MATH. 4548, Autumn 15, MWF 12:40 p.m. QUIZ 1 September 4, 2015 PRINT NAME A. Derdzinski Show all work. No calculators. The problem is worth 10 points. 1. Let f : (, 0) IR be given by f(x) = 1/x 2. Prove that f ( 1) exists and find its value by showing that the difference quotient of f at c = 1 has a specific limit. (Do not use theorems on derivatives and algebraic operations; you may, however, invoke theorems about limits.)
QUIZ 1, 09/04/15 1. Let f : (, 0) IR be given by f(x) = 1/x 2. Prove that f ( 1) exists and find its value by showing that the difference quotient of f at c = 1 has a specific limit. (Do not use theorems on derivatives and algebraic operations; you may, however, invoke theorems about limits.) ANSWER The derivative f ( 1) exists and equals 2, since, at c = 1, f(x) f(c) x c = 1 x 2 1 1 x + 1 = 1 x x 2 2 as x 1.
MATH. 4548, Autumn 15, MWF 12:40 p.m. page 1 of 2 TEST 1 SEPTEMBER 18, 2015 PRINT NAME A. Derdzinski Show all work. No calculators. Problems worth 25 points each. 1. Determine if the function f : [0, ) IR given by f(x) = x is differentiable at 0. Prove your answer using the definition of differentiability. 2. Prove that f : [0, 2] IR given by f(x) = x 5 + x 3 + x is strictly increasing, maps [0, 2] bijectively onto [0, 42], and its inverse function g : [0, 42] [0, 2] is differentiable. Find g (3). You may use theorems established in class.
MATH. 4548, Autumn 15, MWF 12:40 p.m., Test 1 page 2 of 2 NAME 3. Let f : [1, 7] IR be given by f(x) = x + 4/x. Being continuous, f assumes its minimum value m somewhere in [1, 7]. Find m and all c [1, 7] at which f(c) = m. 4. Evaluate L = lim x x 3 ln x. You may use theorems proved in class.
MATH. 4548, Autumn 15, MWF 12:40 p.m., Test 1 ANSWERS 1. Determine if the function f : [0, ) IR given by f(x) = x is differentiable at 0. Prove your answer using the definition of differentiability. Answer: NO, it is not. The difference quotient [f(x) f(0)]/(x 0) = f(x)/x = 1/ x has no (finite) limit as x 0, since it is not bounded on any deleted neighborhood of 0 intersected with the domain of f. Specifically, for any α > 0, one has 1/ x > α whenever 0 < x < 1/α 2. More precisely, the above argument shows that 1/ x as x 0 +. 2. Prove that f : [0, 2] IR given by f(x) = x 5 + x 3 + x is strictly increasing, maps [0, 2] bijectively onto [0, 42], and its inverse function g : [0, 42] [0, 2] is differentiable. Find g (3). You may use theorems established in class. Answer: That f is strictly increasing follows since f (x) = 5x 4 + 3x 2 + 1 is positive for all x > 0. Being continuous, f assumes in [0, 2] all values between 0 = f(0) and 42 = f(2), so that f is a bijection [0, 2] [0, 42], and g = f 1 is differentiable since f 0 everywhere. Finally, g (3) = g (f(1)) = 1/f (1) = 1/9.
3. Let f : [1, 7] IR be given by f(x) = x + 4/x. Being continuous, f assumes its minimum value m somewhere in [1, 7]. Find m and all c [1, 7] at which f(c) = m. Answer: From the Interior Extremum Theorem, f (c) = 0 at any c [1, 7] at which f(c) = m, and as f (x) = 1 4/x 2, only one such c exists: c = 2. The minimum value is not assumed at either endpoint: f(1) = 5 AND f(7) = 53/7 are both greater than f(2) = 4. Thus, m = f(2) = 4. Another proof: We have f(x) 4 = (x 2) 2 /x, which is nonnegative for all x > 0, and equals 0 only for x = 2. 4. Evaluate L = lim x x 3 ln x. You may use theorems proved in class. Answer: L = 0, from l Hospital s rule: ln x 1/x lim x x 3 = lim x 3x 2 = lim 1 x 3x 3 = 0. More precisely, according to the second version of l Hospital s rule, the existence of the second (that is, third) limit implies the existence of the first one, and both limits must be the same.
MATH. 4548, Autumn 15, MWF 12:40 p.m. page 1 of 2 TEST 1 (make-up) SEPTEMBER 25, 2015 NAME M aura Danko A. Derdzinski Show all work. No calculators. Problems worth 25 points each. 1. Determine if the function f : [0, ) IR given by f(x) = x x is differentiable at 0. Prove your answer using the definition of differentiability. 2. Prove that f : [0, 2] IR given by f(x) = x 5 + 2x 3 + 3x is strictly increasing, maps [0, 2] bijectively onto [0, 54], and its inverse function g : [0, 54] [0, 2] is differentiable. Find g (6). You may use theorems established in class.
MATH. 4548, Autumn 15, MWF 12:40 p.m., Test 1 page 2 of 2 NAME 3. Let f : [1, 7] IR be given by f(x) = x + 9/x. Being continuous, f assumes its minimum value m somewhere in [1, 7]. Find m and all c [1, 7] at which f(c) = m. 4. Evaluate L = lim x x 4 ln x. You may use theorems proved in class.
MATH. 4548, Autumn 15, MWF 12:40 p.m. QUIZ 2 September 30, 2015 PRINT NAME A. Derdzinski Show all work. No calculators. The problem is worth 10 points. 1. Find the extrema of the function f : [0, 2] IR given by f(x) = x 3 3x + 7. Prove your answer.
