AB2.15: Theorems of Real Analysis

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1 AB2.15: Theorems of Real Analysis Recall that a set of real numbers S R is called bounded if there is a real number M > 0 such that x < M for every x S. Definition 1. A function f : A B is called bounded provided there is a real number M > 0 such that f(x) < M for every x A. EXAMPLE 1 Show that the function f(x) = x 2 is unbounded on R but is bounded on each bounded interval I. Solution. If f(x) = x 2 is bounded on R then there exists a real number M > 0 such that x 2 < M for every x R; however, f(m + 1) = (M + 1) 2 > M + 1 > M, and this contradiction shows that f(x) = x 2 is bounded on R because there can be no bound for f(x) on R. If we restrict f(x) to a bounded interval I = [a, b] then f(x) M x [a, b], M = max{a 2, b 2 }, and, by Definition 1, f(x) = x 2 is bounded on I. A function f : A B is called bounded above if there is a real number M 1 such that f(x) M 1 for every x A. f : A B is called bounded below if there is a real number M 2 such that f(x) M 2 for every x A. Then a function f : A B is bounded if and only if it is bounded above and bounded below. A function f : A B is called bounded at x 0 A if there is an open interval J containing x 0 such that f is bounded on J A. If f is bounded on A then f is bounded at each x 0 A For example, f(x) = x 2 is bounded on every bounded interval I = [a, b] and hence is bounded at each real number x 0 THEOREM 1. If f(x) is bounded at each point x 0 A and A is a compact set then f(x) is bounded on A. In particular, if f(x) is bounded at each point x 0 [a, b] of a closed bounded interval then f(x) is bounded on [a, b]. Limit of a function Definition 2. The limit of f as x approaches a is a real number L, lim f(x) = L, provided that given any ɛ > 0 there is a δ > 0 such that if 0 < x a < δ then f(x) L < ɛ.

2 EXAMPLE 2 Show that lim(2x 1) = 5. x 3 Solution. Let ɛ > 0 be given. Find δ > 0 such that if 0 < x 3 < δ then (2x 1) 5 < ɛ, that is, x 3 < ɛ 2 Taking δ = ɛ 2, we observe that if 0 < x 3 < δ = ɛ 2 then 2 x 3 = 2x 6 = (2x 1) 5 < 2δ = ɛ. then THEOREM 2. If lim f(x) = L 1, lim g(x) = L 2 lim [f(x) ± g(x)] = L 1 ± L 2. lim f(x)g(x) = L 1L 2. lim 1/g(x) = 1/L 2, L 2 0. Prove the first equality. The limit of f as x approaches a is a number L 1, provided that given any ɛ > 0 there exists a δ 1 > 0 such that if 0 < x a < δ 1 then f(x) L 1 < ɛ/2. Similarly, the limit of g as x approaches a is a number L 2, provided that given any ɛ > 0 there exists a δ 2 > 0 such that if 0 < x a < δ 2 then g(x) L 2 < ɛ/2. Take δ = min{δ 1, δ 2 }; then both inequalities f(x) L 1 < ɛ/2 and g(x) L 2 < ɛ/2 hold if 0 < x a < δ. So [f(x) ± g(x)] (L 1 ± L 2 ) = [f(x) L 1 ] ± [g(x) L 2 ] f(x) L 1 + g(x) L 2 < ɛ/2 + ɛ/2 = ɛ. Continuity Definition 3. A function f(x) is called continuous at x 0 if lim f(x) = f(x 0 ). x x 0 Equivalently, given any ɛ > 0 there is a δ > 0 such that if x a < δ then f(x) f(x 0 ) < ɛ.

