University of Toronto MAT137Y1 Calculus! Test 2 1 December 2017 Time: 110 minutes Please complete this cover page with ALL CAPITAL LETTERS. Last name...................................................................................... First name...................................................................................... Email................................................................. @MAIL.UTORONTO.CA Student number................................................................................. UTOR ID....................................................................................... Instructions: (READ CAREFULLY!) This exam booklet contains 12 pages including this one. It consists of 8 questions. The maximum score is 46 points. SHOW YOUR WORK FOR EVERY QUESTION. We may disallow answers that have no supporting work. If you need scratch paper, use the backs of the pages. We will only read and grade what you write on the front of each page. If you need extra space for a question, you may use Pages 11 and 12 for this purpose. If you do so, clearly indicate it on the corresponding problem page. No aids of any kind are allowed or needed. In particular, no calculators and no extra scrap paper. Do not write or draw anything on the QR code at the top right corner of any page. Do not turn over this page until the invigilators instruct you to do so. Good luck! 1
1. [6 points] (a) Given f(x) = x 100 + cos(2x) + 3, calculate f (0). Your answer: f (0) = 4 Solution: We have f (x) = 100x 99 2 sin(2x) f (x) = 100 99x 98 4 cos(2x). So f (0) = 4. (b) Given g(x) = e x+ x 2 +3, calculate g (1). Your answer: g (1) = 3 2 e3 Solution: We have ( ) g (x) = e x+ x 2 +3 x 1 +. x2 + 3 So g (1) = 3 2 e3. (c) Find the minimum value of the function h(x) = (x + 1) 2 e sin(x3). Hint: Stop. Think. Your answer: The minimum value is 0 Solution: Observe that h(x) 0 for all x R because (x + 1) 2 0 and e sin(x3) > 0 for all x R. We have h(x) = 0 if and only if x = 1. Therefore the minimum value of h(x) is 0. 2
2. [4 points] (a) Let f(x) = arctan x. Write the equation of the line tangent to the graph of f at the point with x coordinate 1. Your answer: y = 1 2 x + π 2 4 Solution: We have f (x) = 1 x 2 + 1. Now note that f(1) = π/4 and f (1) = 1/2. Therefore an equation of the tangent line is as required. y = 1 2 (x 1) + π 4 = 1 2 x + π 2, 4 (b) Let g(x) = sin (2x + π). Calculate g (403) (x). Your answer: g (403) (x) = 2 403 cos(2x + π) Solution: The first few derivatives are g (x) = 2 cos(2x + π) g (x) = 2 2 sin(2x + π) g (x) = 2 3 cos(2x + π) g (4) (x) = 2 4 sin(2x + π) Thus, g (4) (x) = 2 4 g(x). By applying this repeatedly (or by induction) we see that g (4k) (x) = 2 4k sin(2x + π) for every positive integer k. In particular and taking three more derivatives: g (400) (x) = 2 400 sin(2x + π) g (403) (x) = 2 403 cos(2x + π). 3
3. [4 points] Let f be the function with domain (0, ) defined by f(x) = x x. Find a formula for its first and second derivatives. { f (x) = x x (ln x + 1) Your answer: f (x) = x ( ) x (ln x + 1) 2 + 1 x Solution 1: We can write f(x) = e x ln x. Then f (x) = e x ln x (ln x + 1) = x x (ln x + 1), [ ] d f (x) = dx (xx ) (ln x + 1) + x x (ln x + 1) = x x (ln x + 1) 2 + x x 1x ( = xx (ln x + 1) 2 + 1 ), x as required. Solution 2: We can use logarithmic differentiation. First, we write ln f(x) = x ln x and then we take derivative with respect to x on both sides f (x) f(x) = ln x + 1 Therefore: Then we take one more derivative: f (x) = f(x) [ln x + 1] f (x) = f (x) [ln x + 1] + f(x) 1 x = f(x) [ln x + 1] 2 + f(x) 1 [ x = x x (ln x + 1) 2 + 1 ] x 4
4. [5 points] We have a square piece of cardboard with side length 6 metres (figure on the left). We cut off a square (shaded in the picture) of the same size from each corner, and fold it into a box with an open top (figure on the right). What is the largest possible volume of the box? Solution: Let x be the length of a side of a corner square. Then 0 x 3 and the volume of the box is given by f(x) = x(6 2x) 2, 0 x 3. Since f is continuous we know that it has a maximum and a minimum on [0, 3] by the Extreme Value Theorem. The maximum must be at an endpoint (0 or 3) or at a critical point. To find the critical points: f (x) = 12(x 2 4x + 3) = 12(x 1)(x 3). We have only one critical point in (0, 3), namely x = 1. We compare the three candidates: f(0) = 0 f(1) = 16 f(3) = 0 The largest value is 16, so it must be the maximum. The largest possible volume of the box is 16m 3. 5
