = x. f (f 1 (x)) (f 1 ) (x) = 1.. Isolating the derivative of the inverse completes the proof. L13-2

29 MATH 1110 Calculus I - Fall October 2017 Lecturer: Matt Hin Lecture : Derivatives of Inverse Functions and Logarithms We saw earlier that (f 1 ) (b) = 1 f (f 1 (b)). Let s use this identity to fill out our portfolio of derivatives for logarithms and exponentials: d dx (ln x) = 1 x d dx (ax ) = a x ln a, a > 0 d dx (log a(x)) = 1 x ln a. With knowledge of the derivative of logarithms, we can find derivatives of some rational functions more easily. Logarithmic differentiation is a three step process: (1) take the logarithm of the function and separate the function into several terms while respecting domains, (2) implicitly differentiate and isolate the derivative, and (3) substitute any dependent variable and combine terms. Example. Compute the derivative of First we take the logarithm of the equation: y = (x2 + 4)(x + 3) 1/2. x 1 Now we implicitly differentiate: ln y = ln (x2 + 4)(x + 3) 1/2 x 1 Multiplying through by y and substitution yields: = ln(x 2 + 1) + 1 ln(x + 3) ln(x 1). 2 ( ) 1 dy 1 y dx = x 2 2x x x 1. dy dx = (x2 + 4)(x + 3) 1/2 x 1 Example. Compute the derivative of f(x) = x x for x > 0. Theorem 7 (The Number e as a Limit). ( 2x x x + 6 f (x) = d ( e x ln x ) = x x (1 + ln x). dx 1 x 1 ). e = lim x 0 (1 + x) 1/x. L14-1

30 Proof: Let f(x) = ln x. Then we know that f (x) = 1 x for x > 0 and so f (1) = 1. Thus by the definition of a derivative: f f(1 + h) f(1) 1 ( (1) = lim = lim ln(1 + h) = lim ln(1 + h) 1/h) = ln h 0 h h 0 h h 0 ( lim (1 + h)1/h h 0 ) = 1. Thus, lim (1 + x 0 x)1/x = e. 3.7: More About Inverse Trig Functions y = arcsin x, D = [ 1, 1], R = [ π/2, π/2] y = arccos x, D = [ 1, 1], R = [0, π] y = arctan x, D = (, ), R = ( π/2, π/2) y = arccscx, D = (, 1] [1, ), R = [ π/2, 0) (0, π/2] y = arcsecx, D = (, 1] [1, ), R = [0, π/2) (π/2, π] y = arccotx, D = (, ), R = (0, π) Note that arcsin x, arctan x, and arccscx are also odd functions. In order to compute the derivatives of these inverse trig functions, we perform the usual steps but rely on right triangles to compute the compositions: Example. d dx (arcsin x) = 1 cos(arcsin x) = 1, x < 1 1 x 2 d dx (arctan x) = 1 sec 2 (arctan x) = x 2, d dx (arcsecx) = 1 sec y tan y = 1 x x 2 1, x > 1 The other derivatives can be computed in a similar fashion. Alternatively we can use the observation that the sum of the cofunctions is always π/2. Hence, d dx (arccos x) = 1, x < 1 1 x 2 d dx (arccotx) = x 2. d dx (arccscx) = 1 x x 2 1, x > 1 L14-2

31 MATH 1110 Calculus I - Fall October 2017 Lecturer: Matt Hin Lecture : Related Rates Related rates problems are those that feature some relationship between derivatives of expressions. For example, from the volume of a sphere V = 4 3 πr3, we can relate how a change in the radius will induce a change in the volume: Example (Related Rates Problem Strategy). dv dt = 4πr2 dr dt. (1) Draw a picture, label variables and constants. (2) Identify numerical information given. (3) Identify desired quantity from problem. (4) Determine an equation that relates the variables. (5) Differentiate with respect to the independent variable. (6) Evaluate for desired quantity. Example (Water Tank). Water runs into a conical tank at a rate of 9 cubic feet per min. The tank is oriented nose down and has a height of 10 feet with a base radius of 5 feet. How fast is the water level rising when the water is 6 feet deep? We know the following numerical info: We have the following relationships: dv dt = 9 when h = 6 Thus the related rates equation we have is: V = 1 3 πr2 h, x = y 2, = V = π 12 y3. dv dt = π dh h2 4 dt = 9 = π dh (6)2 4 dt Hence, dh dt = 1 π = Example (Drone VTOL). A drone is rising straight up from eye level of its operator 15 meters away. At the moment the operator s elevation angle is π/4, the angle is increasing at the rate of 0.14 radians per minute. How fast is the drone rising at that moment? L15-1

32 We know: Relationship: Therefore for θ = π/4 and dθ dt = 0.14, y = 15 tan θ dθ dt = 0.14 when θ = π 4. hence dy dt = 4.2. dy dt = 15(sec2 θ) dθ dt. Example (Pulley Draw). A rope is slung over a pulley and attached to a weight at one end. The other end is held 5 feet above ground in the hand of a worker. Suppose the pulley is 25 feet above ground, the rope is 45 feet long, and the worker is walking rapidly away from the pulley at the rate of 4 feet/sec. How fast is the weight being raised when the worker is 21 feet away? We know: dx = 4 dt when x = 21. Relationship: x 2 = z 2 and 20 h + z = 45, so x 2 = (25 + x) 2 Hence, our related rates equation is: 2x dx dt = 2(25 + h)dh dt = dh dt = 2.9. L15-2