3 Definition 3 can be summarized in three stages: (i) x 0 is in the domain of f; that is, f(x 0 ) exists; (ii) x x0 lim f(x) exists; (iii) x x0 lim f(x) and f(x 0 ) are equal. If any of these three conditions fails, f is not continuous at x 0. We say that a function f(x) is discontinuous at x 0 or has a discontinuity (break) at x 0. If condition (ii) holds but (i) (and hence (iii)) fails, or if (ii) and (i) hold but (iii) fails then f is said to have a removable discontinuity at x 0. EXAMPLE 3 Prove that f(x) = x2 4 x 2 has a removable discontinuity at x 0 = 2. Solution. The function f(x) = (x 2 4)/(x 2) = x + 2, x 2 is not defined at x 0 = 2 but it is defined at every other real number and hence in a deleted neighborhood 0 < x 2 < δ of x = 2. Hence x = 2 is not in the domain of f and f(2) does not exist. On the other hand lim f(x) = lim x 2 x 2 (x2 4)/(x 2) = lim(x + 2) = 4, x 2 which means that f has a removable discontinuity at x 0 = 2: if we define f(2) = 4 then f(x) has been made continuous at x = 2. EXAMPLE 4 Prove that has a jump discontinuity at x = 0 and f(x) = sgn x, f(0) = 0 f(x) = sgn x, f(0) = 0 has a removable discontinuity at x = 0. Solution. The function f(x) = sgn x, f(0) = 0 is defined at x 0 = 0 and at every other real number and hence in a deleted neighborhood 0 < x < δ of x = 0. However, lim f(x) = 1, lim x +0 f(x) = 1 x 0 which means that lim x ±0 f(x) and f(0) are not equal and lim x 0 f(x) does not exist. Therefore f(x) = sgn x is not continuous and has a jump discontinuity at x = 0 with a jump 2 because

4 lim f(x) lim f(x) = 2. x +0 x 0 On the other hand, f(x) = sgn x has a removable discontinuity at x 0 = 2: if we define f(0) = 1 then f(x) has been made continuous at x = 1 because lim x +0 f(x) = lim x 0 f(x) = lim x 0 f(x) = f(0) = 1. Definition 4. A function f(x) has a simple discontinuity (discontinuity of the first kind) at x = x 0 if the discontinuity is either removable or a jump discontinuity. Every other discontinuity is called a discontinuity of the second kind. So if both limits lim f(x) and lim f(x) exist but x +x 0 x x 0 lim f(x) lim f(x), x +x 0 x x 0 then f(x) has a jump discontinuity. If at least one of the two one-sided limits at x = x 0 fails to exist, f(x) has a discontinuity of the second kind. THEOREM 3. If f(x) and g(x) are each continuous at x = x 0 then f ± g, fg are continuous at x = x 0 and f/g is continuous at x 0 if g(x 0 ) 0. Prove the continuity for f ± g. For f(x) and g(x) that are continuous at x = x 0, we have Then, by Theorem 2, lim f(x) = L 1 = f(x 0 ), lim g(x) = L 2 = g(x 0 ). which proves the continuity of f ± g. Thus, every polynomial lim [f(x) ± g(x)] = L 1 ± L 2 = f(x 0 ) ± g(x 0 ), f(x) = a 0 x n + a 1 x n a n 1 x + a n is continuous at each real number x = x 0, and every rational function f(x) = a 0x n + a 1 x n a n 1 x + a n b 0 x m + b 1 x m b m 1 x + b m (a ratio of two polynomials) is continuous at each real number x = x 0 except the finitely many zeros of the denominator polynomial q(x) = b 0 x m + b 1 x m b m 1 x + b m. THEOREM 4. If f(x) is continuous at x 0 and g(x) is continuous at f(x 0 ) then g(f(x)) is continuous at x = x 0. PROOF. Let ɛ > 0 be given. Since g(x) is continuous at f(x 0 ) there is a η > 0 such that if y f(x 0 ) < η then g(y) g(f(x 0 )) < ɛ. But since f(x) is continuous at x 0 there is a δ > 0 such that if x x 0 < δ then f(x) f(x 0 ) < η. Consequently, for each x satisfying x x 0 < δ we have g(f(x)) g(f(x 0 )) < ɛ. Therefore g(f(x)) is continuous at x = x 0.

