TOPIC 3: CONTINUITY OF FUNCTIONS

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1 TOPIC 3: CONTINUITY OF FUNCTIONS. Absolute value We work in the field of real numbers, R. For the study of the properties of functions we need the concept of absolute value of a number. Definition.. Let a be a real number. Then the absolute value of a is the real number a defined by { a, if a a =. a, if a The absolute value a can be interpreted as its distance from zero on the number line. alternative definition is a = a. An.. Properties. Let a, b R. () a, and a = if and only if a =. () a a a. (3) ab = a b. (4) If a b, then a b. (5) (The Triangle Inequality) a + b a + b. (6) Let p any positive number. Then (a) a p if and only if p a p. (b) a p if and only if a p or a p. Example.. Prove the Triangle Inequality. Solution: a + b = (a + b) = a + ab + b = a + ab + b a + a b + b = ( a + b ), from which we conclude a + b a + b. Example.3. Solve the inequality 5 < x 9. Solution: From (6b), 5 < x if and only if x > 5 or x < 5; hence, x > 3 or x <, that gives the set (, ) (3, + ). From (6a), x 9 if and only if 9 x 9; hence 4 x 5, that is, x [ 4, 5]. Both inequalities hold simultaneously if and only if x belongs to (, ) (3, + ) and to [ 4, 5]. Hence, the solution set is [ 4, ) (3, 5].. Functions A function f consists of two sets, D and R, called the domain and the range of f, and a rule that assigns to each element of D exactly one element of R. This is expressed as f : D R. The graph of f is the set G of ordered pairs (x, y) such that x is in the domain of the function and y is the corresponding element in the range. The value of the function at x D is the element y R such that (x, y) G and will be denoted y = f(x). Thus, the range is the set R = {f(x) x D}, and the graph is G = {(x, f(x)) x D}.

2 y TOPIC 3: CONTINUITY OF FUNCTIONS To stress the dependence of the above sets of the function f, we most often write D(f), R(f) and G(f). We are specially interested in real functions of one variable with D, R R and of two variables, with D R R, R R. We concentrate first in functions of one variable. Example.. () Let f(x) = x. The domain is D = R and the range is R = [, + ). The graph is a parabola passing through (, ) that opens upward () Let g(x) = x. The domain is D(g) = [, + ) and the range is R(g) = [, + ). The graph is shown below (3) Let h(x) = /x. The domain is D(h) = R {} and the range is R(h) = R {}. The graph is shown below x

3 TOPIC 3: CONTINUITY OF FUNCTIONS 3 (4) Let l(x) = ln x. The domain is D(l) = (, + ) and the range is R(l) = R. The graph is shown below. 3 4 (5) Let m(x) = x. The domain is D(m) = R and the range is R(m) = [, + ). The graph is shown below (6) Let n(x) = x( x). The domain is D(n) = [, ] and the range is R(n) = [, ]. The graph is shown below The domain of a function is clear from the statement of the function. The range, however, is not always easily determined. The graph of the function helps to visualize the range. If you project lights horizontally toward the y axis from a great distance away, the shadow of the graph on the y axis gives a picture of the range.

4 4 TOPIC 3: CONTINUITY OF FUNCTIONS A function can be defined by different formulas for different intervals. As an example, the graph of the function x, if x ; f(x) = x, if < x ; x, if x >. is shown below. 3 3 Example.. It costs + /x euros per liter to manufacture x liters of a detergent. What is the total cost of manufacturing liters? What is the profit in manufacturing x liters of detergent if it is sold at euros per liter? Solution: The cost of manufacturing liters is ( + /) = euros. Let c(x) be the cost of manufacturing x liters; then c(x) = x( + /x) = x +. The profit Π(x) is equal to sales minus cost thus, Π(x) = x c(x) = x. In particular, the profit of manufacturing liters is Π() = 99 euros... Operations with functions. Consider functions f, g : D R R and λ R. () The sum of f and g is the function defined as (f + g)(x) = f(x) + g(x). () The product of function f by a scalar λ is defined as the function (λf)(x) = λf(x). (3) The product of f and g is the function defined as (fg)(x) = f(x)g(x). (4) The quotient of f and g is the function defined as (f/g)(x) = f(x)/g(x) whenever g(x)... Composition of functions. Let functions f : D(f) R and g : D(g) R such that R(f) D(g). Observe that we have imposed that g is well defined at some points of the range of f. Definition.3. The composition of functions f and g is the function g f : D R defined as (g f)(x) = g(f(x)). Example.4. If f(x) = 4 x and g(x) = x, find g f, f g and their domains and ranges. Solution: By definition Observe that g f f g. Finally (g f)(x) = g(f(x)) = g( 4 x ) = 4 x, (f g)(x) = f(g(x)) = f(x) = 4 (x) = x. D(g f) = [, ], D(f g) = [, ].

