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The answer key for the Math 444 Group E13 Final Exam at the University of Illinois at Urbana-Champaign in Fall 2006. The exam covers topics such as continuity, differentiability, and boundedness of functions on closed bounded intervals. The exam consists of four questions, each worth a different number of points. detailed solutions to each question, including explanations and proofs. The exam allows the use of textbooks but prohibits the use of calculators, mobile phones, and other electronic devices.
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University of Illinois at Urbana-Champaign Fall 2006 Math 444 Group E
Final Exam : Answer Key
You are allowed to use your textbook, but no other kind of documentation. Calculators, mobile phones and other electronic devices are prohibited.
NAME โโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
Define a function f : [0, +โ) โ R by setting f (x) = sin(
x). Show that f is continuous on [0, +โ) and differentiable on (0, +โ) ; give a formula for f โฒ(x) for all x > 0. Is f differentiable at 0? (You may use without demonstration the fact that the function x 7 โ sin(x) is differentiable on R and that sinโฒ(x) = cos(x), and the fact that x 7 โ
x is differentiable on (0, +โ) and (
x)โฒ^ = 2 โ^1 x )
Answer. The function g : x 7 โ sin(x) is continuous on R, and the function h 7 โ
x is continuous on R+. Hence f = g โฆ h is continuous on R+, since it is a obained by composition of two continuous functions. Similarly, the Chain Rule ensures that f is differentiable on (0, +โ), and f โฒ(x) = 2 โ^1 x cos(
x).
To see whether f is differentiable at 0 , the simplest thing is to go back to the definition ; one has f (0) = 0, so f (x)โf (0) xโ 0 =^
f (x) x =^
sin(โx) x. Since we know that^ limxโ^0
sin(x) x = cos(0) = 1, we see that^
f (x) x is not bounded in the neighborhood of 0 (it is the product of a function with limit 1 and a function which is not bounded).
Hence f^ (x x)โโf 0 (0) doensโt have a limit at 0 , and this shows that f is not differentiable at 0.
Let f be continuous on [0, +โ) ; for all x > 0 , set g(x) =
x
โซ (^) x
0
f (t)dt.
(a) Show that g is continuous on (0, +โ), and that g has a limit at 0 ; give the value of this limit. (b) Show that g is differentiable on (0, +โ) and that for all x > 0 one has
gโฒ(x) =
f (x) โ g(x) x
Answer. (a) By the fundamental theorem of integration we know, since f is continuous on [0, +โ), that the function x 7 โ F (x) =
โซ (^) x 0 f^ (t)dt^ is differentiable on^ [0,^ +โ)^ ; hence it is continuous on^ [0,^ +โ). So on^ (0,^ +โ) g is the product of two continuous functions, which shows that it is continuous on this interval. Also, with our notations one has g(x) = F^ ( xx ). Since F (0) = 0, and F is differentiable at 0 , we know that limxโ 0 F^ ( xx )exists and is equal to F โฒ(0) = f (0). This is equivalent to saying limxโ 0 g(x) = f (0). (b) The product of two differentiable functions is a differentiable function, so (since F is differentiable, F โฒ(x) = f (x) and g(x) = (^) x^1 ยท F (x)) we see that g is differentiable on (0, +โ) and
gโฒ(x) = โ
x^2
F (x) +
x
F โฒ(x) = โ
g(x) x
f (x) x
f (x) โ g(x) x
Pick two real numbers a, b such that a < b and let f : [a, b] โ R be continuous. We want to show that
sup{f (x) : x โ (a, b)} = sup{f (x) : x โ [a, b]}.
(a) Explain why sup{f (x) : x โ (a, b)} and sup{f (x) : x โ [a, b]} exist. (b) Show that sup{f (x) : x โ (a, b)} โค sup{f (x) : x โ [a, b]}. (c) Assume f (a) = sup{f (x) : x โ [a, b]}. Show that one also has f (a) = sup{f (x) : x โ (a, b)} (look at the sequence (a + (^) n^1 )). Can you prove a similar result when f (b) = sup{f (x) : x โ [a, b]}? (d) Prove the equality sup{f (x) : x โ (a, b)} = sup{f (x) : x โ [a, b]}.