5 EXAMPLE 5 Prove that f(x) = sin 1 x has a discontinuity of the second kind at x = 0. Solution. Consider the function f(x) = sin(1/x) and define f(0) = 0. The rational function 1/x is continuous at each nonzero real number and the function sin x is continuous at each real number. Hence, by Theorem 4, f(x) is continuous at each x 0. By considering the sequence we see that x n = 2 2πn 0 lim f(x) x +0 does not exist and so f(x) has a discontinuity of the second kind at x = 0. Definition 5. A function f(x) is called right-continuous at x 0 if or left-continuous at x 0 if lim x +x 0 f(x) = f(x 0 ). lim x x 0 f(x) = f(x 0 ). Thus f(x) is continuous at x 0 if and only if it is both left- and right-continuous at x 0. Properties of continuous functions LEMMA 1. If f(x) is continuous at x 0 then there exists a δ > 0 such that f is bounded on (x 0 δ, x 0 + δ); that is f(x) is bounded at x 0. PROOF. Since f(x) is continuous at x 0 there is a δ > 0 such that if x x 0 < δ then f(x) f(x 0 ) < 1, so that f(x) < 1 + f(x 0 ) = M for each x (x 0 δ, x 0 + δ). Therefore f(x) is bounded at x 0. THEOREM 5. If f(x) is continuous on the closed bounded interval [a, b] then f(x) is bounded on [a, b]. PROOF. Suppose f(x) : [a, b] R is continuous on the closed bounded interval [a, b]. By Lemma 1 f(x) is bounded at each x 0 [a, b]; thus f(x) is bounded on [a, b] by Theorem 1. COROL- LARY. If f(x) is continuous on R then f(x) is bounded on every bounded itnterval [a, b]. Define M = sup x A f(x), m = inf x A f(x).

6 EXAMPLE 6 Let f(x) = x 2 on (0, 2). f(x) is continuous and bounded on a bounded interval (0, 2). M = 4 and m = 0 but there are no points x 1, x 2 (0, 2) with f(x 1 ) = 4 and f(x 2 ) = 0. EXAMPLE 7 Let f(x) = x x { x x = if x > 0, 1+x 2 2 x2 if x < 0, 1+x 2 on R. f(x) is continuous and bounded on R. M = 1 and m = 1 (f is an odd function) but there are no points x 1, x 2 R with f(x 1 ) = 1 and f(x 2 ) = 1. THEOREM 6 (extreme-value theorem). If f(x) is continuous on the closed bounded interval [a, b] then there exist points x 1, x 2 [a, b] such that f(x 2 ) f(x) f(x 1 ) for all x [a, b]. That is, f(x 1 ) = M, f(x 2 ) = m. PROOF. If f(x) is continuous on the closed bounded interval [a, b] then, by Theorem 5, f(x) is bounded on [a, b], and so M = sup x A f(x), m = inf x A f(x) exist as real numbers. Suppose that value M is not assumed; then f(x) < M x [a, b]. Define g(x) = 1 M f(x), x [a, b]. Clearly, g(x) > 0 x [a, b] and g(x) is continuous on [a, b]. Then, by Theorem 5, g(x) is bounded on [a, b], and so there is a real number k > 0 such that g(x) k x [a, b]. Now for every x [a, b] k g(x) = 1 M f(x), and so M f(x) 1 k > 0, f(x) M 1, x [a, b], k and this contradicts the definition of M as the least upper bound of f(x) on [a, b]. Therefore, value M must be assumed; i.e., there exist a point x 1 [a, b] with f(x 1 ) = M.