5 TOPIC 3: CONTINUITY OF FUNCTIONS 5 and (why?) R(g f) = [, 4], R(f g) = [, ]..3. Inverse function. The function I(x) = x is called the identity function. We will define the inverse of a function f as the function g (if any) such that f g = g f = I. Definition.5. The inverse of the function f : D R is denoted f and satisfies f(f (x)) = x and f (f(x)) = x x. Observe that the inverse has the following properties: () The domain of f is the range of f. () The range of f is the domain of f. Example.6. If f(x) = x and g(x) = x +, f and g are inverse functions. Actually, ( ) ( ) x x g(f(x)) = g = + = x (x + ) f(g(x)) = f(x + ) = = x. Example.7. Find the inverse of f(x) = 4 x for x. Solution: The inverse of f can be determined as follows: () Set y = 4 x. () Switch x and y, x = 4 y, y (describes y = f (x)). (3) Solve for y, y = 4 x (since y we disregard the negative root). Then f (x) = 4 x, with domain (, 4]. The graphs of f and its inverse are shown in the figure below. 4 3 (x,y) f(x) f (x) y=x (y,x) 3 4 Observe that for every point (x, y) in the graph of y = f(x), there is a point (y, x) in the graph of y = f (x). This occurs because whenever f(x) = y, then f (y) = x by the definition of inverse functions. Hence, if we fold the coordinate plane along the line y = x, then the graphs of f and f will coincide. The line y = x is the line of symmetry. Remark.8. Not all functions have inverse functions. Functions that do have inverses are called one to one functions, meaning that to each y value corresponds exactly one x value. One to one functions can be identified by the horizontal line test: If the graph of f is such that no horizontal line intersects the graph in more than one point, then f is one to one and admits inverse.

6 6 TOPIC 3: CONTINUITY OF FUNCTIONS Example.9. The function y = f(x) = 4 x is not one to one since to y = 3 corresponds two different values of x, namely x = and x = : f() = f( ) = 3. Hence, there is no inverse of f in R. This is the reason we considered the function in the region x in the previous example. The function fails to pass the horizontal line test in R but the test is positive in the region x and f admits an inverse in [, )..4. Periodic functions and symmetry. A function f is periodic with period p > if for every x in the domain of f, f(x + p) = f(x) (actually, the period is the smallest p > with this property). As we know, sin(x) and cos(x) are periodic functions of period π, and tan(x) is of period π. In general, if a function f(x) is periodic with period p, then g(x) = f(cx) is periodic with period p/c (c ). This is easily shown as follows: g(x + p/c) = f(c(x + p/c)) = f(cx + p) = f(cx) = g(x). Thus, the function sin x is periodic with period π. A function f is an even function if f( x) = f(x) for every x in the domain of f. The graph of an even function is symmetric with respect to the y axis. Functions f(x) = x, f(x) = x + x 4 and f(x) = cos x are even. A function f is an odd function if f( x) = f(x) for every x in the domain of f. The graph of an odd function has the origin as point of symmetry, that is, the origin is the midpoint of the segment joining pairs of points of the graph of f. Functions f(x) = x, f(x) = x 3 + x 5 and f(x) = sin x are odd. 3. Monotonicity A function f is said to be monotonous increasing if for any x, y D(f) x < y implies f(x) f(y) (strictly increasing if f(x) < f(y)). A function f is said to be monotonous decreasing if for any x, y D(f) x < y implies f(x) f(y) (strictly decreasing if f(x) > f(y)). The definitions can be referred to an interval I. Example 3.. The function f(x) = x is not monotonous in R but it is strictly decreasing in (, ] and strictly increasing in [, ). To show that it is strictly increasing in [, ), let x < y. Then f(y) f(x) = y x = (y x)(y + x) > since y > x. 4. Limits To determine the behavior of a function f as x approaches a finite value c, we use the concept of it. We say that the it of f is L, and write x c f(x) = L, if the values of f approaches L when x gets closer to c. Definition 4.. (Limit when x approach a finite value c). We say that x c f(x) = L if for any small positive ɛ, there is a positive δ such that whenever < x c < δ. f(x) L < ɛ We can split the above definition in two parts, using one sided its. Definition 4.. () We say that L is the it of f as x approaches c from the right, x c + f(x) = L, if for any small positive ɛ, there is a positive δ such that whenever < x c < δ. f(x) L < ɛ