5. [5 points] Let a R. Let f be a function which is differentiable at a. Assume that f is always positive. We define a new function g by g(x) = f(x). Prove that g is differentiable at a and that g (a) = f (a) 2 f(a). Write a proof directly from definition of derivative as a limit without using any of the differentiation rules. Solution: According to the definition of derivative we consider g g(x) g(a) f(x) f(a) (a) = lim = lim x a x a ( f(x) f(a))( f(x) + f(a)) = lim = lim (x a)( f(x) + f(a)) ( f(x) f(a) x a ) ( ) 1 f(x) + f(a) Note that we can divide by f(x) + f(a) because f is always positive. Now since f is differentiable at a we have f(x) f(a) lim = f (a). x a Since differentiability implies continuity and the square root function is continuous we have lim f(x) = lim f(x) = f(a). Plugging in everything we obtain g (a) = lim g(x) g(a) x a as required. ( f(x) f(a) = lim x a ) lim ( ) 1 = f (a) f(x) + f(a) 2 f(a), 6
6. [10 points] Let I be an open interval. Let f be a function with domain I. Let c I. Define the following concepts: (a) f is differentiable at c. We say f is differentiable at c if the following limit exists. f(x) f(c) lim x c x c (b) f is one-to-one on I. Note: one-to-one and injective mean the same thing. Any of the following two is a valid definition: We say f is one-to-one on I if for all x 1, x 2 I, x 1 x 2 implies f(x 1 ) f(x 2 ). We say f is one-to-one on I if for all x 1, x 2 I, f(x 1 ) = f(x 2 ) implies x 1 = x 2. (c) f is decreasing on I. We say f is decreasing on I if for all x 1, x 2 I, x 1 < x 2 implies f(x 1 ) > f(x 2 ). (d) f has a local minimum at x = c. We say f has a local minimum at x = c if c is an interior point and there exists a δ > 0 such that for for all x (c δ, c + δ) we have f(x) f(c). (e) c is a critical point of f. We say c is a critical point of f if c is an interior point, and f is not differentiable at c or f (c) = 0. 7
7. [6 points] (a) State the Mean Value Theorem. Let f be a function which is continuous on [a, b] and differentiable on (a, b). Then there exists a number c (a, b) such that f (c) = f(b) f(a). b a (b) Prove the following theorem. Let a < b. Let f be a differentiable function on (a, b). IF x (a, b), f (x) < 0, THEN f is decreasing on (a, b). Proof. Pick any x 1, x 2 (a, b) such that x 1 < x 2. I need to show that f(x 1 ) > f(x 2 ). I want to use the Mean Value Theorem for the function f on the interval [x 1, x 2 ]. First, I check the hypotheses. Since f is differentiable on (a, b), it follows that f is continuous on [x 1, x 2 ] and differentiable on (x 1, x 2 ). Therefore by the Mean Value Theorem there exists c (x 1, x 2 ) such that f (c) = f(x 2) f(x 1 ) x 2 x 1. By assumption we know that f (c) < 0. Since x 2 x 1 > 0, it follows that f(x 2 ) f(x 1 ) < 0, or f(x 1 ) > f(x 2 ). I have proven that f is decreasing on (a, b), as required. 8
8. [6 points] On the next page, sketch the graph of a function f that satisfies all of the following properties at once: (a) [1 point] Domain f = R (b) [1 point] f is differentiable everywhere. (c) [2 points] The restriction of f to [0, ) is one-to-one, and its INVERSE has a vertical tangent line at 2. (d) [2 points] The restriction of f to (, 0] is one-to-one, and its INVERSE has derivative 2 at 2. Notes : There are no typos. It is possible to satisfy all of the properties at once. One-to-one and injective mean the same thing. If you need to do some rough work, do it on this page, and then sketch the graph of f on the next page. You will be evaluated only on the graph. If there is no graph, you will get no points. If you cannot find a function that satisfies all of the properties at once, then find one satisfying as many of them as you can for partial credit. Solution: Let g be the inverse of f restricted to [0, ). Then g has a vertical tangent line at 2. We have ± = g 1 (2) = f (g(2)) = f (g(2)) = 0. So f has a horizontal tangent line at g(2). Similarly let h be the inverse of f restricted to (, 0]. Then 2 = h (2) = So h has a tangent line whose slope is 1/2 at h(2). 1 f (h(2)) = f (h(2)) = 1 2. We have freedom in choosing the values of g(2) and h(2). The only requirement is that g(2) 0 and h(2) 0. So we can pick say g(2) = 2 and h(2) = 2. Note that f is one-to-one on [0, ) and f is one-to-one on (, 0], but it will not be one-to-one on R. A sketch of the graph of f is given in the next page. 9
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