7 A similar statement for m and x 2 [a, b] can be proved by applying the same argument to f. THEOREM 7 (intermediate-value theorem). If f(x) is continuous on the closed bounded interval [a, b] and k is between f(a) and f(b) then there exist a c [a, b] such that f(c) = k. PROOF. Let f(x) be continuous on the closed bounded interval [a, b] and k be between f(a) and f(b); that is, either f(a) < k < f(b) or f(b) < k < f(a). We will prove the former. Define the function g(x) = f(x) k, x [a, b]. Then Let g(a) = f(a) k < 0, g(b) = f(b) k > 0, x [a, b]. c = sup {x [a, b] : g(x) < 0}. Since g(x) is continuous on the closed bounded interval [a, b] then g(x) < 0 on some interval [a, a + δ 1 ], δ 1 > 0 and g(x) > 0 on some interval [b δ 2, b], δ 2 > 0. Thus a < c < b. Now for every x satisfying c < x b g(x) 0, for otherwise c would not be an upper bound of the set {x [a, b] : g(x) < 0}. Hence By continuity of g, lim g(x) 0. x +c g(c) = lim g(x) : g(c) 0. x +c Choose a sequence x n in [a, b] such that x n < c and g(x n ) < 0 for every integer n. Note that if there no such a sequence then the set {x [a, b] : g(x) < 0} would have an upper bound less than c, in contradiction with the definition of c. Now since g(x) is continuous at c, g(x n ) g(c). Hence g(c) 0. It follows that g(c) = 0 and consequently f(c) = k. EXAMPLE 8 Prove that the function f(x) = x + 1, x (0, 1], f(0) = 0 does not satisfy the conclusion of the intermediate-value theorem.

8 Solution. f is bounded but not continuous on [0, 1] since f fails to be right-continuous at x = 0. We have M = sup x [0,1] f(x) = 2, m = inf x [0,1] f(x) = 0. In fact f(0) = 0 and f(1) = 2 but there is no c (0, 1) with f(c) = 1, for example. It is easy to see that none of the intermediate values x (0, 1] are assumed by f. THEOREM 8. If f(x) is continuous on the closed bounded interval [a, b] and f(x) [a, b] then f has a fixed point: there exists and x 0 [a, b] such that f(x 0 ) = x 0. PROOF. Let f(x) be continuous on the closed bounded interval [a, b] and f(x) [a, b] for every and x [a, b]. If f(a) = a or f(b) = b then the theorem is proved; hence assume that f(a) < a and f(b) < b. Define g(x) = f(x) x for every and x [a, b]. Clearly g(a) > 0a, g(b) < 0, and g(x) is continuous on [a, b]. Now 0 is an intermediate value for g on [a, b]. Now, by the intermediate-value theorem, there exist a c = x 0 [a, b] such that g(x 0 ) = 0. Then f(x 0 ) = x 0. THEOREM 9. If f(x) is one-to-one and continuous on the closed bounded interval [a, b] then f is strictly monotone on [a, b]. PROOF. Let f(x) be one-to-one and continuous on the closed bounded interval [a, b]. Then f(a) f(b). Consider the case f(a) < f(b). If f is not strictly increasing on [a, b] then there exists x 1, x 2 [a, b] such that x 1 < x 2, f(x 1 ) f(x 2 ). Equality here violates the assumption that f(x) is one-to-one and continuous, and so we must have f(x 1 ) > f(x 2 ). There are two possibilities: (a) f(x 1 ) > f(b). Choose k (f(b), f(x 1 )). Then k (f(a), f(x 1 )), and by the intermediatevalue theorem there is a c 1 (a, x 1 ) with f(c 1 ) = k and there is a c 2 (x 1, b) with f(c 2 ) = k. But c 1 c 2 and this contradicts the assumption that f(x) is one-to-one. (b) f(x 1 ) < f(b). Then f(x 2 ) < f(b). Choose k (f(x 2 ), f(x 1 )). Then k (f(x 2 ), f(b)), and by the intermediate-value theorem there is a c 1 (x 1, x 2 ) with f(c 1 ) = k and there is