7 TOPIC 3: CONTINUITY OF FUNCTIONS 7 () We say that L is the it of f as x approaches c from the left, x c f(x) = L, if for any small positive ɛ, there is a positive δ such that whenever < c x < δ. Theorem 4.3. x c f(x) = L if and only if x c f(x) L < ɛ x c f(x) = L and f(x) = L. + We can also wonder about the behavior of the function f when x approaches + or. Definition 4.4. (Limits when x approaches ± ) () x + f(x) = L if for any small positive ɛ, there is a positive value of x, call it x, such that f(x) L < ɛ whenever x > x. () x f(x) = L if for any small positive ɛ, there is a negative value of x, call it x, such that f(x) L < ɛ whenever x < x. If the absolute values of a function become arbitrarily large as x approaches either a finite value c or ±, then the function has no finite it L but will approach or +. It is possible to give the formal definitions. For example, we will say that x c f(x) = + if for any large positive number M, there is a positive δ such that f(x) > M whenever < x c < δ. Please, complete the remainder cases. Remark 4.5. If the point c belongs to the domain of the function f, then x c f(x) = f(c). Example 4.6. Consider the following its. () x x + 7 = 3. x 6 () x ± x x + 7 =, because the leading term in the polynomial gets arbitrarily large. (3) x + x3 x =, because the leading term in the polynomial gets arbitrarily large for large values of x, but x x3 x = because the leading term in the polynomial gets arbitrarily large in absolute value, and negative. =, since for x arbitrarily large in absolute value, /x is arbitrarily small. x (4) x ± (5) x x does not exists. Actually, the one sided its are: x + x = +. x x =. The right it is infinity because /x becomes arbitrarily large when x is small and positive. The left it is minus infinity because /x becomes arbitrarily large in absolute value and negative, when x is small and negative. (6) x + x sin x does not exist. As x approaches infinity, sin x oscillates between and. This means that x sin x changes sign infinitely often when x approaches infinity, whilst taking arbitrarily large absolute values. The graph is shown below.

8 8 TOPIC 3: CONTINUITY OF FUNCTIONS x, if x ; (7) Consider the function f(x) = x, if < x ; x f(x) = f() =, but x f(x) x, if x >. does not exist since the one sided its are different. x (8) x x f(x) = x =, x + x + f(x) = x x x =. does not exist, because the one sided its are different. x x + x = x x + x =, x x x = x = (when x is negative, x = x). x x In the following, f(x) refer to the it as x approaches +, or a real number c, but we never mix different type of its. 4.. Properties of its. f and g are given functions and we suppose that all the its below exist; λ R denotes an arbitrary scalar. () Product by a scalar: λf(x) = λ f(x). () Sum: (f(x) + g(x)) = f(x) + g(x). (3) Product: f(x)g(x) = ( f(x))( g(x)). (4) Quotient: If g(x), then f(x) f(x) = g(x) g(x). Theorem 4.7 (Squeeze Theorem). Assume that the functions f, g and h are defined around the point c, except, maybe, for the point c itself, and satisfy the inequalities Let x c g(x) = x c h(x) = L. Then Example 4.8. Show that x x sin g(x) f(x) h(x). ( ) =. x f(x) = L. x c Solution: We use the theorem above with g(x) = x and h(x) = x. Notice that for every x, sin (/x) thus, when x > x x sin (/x) x,

9 TOPIC 3: CONTINUITY OF FUNCTIONS 9 and when x < x x sin (/x) x. These inequalities mean that x x sin (/x) x. Since x = x =, x x we can use the theorem above to conclude that x x sin x =. 4.. Techniques for evaluating f(x) g(x). () Use the property of the quotient of its, if possible. () If f(x) = and g(x) =, try the following: (a) Factor f(x) and g(x) and reduce f(x) g(x) to lowest terms. (b) If f(x) or g(x) involves a square root, then multiply both f(x) and g(x) by the conjugate of the square root. Example 4.9. x 9 x 3 x + 3 = x 3 + x + x = x x x x (x 3)(x + 3) x + 3 ) ( + + x + + x = x (3) If f(x) and g(x) =, then either f(x) g(x) = x 3 (x 3) =. x x( + + x) = x + + x =. f(x) does not exist or g(x) = + or. (4) If x approaches + or, divide the numerator and denominator by the highest power of x in any term of the denominator. Example An important it. Recall that Example 4.. Find the following its: tan x sin x () = x x x x cos x = sin x x x sin 3x (z=3x) sin z () = x x z z x 3 x x x 4 + = x x 3 x + = + =. x 4 3 sin x x x =. x sin z = 3 = 3. z z 5. Asymptotes cos x = =. An asymptote is a line that the graph of a function approaches more and more closely until the distance between the curve and the line almost vanishes. Definition 5.. Let f be a function () The line x = c is a vertical asymptote of f if x c f(x) =. () The line y = b is a horizontal asymptote of f if x + f(x) = b or x f(x) = b. (3) The line y = ax + b is an oblique asymptote of f if (a) (b) x + x f(x) x f(x) x = a and (f(x) ax) = b, or x + = a and (f(x) ax) = b. x