Answer. (a) Since f is a continuous function on the closed bounded interval [a, b], the Boundedness Theorem ensures that there exists m, M such that m โค f (x) โค M for all x โ [a, b]. This shows that the set {f (x) : x โ [a, b]} is bounded, so it has a supremum (because of the completeness property of the real numbers). Since {f (x) : x โ (a, b)} is a subset of {f (x) : x โ [a, b]} and the latter set is bounded, we see that sup{f (x) : x โ (a, b)} exists too. (b) For any x โ (a, b) one has f (x) โค sup{f (x) : x โ [a, b]}, so sup{f (x) : x โ [a, b]} is an upper bound of {f (x) : x โ (a, b)}. By definition of a supremum (least upper bound), this implies that sup{f (x) : x โ [a, b]} โค sup{f (x) : x โ [a, b]}. (c) There exists N such that an = a + (^1) n โค b for all n โฅ N. Thus we can consider the sequence (f (an))nโฅN ; since f is continuous at a, this sequence converges to f (a). Since one has f (an) โค sup{f (x) : x โ (a, b)}, we also have lim f (an) = f (a) โค sup{f (x) : x โ (a, b)}. Thus if f (a) = sup{f (x) : x โ [a, b]} then we get sup{f (x) : x โ [a, b]} โค sup{f (x) : x โ (a, b)}, and the result of question (b) ensures that in factsup{f (x) : x โ [a, b]} = sup{f (x) : x โ (a, b)} in that case. Considering the sequence bn = b โ (^1) n , we obtain in the same way that if f (b) = sup{f (x) : x โ [a, b]} then sup{f (x) : x โ [a, b]} = sup{f (x) : x โ (a, b)}. (d) If the sup for f on [a, b] is obtained at either a or b then the result of question (c) shows that sup{f (x) : x โ [a, b]} = sup{f (x) : x โ (a, b)}. There must be a sup for the continuous function f on [a, b], and actually it is a maximum ; so if we are not in the case above then there exists c โ (a, b) such that f (c) = sup{f (x) : x โ [a, b]}. By definition, one has f (c) โค sup{f (x) : x โ (a, b)}, so we again obtain
sup{f (x) : x โ [a, b]} โค sup{f (x) : x โ (a, b)}
Since the maximum for f must be attained at either a, b, or some c โ (a, b), the reasoning above shows that if f is continuous on a closed bounded interval [a, b] then
sup{f (x) : x โ [a, b]} = sup{f (x) : x โ (a, b)}.
Let f : [0, 1] โ [0, 1] be an increasing function (not necessarily continuous). Show that there exists x โ [0, 1] such that f (x) = x. Hint. Consider the set E = {x โ [0, 1] : f (x) > x} ; show that one can assume that 0 โ E. Show that x = sup(E) works.
Answer. If f (0) = 0 there is nothing to prove, so we may assume that f (0) > 0 , and this gives
0 โ E = {x โ [0, 1] : f (x) > x}.
Thus we may assume that E is nonempty. Since E is bounded (it is a subset of [0, 1]), it has a supremum S, which is larger than 0 (because 0 โ E) and smaller than 1 (because 1 is an upper bound for E). For any ฮต > 0 there exists x โ E such that S โ ฮต < x < S. Since f is increasing, f (S) โฅ f (x) > x > S โ ฮต, so f (S) > S โ ฮต for all ฮต > 0. This yields f (S) โฅ S. If f (S) = S we are done ; assume that it is not true and f (S) > S. Then pick a โ [0, 1] such that S < a < f (S). Since f is increasing one has f (a) โฅ f (S) > a, so a โ E, which is impossible because S is the supremum of E. Hence f (S) = S, and we are done.
Recall that if X is a set, one denotes by P(X) the set whose elements are the subsets of X ; in other words, P(X) = {A : A โ X}. Let now X, Y be sets and f : X โ Y be a function. (a) Define a function fห : P(X) โ P(Y ) by setting fห (A) = f (A) for all A โ X. Show that fห is injective if, and only if, f is injective. (b) Similarly, define a function fห : P(Y ) โ P(X) by setting fห (B) = f โ^1 (B) for all B โ Y. Compute fห (โ ) ; show that fห is injective if, and only if, f is surjective.
Answer. (a) Assume that fห is injective, and let x, y โ X be such that f (x) = f (y). Then fห ({x}) = {f (x)} = fห ({y}), so since fห is injective we obain {x} = {y}, in other words x = y. Thus if fห is injective then f is injective. Converesly, assume that f is injective and A, B โ X are such that f (A) = f (B). Then pick a โ A. One has f (a) โ f (A) = f (B), so there exists b โ B such that f (b) = f (a). Since f is injective, this is only possible if b = a, hence a โ B. Thus A โ B ; similarly, one sees that if B โ A. This shows that A = B ; hence if f is injective then fห is injective too. We have just proved that f is injective if, and only if, fห is injective.
(b) This one is perhaps a bit more complicated. Assume that fห is injective ; one has fห (โ ) = โ , so for all y โ Y one has fห ({y}) 6 = โ . This exactly means that for all y โ Y f โ^1 ({y}) = {x โ X : f (x) = y} is nonempty, in other words that f is surjective. Converesly, assume that f is surjective, and A, B โ Y are such that fห (A) = fห (B). Then pick a โ A ; since f is surjective, there exists x such that f (x) = a. By definition, x โ f โ^1 (A), and since f โ^1 (A) = f โ^1 (B) we also have x โ f โ^1 (B), which means that f (x) = a โ B. This is true for all a โ A, so A โ B. Since A, B play symmetric roles here, one obtains similarly that B โ A. Hence A = B, hence fห is injective. This shows that is f is surjective then fห is injective. We have thus proved that fห is injective if, and only if, f is surjective.