9 a c 2 (x 2, b) with f(c 2 ) = k. But c 1 c 2 and this contradicts the assumption that f(x) is one-to-one. It follows that f is strictly increasing on [a, b]. Finally for the case f(a) > f(b) the above argument applied to f shows that f is strictly increasing on [a, b] and hence f is strictly decreasing on [a, b]. [a, b]. THEOREM 10. If f(x) is one-to-one and continuous on [a, b] then f 1 is continuous on [m, M] where M = sup x [a,b] f(x), m = inf x [a,b] f(x). PROOF. We know that the range of f 1 is the closed bounded interval [m, M] and so f 1 is a well-defined one-to-one function defined on [m, M] with the range [a, b]. Let y 0 [m, M]. It suffices to prove that f 1 is continuous at y 0. Choose a sequence y n in [m, M] such that y n y 0. It is sufficient to show that Let f 1 (y n ) f 1 (y 0 ), n. x 0 = f 1 (y 0 ), x n = f 1 (y n ) n = 1, 2,..., and suppose that x n does not converge to x 0. Then for some ɛ > 0 x n x 0 ɛ for infinitely many integers n. Extract a subsequence x n of x n such that x n x 0 ɛ n. Now x n is a bounded sequence and by the Bolzano Weierstrass theorem x n has a convergent subsequence, say x n. Then By the continuity of f at c, x n x 0 ɛ n, x n c [a, b], c x 0. f( x n) f(c). But {f( x n)} is a subsequence of {f(x n ) = y n } and y n y 0 = f(x 0 ). Therefore, f(c) = f(x 0 ). Since this contradicts the assumption that f is a one-to-one function, we must have x n x 0 ; that is, f 1 (y n ) f 1 (y 0 ), n, which means that f 1 is continuous on [m, M]. Derivative. Differentiable functions Definition 6. The derivative of a function f(x) at x 0 is f (x 0 ) = lim f(x) f(x 0 + h) f(x 0 ). h 0 h

10 Whem the limit exists we say that f(x) is differentiable at x 0. or THEOREM 11. If f(x) is differentiable at x 0 then f(x) is continuous at x 0. PROOF. Recall that a function f(x) is continuous at x 0 if Assume that f(x) is differentiable at x 0. Then lim [f(x) f(x 0 )] = lim x x 0 f(x) f(x 0 ) x x0 x x 0 lim f(x) = f(x 0 ), x x 0 lim [f(x) f(x 0 )] = 0. x x 0 (x x 0 ) = lim x x0 f(x) f(x 0 ) x x 0 x x0 lim (x x 0 ) = f (x 0 ) 0 = 0. THEOREM 12 If f(x) has a local extremum (maximum or minimum) then either f (x 0 ) = 0 or f (x 0 ) does not exist. PROOF Let f(x) have a local maximum. We have If f (x 0 ) exists then necessarily By the above f(x) f(x 0 ) x x 0 0, x 0 < x < x 0 + δ. f(x) f(x 0 ) x x 0 0, x 0 δ < x < x 0. f(x) f(x 0 ) lim (x x 0 ) = f f(x) f(x 0 ) (x 0 ) = lim x +x 0 x x 0 x x 0 x x 0 f(x) f(x 0 ) f(x) f(x 0 ) lim (x x 0 ) 0, lim 0. x +x 0 x x 0 x x 0 x x 0 It follows that f (x 0 ) = 0. If f (x 0 ) does not exist, we have several obvious examples when f(x) has a local extremum. Note that the converes is not true: If f (x 0 ) = 0 or f (x 0 ) does not exist then f(x) may not have a local extremum (maximum or minimum). To see this consider the function f(x) = x 3 at x = 0. THEOREM 13 (Rolle s theorem). If f(x) is continuous on [a, b] and differentiable on (a, b) and f(a) = f(b) then there exists an x 0 (a, b) such that f (x 0 ) = 0. PROOF If f(x) = f(a) for every x (a, b) then f is constant on [a, b] and so f (x) = 0 for every x (a, b). Hence we can assume that there is some x (a, b) for which f(x) f(a). By

