Math 30G – Calculus – Sample Exam Fall 2012 The Harvey Mudd College Honor Code applies to this exam. That means it is a violation of the Honor Code to make this exam accessible in any format to any person not currently enrolled in this class this semester. DO NOT RE-DISTRIBUTE THIS SOLUTION FILE 1 20 2 20 3 10 4 10 5 10 Total 70 1 Problem 1 (2 pts each). True or False (with no explanations needed, no partial credit). (a) F If f and g are differentiable, then d [f (x)g(x)] = f 0 (x)g 0 (x). dx (b) T If f and g are differentiable, then d [f (g(x))] = f 0 (g(x))g 0 (x). dx (c) T There exists a function f : R → R such that f is continuous nowhere but |f | is continuous everywhere. (d) F If a function f is continuous on [0, ∞), then it has an absolute minimum on this interval. (e) F The slope of the tangent line at a point (a, f (a)) of the graph of a differentiable function f is equal to f 01(a) . (f) F The sum of the series ∞ X 22k k=0 Suppose we know that ∞ X 5k is 1 . 5 ak converges to 0.8. We are given no other information about k=1 the infinite series. For the remaining parts, determine if the following statements are • true (i.e., must be true), • or false (i.e., must be false). (g) F (h) T (i) F (j) T lim ak = 0.8. k→∞ lim ak = 0. k→∞ ak+1 > 1. k→∞ ak lim lim Sn = 0.8, where Sn = a1 + a2 + · · · + an . n→∞ 2 Problem 2 (5 pts each, 20 pts total). Short answer. (a) The function ( if x < 1 x + b if x ≥ 1 f (x) = is continuous for every x if b = x2 −4 x−2 2 . (b) Assume we have two differentiable functions f and g. We are given that f 0 (x) = g 0 (x), f (1) = 0, and g(1) = 5. Are f and g related? Can you find an explicit relationship between f and g? From a result proven in class, we know that since f 0 (x) = g 0 (x), it follows that f (x) = g(x) + C for some constant C. We also know that 0 = f (1) = g(1) + C = 5 + C =⇒ C = −5. So, the relationship between f and g is f (x) = g(x) − 5 . √ 3 1+h−1 by relating it to the value of a derivative. (This means you should h→0 h not use l’Hˆopital’s rule here.) √ Consider the function f (x) = 3 x. This function is differentiable at x = 1, and the 2 derivative at 1 is f 0 (1) = 13 1− 3 = 13 . By the definition of derivative, we also know that (c) Find lim f (1 + h) − f (1) f (1) = lim = lim h→0 h→0 h 0 √ √ √ 3 3 1+h− 31 1+h−1 = lim . h→0 h h Thus, it must be that the value of this limit is 1 since f 0 (1) = 13 . 3 1 = 0 means the following (fill in blanks appropriately): n→∞ n2 (d) To say that lim For every we have > 0 , there is an index N such that for all integers n ≥ N , 1 − 0 < . n2 (If you just wrote n12 , that is correct also. If you did not include the word “integers” above, that is fine also.) To satisfy the above definition, if = 1 , 100 then the smallest integer N can be is 3 11 .. Problem 3 (10 pts). (a) Suppose f (x) is a differentiable function. Show that at any x, f (x + h) − f (x − h) = f 0 (x). h→0 2h lim Hint: You may find it helpful to add and subtract a value to the numerator in the limit. Assume f is a differentiable function. Then we have f (x + h) − f (x) + f (x) − f (x − h) by adding and subtracting f (x + h) − f (x − h) = lim f (x) in numerator h→0 h→0 2h 2h f (x + h) − f (x) f (x − h) − f (x) = lim − h→0 2h 2h f (x + h) − f (x) 1 f (x + (−h)) − f (x) by limit 1 + lim = lim properties 2 h→0 h 2 h→0 −h 1 1 by definition of derivative = f 0 (x) + f 0 (x) 2 2 = f 0 (x). lim (b) Use the limit given in part (a) to find f 0 (x) for the function f (x) = x2 . For the function f (x) = x2 , we have f (x + h) − f (x − h) (x + h)2 − (x − h)2 = lim h→0 h→0 2h 2h 2 x + 2xh + h2 − (x2 − 2xh + h2 ) = lim h→0 2h 4xh = lim h→0 2h = lim 2x lim h→0 = 2x . 4 Problem 4 (10 pts). Let f be a function having derivatives of all orders for all real numbers. The third-degree Taylor polynomial for f at the point a = −2 is given by 3 1 T3 (x) = 2 − (x + 2)2 − (x + 2)3 . 8 12 (a) Find f (−2), f 0 (−2), and f 00 (−2). Using the formula for the third-order Taylor polynomial for f at a = −2, we have 3 X f (k) (−2) k=0 k! 3 1 (x + 2)k = T3 (x) = 2 − (x + 2)2 − (x + 2)3 . 8 12 It follows that f (−2) = 2 f 0 (−2) = 0 f 00 (−2) = − 3 4 (since no (x + 2) term exists) f 00 (−2) 3 since =− . 2! 8 (b) Determine whether f has a local minimum, a local maximum, or neither at the point −2. Justify your answer. Since f 0 (−2) = 0 and f 00 (−2) = − 43 < 0, we know that f has a local maximum at x = −2 by the second derivative test. (c) Use T3 (x) to find an approximation for f (−1). For an approximation to f (−1), we have 3 1 3 1 37 f (−1) ≈ T3 (−1) = 2 − (−1 + 2)3 − (−1 + 2)3 = 2 − − = . 8 12 8 12 24 (d) Find a best linear approximation to the function f at a = −2. The best linear approximation to f at the point a = −2 is given by the tangent line to f at the point −2. Furthermore, this tangent line is contained in the third-order polynomial and is T1 (x) = 2 . 5 Problem 5 (10 pts). Use the mean value theorem to prove that | sin y − sin z| ≤ |y − z| for all y, z ∈ R. Let [y, z] be any interval of the real line. Since the function f (x) = sin x is continuous and differentiable over all of R, it is certainly differentiable over (y, z). By the mean value theorem, there exists some value c ∈ (y, z) such that f 0 (c) = f (y) − f (z) sin y − sin z = . y−z y−z Therefore, we have sin y − sin z = f 0 (c)(y − z) = (cos c)(y − z) since f 0 (x) = cos x, which implies that |sin y − sin z| = |(cos c)(y − z)| = |cos c| |y − z| ≤ |y − z| since | cos x| ≤ 1 for all x (including x = c). Thus, we have shown that | sin y − sin z| ≤ |y − z| for all y, z ∈ R. 6