33 MATH 1110 Calculus I - Fall October 2017 Lecturer: Matt Hin Lecture : Linearization Sometimes a function can be very difficult to analyze and compute values for in this case, we can use a linear approximation to use as a surrogate. We ve seen linearizations before in the context of derivatives: the standard linear approximation is the tangent line to the graph y = f(x) at the point x = a. Note that the tangent line to the graph has a slope equal to the derivative of the graph at the point x = a. Furthermore, the tangent line has a point (x, y) = (a, f(a)). Thus, with the point-slope information, we have: Hence, y = k + m(x h) = y = f(a) + f (a)(x a). Definition 9. If f is differentiable at x = a, then the approximating function L(x) = f(a) + f (a)(x a) is the linearization of f at a. The approximation f(x) = L(x) of f by L is the standard linear approximation of f at a. The point x = a is the center of the approximation. In general, this approximation is only good near the point that the linearization is created around. Linear approximations act as a decent surrogate for the function over an interval for which the approximation is good. Example. Determine the linearization of f(x) = x 2 at x = 0 and x = 1. First thing we need for a linearization is the derivative of f. Since f is differentiable, we can compute the derivative of f for all x and then evaluate the derivative at each relevant point. Thus, we find f (x) = 2x and the linearizations: L(x) = f(0) + f (0)(x 0) = L(x) = 0 and L(x) = f(1) + f (1)(x 1) = L(x) = 1 + 2(x 1). Example. Determine the linearization of f(x) = 1 + x at x = 0. Since f(x) is differentiable, we find: f (x) = 1 2 (1 + x) 1/2. Thus, L(x) = f(0) + f (0)(x 0) = L(x) = 1 + x 2. We can check the accuracy of this approximation: L16-1

34 x Approx True Abs Err e e e-2 Example. Determine the linearization of f(x) = cos 2 x at x = π/4. Since f(x) is differentiable at x = π/4, we find: f (π/4) = 2 cos(π/4) sin(π/4) = 1 Thus, L(x) = f(π/4) + f (π/4)(x π/4) = L(x) = 2 2 (x π/4). Example. Approximate f(x) = ln x at x = We note that x = 2.99 is close to a = 3, we can expand a linearization of f around a = 3. Since f(x) is differentiable, we find: Thus, f (x) = 1 x, L(x) = f(3) + f (3)(x 3) = L(x) = ln 3 + x 3. 3 Therefore, the approximation is given by L(2.99) = ln = ln L16-2

35 MATH 1110 Calculus I - Fall October 2017 Lecturer: Matt Hin Lecture : Extreme Values of Functions on Closed Intervals Extreme values or extrema are maximal or minimal values of a function. There are two types of extrema: global and local. Definition 10. Let f be a function with domain D. Then f has an absolute/global maximum value on D at a point c if for all x D, f(x) f(c). Similarly, f has an absolute/global minimum value on D at a point c if for all x D, f(x) f(c). A local maximum/minimum of the function f at a point c occurs when the maximum/minimum condition holds for an open interval in D. Theorem 8 (Extreme Value Theorem). If f is continuous on a closed interval [a, b], then f attains both a global maximum M and global minimum m in [a, b]. That is, there are numbers x 1 and x 2 in [a, b] such that f(x 1 ) = m, f(x 2 ) = M, and m f(x) M for every other x [a, b]. Theorem 9 (First Derivative for Local Extrema). If f has a local maximum or minimum at an interior point c of its domain, and if f is defined at c, then f (c) = 0. This property for local extrema gives us the ability to locate these special points by determining where f (x) = 0. We call these points critical points. Not all critical points are local extrema however. For example, x 3 has a zero derivative at x = 0, but x = 0 is not a local extremum. By the EVT, we can generate a procedure to determine the global extrema of a continuous function on a closed domain however. 1. Find all critical points of f on an interval. 2. Evaluate f at all critical points and endpoints. 3. Take the largest and smallest of these points. Example. Find the extrema values of f(x) = 10x(2 ln x) on I = [1, e 2 ]. We build our candidate list of extrema, which will consist of endpoints of I and any critical points on the interior of I. We find the critical points: f (x) = 0 = 10(1 ln x) = 0 = x = e. Thus, the candidate list are x = 1, e, e 2, We find that f(1) = 20, f(e) = 10e, and f(e 2 ) = 0. Thus, x = e is the global maximum and x = e 2 is the global minimum. Example. Find the extrema values of f(x) = x 3 x on I = [ 2, 2]. We build our candidate list of extrema, which will consist of endpoints of I and any critical points on the interior of I. We find the critical points: f (x) = 0 = 3x 2 1 = 0 = x = ± 1 3. Thus, the candidate list are x = ± 1 3, ±2, We find that f(±2) = ±6 and f(±1/ 3) = 2 3/9. Thus x = 2 is global minimum, x = 2 is global maximum, while the critical points are local extrema. L17-1