11 the extreme-value theorem, f assumes its absolute-maximum and absolute-mainimum values on [a, b]; that there exists x 1, x 2 [a, b] such that f(x 1 ) f(x) f(x 2 ) x [a, b]. By our assumption that f is not constant on [a, b] together with the fact that f(a) = f(b) it follows that there is x 0 (a, b) such that f has an absolute extremum at x 0, and f(x 0 ) is a local extremum of f. Also f (x 0 ) exists by the condition that f(x) is differentiable on (a, b). Thus, by Theorem 12, f (x 0 ) = 0. THEOREM 14 (Cauchy mean-value theorem). If f(x) and g(x) are each continuous on [a, b] and differentiable on (a, b) then there exists an x 0 (a, b) such that f (x 0 )[g(b) g(a)] = g (x 0 )[f(b) f(a)]. PROOF Let F (x) = f(x)[g(b) g(a)] g(x)[f(b) f(a)]. F is continuous on [a, b] and differentiable on (a, b). Also F (b) = f(b)[g(b) g(a)] g(b)[f(b) f(a)] = f(a)g(b) g(a)f(b) = f(a)[g(b) g(a)] g(a)[f(b) f(a)] = F (a) By Rolle s theorem, there exists an x 0 (a, b) such that F (x 0 ) = 0. But F (x) = f (x)[g(b) g(a)] g (x)[f(b) f(a)]. Hence f (x 0 )[g(b) g(a)] = g (x 0 )[f(b) f(a)].

12 PROBLEM 15.1 Show that the function f(x) = 1/x is unbounded for 0 < x < but is bounded on (a, ) for every a > 0. Solution. Take M > 0 and x > 0 such that x < 1/M; then 1/x > M so that M is not a bound for f(x) and, by Definition 1, f(x) = 1/x is unbounded for 0 < x <. Note that f(x) = 1/x is bounded at each x 0 (0, 1) (and at each x 0 > 0) but is not bounded on (0, 1). PROBLEM 15.2 Show that the function is bounded for all real x. Solution. Since we have f(x) = sin x, x 0, f(0) = 1 x sin x x x R, f(x) = sin x x and we can choose, e.g., M = 2 as a bound for f(x) so that, by Definition 1, f(x) = sin x/x is bounded on R. Note that 1/x is unbounded on R and sin x is bounded on R; however, their product, f(x) = sin x/x, is bounded on R (and at each point x 0 ). PROBLEM 15.3 Show that the function f(x) = x sin x 1, is unbounded for all real x (on R). Solution. Take an arbitrary M > 0 and choose integers k 1 and k 2 such that 2πk 1 + π 2 > M, 2πk 2 + π 2 < M. Now, Hence, sin(2πk 1 + π 2 ) = 1, sin(2πk 2 π 2 ) = 1, f(2πk 1 + π 2 ) = 2πk 1 + π 2 > M, f(2πk 2 π 2 ) = 2πk 2 + π 2 < M. Thus f(x) is neither bounded above nor bounded below and thus unbounded on R. PROBLEM 15.4 Prove that x 2 4 lim x 2 x 2 = 4.

13 Solution. The function f(x) = x2 4 x 2 = (x 2)(x + 2) x 2 = x + 2, x 2 is not defined at x = 2 but it is defined at every other real number and hence in a deleted neighborhood 0 < x 2 < δ of x = 2. Let ɛ > 0 be given. Find δ > 0 such that if 0 < x 2 < δ then f(x) 4 < ɛ, that is, Taking, for example (or δ = ɛ/2) we observe that if then PROBLEM 15.5 (x + 2) 4 = x 2 < ɛ (x 2). δ = ɛ 0 < x 2 < δ ɛ x 2 = (x + 2) 4 = f(x) 4 < δ ɛ. Prove that lim x 1 x 2x + 1 = 1. Solution. The function f(x) = x 2x + 1 is defined at every real number x 1/2 and hence in a deleted neighborhood 0 < x+1/2 < δ of x = 1/2 and in a deleted neighborhood 0 < x+1 < η of x = 1 provided that 0 < η 1/2. Let ɛ > 0 be given. Find δ > 0 such that if 0 < x + 1 < δ then f(x) 1 < ɛ. That is, we must prove that for such δ Assume that x 2x x 1 = 2x + 1 = x x + 1 < ɛ x + 1 < 1 ( 4 : x 5 ) 4, 3. 4 (x 1/2). Then 2x + 1 > 2x 1 > = 1 2,