QUIZ 2, 09/30/15 1. Find the extrema of the function f : [0, 2] IR given by f(x) = x 3 3x + 7. Prove your answer. ANSWER Since f is differentiable, all extremum values are assumed at the endpoints and/or interior points at which f vanishes. We have f (x) = 3(x 2 1), and so f has only one root in (0, 2), at x = 1. Comparing the values f(0) = 7, f(1) = 5, f(2) = 9, we see that maxf = 9 and minf = 5.
MATH. 4548, Autumn 15, MWF 12:40 p.m. QUIZ 2 (early) September 29, 2015 NAME Sidy H. Ba A. Derdzinski Show all work. No calculators. The problem is worth 10 points. 1. Find the extrema of the function f : [0, 2] IR given by f(x) = x 5 4x + 7. Prove your answer.
MATH. 4548, Autumn 15, MWF 12:40 p.m. QUIZ 3 October 12, 2015 PRINT NAME A. Derdzinski Show all work. No calculators. The problem is worth 10 points. 1. Let G : IR IR be given by G(x) = x 2 0 cos t 2 dt. Evaluate G (x). YOUR ANSWER: G (x) =
QUIZ 3, 10/12/15 1. Let G : IR IR be given by G(x) = x 2 0 cos t 2 dt. Evaluate G (x). ANSWER We have G(x) = F (x 2 ) for F : IR IR given by F (y) = y 0 cos t 2 dt. From the fundamental theorem, F (y) = cos y 2, while the chain rule implies that G (x) = 2xF (x 2 ), and so G (x) = 2x cos x 4.
MATH. 4548, Autumn 15, MWF 12:40 p.m. page 1 of 2 TEST 2 OCTOBER 23, 2015 PRINT NAME A. Derdzinski Show all work. No calculators. Problems worth 25 points each. 1. Suppose that f : IR IR and f, f, f all exist and are continuous on IR. Prove that, for any x IR, x 0 (x t) 2 f (t) dt = x 2 f (0) 2xf (0) 2f(0) + 2f(x). You may use theorems established in class, except either version of Taylor s Theorem. 2. Evaluate L = e 1 ln x x dx. YOUR ANSWER: L =
MATH. 4548, Autumn 15, MWF 12:40 p.m., Test 2 page 2 of 2 NAME 3. Find and explicit formula, not containing the integral symbol, for the function F : IR IR such that F (0) = 0 and F (x) = 3x 2 sgn x. Prove your answer. YOUR ANSWER: F (x) = 4. Use the third-order Taylor polynomial P 3 of the sine function to provide an approximate value of sin 1 10. Write the result as a fraction. YOUR ANSWER: sin 1 10 P 3( 1 10 ) =
MATH. 4548, Autumn 15, MWF 12:40 p.m., Test 2 ANSWERS 1. Suppose that f : IR IR and f, f, f all exist and are continuous on IR. Prove that, for any x IR, x 0 (x t) 2 f (t) dt = x 2 f (0) 2xf (0) 2f(0) + 2f(x). You may use theorems established in class, except either version of Taylor s Theorem. Answer: Repeated integration by parts gives x 0 (x t) 2 f (t) dt = x 2 f (0) + 2 = x 2 f (0) 2xf (0) + 2 x 0 x 0 (x t)f (t) dt f (t) dt = x 2 f (0) 2xf (0) 2f(0) + 2f(x). 2. Evaluate L = e 1 ln x x dx. Answer: From the fundamental theorem, L = F (e) F (0) for F (x) = (ln x) 2 /2, and so L = 1/2.
3. Find and explicit formula, not containing the integral symbol, for the function F : IR IR such that F (0) = 0 and F (x) = 3x 2 sgn x. Prove your answer. Answer: F (x) = x 3. 4. Use the third-order Taylor polynomial P 3 of the sine function to provide an approximate value of sin 1 10. Write the result as a fraction. Answer: and so We have P 3 (x) = x 1 6 x3, sin 1 10 P 3( 1 10 ) = 599 6000.
MATH. 4548, Autumn 15, MWF 12:40 p.m. QUIZ 4 November 6, 2015 PRINT NAME A. Derdzinski Show all work. No calculators. The problem is worth 10 points. 1. The sequence f n : (0, 1] IR given by f n (x) = 1 nx is, obviously, poinwise convergent to the zero function. Determine if the convergence is uniform on (0, 1]. Prove your answer. (You may use theorems established in class.) YOUR ANSWER: UNIFORM NOT UNIFORM (circle one)
QUIZ 4, 11/06/15 1. The sequence f n : (0, 1] IR given by f n (x) = 1 nx is, obviously, poinwise convergent to the zero function. Determine if the convergence is uniform on (0, 1]. Prove your answer. (You may use theorems established in class.) The convergence is not uniform. ANSWER In fact, if you fix any ε > 0, you will not have f n (x) = f n (x) 0 < ε for all x (0, 1] and all but finitely many n since, whatever n > 1/(2ε) you consider, choosing x = 1/(2nε) you will have x (0, 1] and f n (x) = 2ε. Another argument: we know that a sequence f n : A IR will certainly not converge uniformly on A to f if there exist sequences n k of positive integers and x k of elements of A, with k = 1, 2, 3,..., such that, for some fixed ε 0 > 0 and all k, n k k and f nk (x k ) f(x k ) ε 0. In our case, we may choose n k = k, x k = 1/k and ε 0 = 1. Yet another argument: we know that for the convergence f n f to be uniform on A, the functions f n f must be bounded on A for all but finitely many n. Here f = 0 and our f n are all unbounded, since f n (x) as x 0+.