10 TOPIC 3: CONTINUITY OF FUNCTIONS Notice that a horizontal asymptote is a particular case of oblique asymptote with a =. Example 5.. Determine the asymptotes of f(x) = ( + x)4 ( x) 4. Solution: Since the denominator vanishes at x =, the domain of f is R {}. Let us check that x = is a vertical asymptote of f: On the other hand ( + x) 4 x ± ( x) 4 = + ( + x) 4 x + ( x) 4 = x 4 + lower order terms x + x 4 + lower order terms =, hence y = is a horizontal asymptote at +. In the same way, y = is a horizontal asymptote at. There is no other oblique asymptotes. Example 5.3. Determine the asymptotes of f(x) = 3x3 x. Solution: The domain of f is R {}. Let us check if x = is a vertical asymptote of f. 3x 3 x ± x = x ±(3x x ) = 3x x ± Thus, x = is a vertical asymptote of f. On the other hand x ± x =. 3x 3 x ± x = (3x x ± x ) = ± thus, there is no horizontal asymptote. Let us study now oblique asymptotes: a = b = f(x) x ± x = 3x 3 x ± x 3 = (f(x) 3x) = x ± x ± (3 x ) x ± 3 = 3, ( 3x 3 ) 3x We conclude that y = 3x is an oblique asymptote both at + and. x 6. Continuity = ( x ) x ± =. The easiest its to evaluate are those involving continuous functions. Intuitively, a function is continuous if one can draw its graph without lifting the pencil from the paper. Definition 6.. A function f : R R is continuous at c if c D(f) and f(x) = f(c). x c Hence, f is discontinuous at c if either f(c) is undefined or x c f(x) does not exist or x c f(x) f(c). 6.. Properties of continuous functions. Suppose that the functions f and g are both continuous at c. Then the following functions are also continuous at c. () Sum. f + g. () Product by a scalar. λf, λ R. (3) Product. f g. (4) Quotient. f/g, whenever g(c). 6.. Continuity of a composite function. Suppose that f is continuous at c and g is continuous at f(c). Then, the composite function g f is also continuous at c.

11 TOPIC 3: CONTINUITY OF FUNCTIONS 6.3. Continuity of elementary functions. A function is called elementary if it can be obtained by means of a finite number of arithmetic operations and superpositions involving basic elementary functions. The functions y = C = constant, y = x a, y = a x, y = ln x, y = e x, y = sin x, y = cos x, y = tan x, y = arctan x are examples of elementary functions. Elementary functions are continuous in their domain. Example 6.. () The function f(x) = 4 x is the composition of the functions y = 4 x and f(y) = y /, which are elementary, thus f is continuous in its domain, that is, in D = [, +]. () The function g(x) = 4 x is the composition of function f above and function g(y) = /y, thus it is elementary and continuous in its domain, D(g) = (, +) Continuity theorems. Continuous functions have interesting properties. We shall say that a function is continuous in the closed interval [a, b] if it is continuous at every point x [a, b]. Theorem 6.3 (Bolzano s Theorem). If f is continuous in [a, b] and f(a) f(b) <, then there exists some c (a, b) such that f(c) =. Example 6.4. Show that the equation x 3 + x = admits a solution, and find it with an error less than.. Solution: With f(x) = x 3 + x the problem is to show that there exists c such that f(c) =. We want to apply Bolzano s Theorem. First, f is continuous in R. Second, we identify a suitable interval I = [a, b]. Notice that f() = < and f() = > thus, there is a solution c (, ). Now, to find an approximate value for c, we use a method of bracket and halving as follows: consider the interval [.5, ]; f(.5) = /8 + / < and f() >, thus c (.5, ). Choose now the interval [.5,.75]; f(.5) < and f(.75) = 7/64 + 3/4 > thus, c (.5,.75). Let now the interval [.65,.75]; f(.65).3 and f(.74) > thus, c (.65,.75). The solution is approximately c =.6875 with a maximum error of.65. Theorem 6.5 (Weierstrass Theorem). If f is continuous in [a, b], then there exist points c, d [a, b] such that f(c) f(x) f(d) for every x [a, b]. The theorem asserts that a continuous function attains over a closed interval a minimum (m = f(c)) and a maximum value (M = f(d)). The point c is called a global minimum of f on [a, b] and d is called a global maximum of f on [a, b]. Example 6.6. Show that the function f(x) = x + attains over the closed interval [, ] a minimum and a maximum value. Solution: The graph of f is shown below

12 TOPIC 3: CONTINUITY OF FUNCTIONS We can see that f is continuous in [, ], actually f is continuous in R, and f attains the minimum value at x =, f() =, and the maximum value at x =, f() = 5.

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