14 so that Hence which means that for Taking 1 2x + 1 < 2. 1 f(x) 1 = x + 1 < 2 x + 1 2x + 1 f(x) 1 < ɛ 0 < x + 1 < δ = ɛ 2 δ = min{ ɛ 2, 1 4 } we observe that both inequalities x + 1 < 1 4 Note that the initial restriction and f(x) 1 < ɛ will be satisfied. x + 1 < 1 4 could be replaced by for any fixed positive x + 1 < η η < 1 2. PROBLEM 15.6 Prove that Solution. The function lim x = a, a > 0. f(x) = x is defined at every real number x 0 and hence in a deleted neighborhood 0 < x a < δ of x = a > 0. Now x + a > a so that Take ɛ > 0 and we obtain x x a a = < x a 1 x + a a δ = min{a, ɛ a} f(x) a = x a < x a 1 a < δ a ɛ a a = ɛ. PROBLEM 15.7

15 Prove that f(x) = x sin 1 x has a removable discontinuity at x = 0. Solution. Consider the function f(x) = x sin(1/x) and define f(0) = 0. The rational function 1/x is continuous at each nonzero real number and the functions sin x and x are continuous at each real number. Hence, by Theorem 3, f(x) is continuous at each x 0. The limit below exists, and lim f(x) = 0 x +0 because sin x 1 and lim x = 0. Therefore we have x 0 (i) f(0) defined as f(0) = 0 exists; (ii) lim f(x) exists; x 0 (iii) lim f(x) and f(0) are equal (to zero). x 0 So f(x) has a removable discontinuity at x = 0 and is continuous at x = 0, (and thus at each real number). PROBLEM 15.8 Find and classify all discontinuity points of the function f(x) = x 2 if x < 1, 2x + 3 if 1 x 0, x 1 if 0 < x < 2, x 3 7 if 2 x < 3, (x 3)/(x 4) if 3 x < 4, 0 if x 4, PROBLEM 15.9 Find for M = sup x [a,b] f(x), m = inf x [a,b] f(x) x 2 + 2x + 3 if x [0, 4], 2 x 1 if x [ 2, 2], f(x) = e 1/x if x [0, ), 1 x 2 if x ( 2, 1], PROBLEM Prove that the polynomial p(x) = a 0 x n + a 1 x n a n 1 x + a n, a 0 0, n = 2l 1, l = 1, 2,...,

16 has at least one real root x 0 R: p(x 0 ) = 0. Solution. Consider, for example We have p(x) = x 3 + 2x 2 + x + 1. p( 3) = = 11 < 0, p( 1) = = 1 > 0. Thus, by the intermediate-value theorem, there is an x 0 ( 3, 1) at which p(x 0 ) = 0 because 11 < 0 < 1. For a polynomial p(x) = a 0 x n + a 1 x n a n 1 x + a n, a 0 > 0, n = 2l 1 (l = 1, 2,...), one can choose a sufficiently large by modulus negative x 1 such that a 0 x n > a 1 x n a n 1 x + a n, x < x 1 and a sufficiently large positive x 2 such that a 0 x n > a 1 x n a n 1 x + a n, x > x 2. Thus, by the intermediate-value theorem, there is an x 0 (x 1, x 2 ) at which p(x 0 ) = 0.