Calculus I Practice Final Exam B

Transcription

1 Calculus I Practice Final Exam B This practice exam emphasizes conceptual connections and understanding to a greater degree than the exams that are usually administered in introductory single-variable calculus courses. It is designed to guide students who are taking such courses to a deeper mastery of the material. While a number of questions here are fairly typical for actual examinations, you should not infer from the expression practice exam that exams encountered in introductory single-variable calculus courses will ask the same types of questions. Multiple Choice. Determine whether the given statements about a function f are true or false. f(a+h) f(a) Statement I: If lim exists, then lim f(x) exists as well. h h x a Statement II: If f has an inflection point at x = a, then f (a) =. Statement III: If f is continuous on [a,b], then f(x)dx = f(b) f(a). A. All three statements are true. B. I is true, the other two are false. C. They are all false. D. I and II are true, III is false. 2. The velocity of a particle moving along a horizontal line is given by v(t) = sin (t). Find the distance travelled by the particle from time t = to t = 2π. A. B. C. 2 D. 4 b a x 7 3. Algebraically calculate the exact limit: lim x 7 x 7 A. / B. 2 7 C D. 2 7 E. Does not exist. x 7 4. Algebraically calculate the exact limit: lim x x 7 A. B. / C. D. 2 7 E. Does not exist. 5. Express the quantity lim n n A. x3 dx n 3 C. dx x 3 n n (k n )3 k= as a definite integral. n B. x3 dx n 4 D. x 3 dx 26 R. Boerner, ASU School of Mathematical and Statistical Sciences

2 6. When you increase the side of a cube by 2%, by what percentage does the volume approximately increase? A. 2 % B. 2% C.4% D. 6% E. 8% 3 7. Given are plots of f(x), f (x) and f (x). Determine which is which. A. is f(x), 2 is f (x), 3 is f (x). B. is f(x), 2 is f (x), 3 is f (x). C. is f (x), 2 is f(x), 3 is f (x). D. is f (x), 2 is f (x), 3 is f(x). 8. Differentiate f(x) = ln(sin x). A. f (x) = B. f (x) = x cos x C. f (x) = x cos x D. f (x) = ln(cos x) 26 R. Boerner, ASU School of Mathematical and Statistical Sciences

3 9. Find the second derivative d2 y dx 2 of the implicit curve y3 = xy + at (,). A. B. C. 3 D. 3 x 3. Differentiate F(x) = t 8 dt. E. None of the others. A. 8t 7 B. 4x 2 C. 8t 7 4x 2 D. 3x 26 E. 8x 2. Use the right Riemann sum with n = 2 rectangles to approximate decimal places. dx x 4 +. Round to 3 A..867 B..72 C..97 D If f (x) > 3 for all x and f(2) = 5, what can you conclude about f(4)? You must select the best answer to receive credit for this question. A. f(4) > 6 B. f(4) > 8 C. f(4) > D. f(4) > 2 E. f(4) > 7 3. Based on the table of values for the differentiable, invertible function f and its derivative, evaluate (f ) (2). x 2 3 A. f(x) f (x) B. 7 C. 4 D. Cannot be determined based on the given information. 4. Evaluate the limit: lim x arctan( ln x) A. B. π 2 C. π 2 D. 5. None of the above. 26 R. Boerner, ASU School of Mathematical and Statistical Sciences

4 Free Response:. The following is a graph of the derivative of a twice differentiable function f defined on [,5]. Its intercepts and extrema are marked. Additionally, you know that f() = 2 and f is positive at the end points x= and x=5. The critical points of f are at x= The inflection points of f are at x= f has its absolute max y= at x= f is concave up on f has its absolute min y= at x= f is concave down on f is increasing on f(5) f() = f is decreasing on f has a relative but not absolute maximum of y= at x= The average rate of change of f on [,5] is There are c values that satisfy the conclusion of the mean value theorem applied to f on [,5]. 26 R. Boerner, ASU School of Mathematical and Statistical Sciences

5 2. Show that of all rectangles with a given area, the square has the smallest perimeter. 3. Given the table of values for a differentiable function f that has a continuous derivative, a. approximate f () and show appropriate work. x f(x) b. use the result from a. and linear approximation of f at x= to estimate f(.2). 4. Evaluate the definite integral exactly and simplify as much as possible. Give the exact answer, not a decimal approximation. 5. Algebraically calculate the exact limits: a. lim arctan e x x b. lim x + c. lim x arctan e x arctan e x π 4 sec x tan x dx 6. Evaluate the definite integral and explain your reasoning. (x 7 + 2x 5 + 4x 3 + 6x)(3 x 6 + 5x 4 + 7x 2 + 9)dx 26 R. Boerner, ASU School of Mathematical and Statistical Sciences

6 Answers Multiple choice: B 2D 3D 4C 5D 6D 7D 8E 9A D B 2C 3C 4C Free Response:. The critical points of f are at x=,2,4 The inflection points of f are at x=,3 f has its absolute max y=3 at x=2 f is concave up on (,) (3,5) f has its absolute min y= at x=4 f is concave down on(,3) f is increasing on [,2) (4,5] f(5) f() =.5 f is decreasing on(2,4) The average rate of change of f on [,5] is. f has a relative but not absolute maximum of y=2.5 at x=5 There are 3 c values that satisfy the conclusion of the mean value theorem applied to f on [,5]. 2. Show that of all rectangles with a given area, the square has the smallest perimeter. Suppose A is the area, P is the perimeter of a rectangle, and x, y are its sides. Then A = xy and P = 2x + 2y. We are seeking the smallest perimeter, given the area. Therefore, we use the A equation to express one of the two sides: y = A and use that to express the perimeter as a function of x x alone: P(x) = 2x + 2A. Taking the derivative, we find x P (x) = 2 2A x 2. Since x >, P (x) is never undefined. Therefore, the only way P(x) can have a critical 26 R. Boerner, ASU School of Mathematical and Statistical Sciences

7 point is where P (x) =. We solve that equation: 2 = 2A x 2 x2 = A x = ± A. Since x can t be negative, x = A is the only critical point. We can confirm that P(x) has a relative minimum at that location by using the second derivative test: P (x) = 4A x 3 This function is positive for all x >, hence P ( A) >. This means that P has a relative minimum at x = A. Now we need to explain why the relative minimum is an absolute minimum. There is a theorem we learned in class: if a differentiable function is defined on an interval and has exactly one relative minimum, then that relative minimum must be absolute. In this case, the function P(x) is defined on (, ), an interval, and it has only one relative minimum, so the theorem applies and the relative minimum must be absolute. 3. Since the absolute minimum of the perimeter happens at x = A, and since for that x value, y = A = A = A, we conclude that the absolute minimum of the perimeter occurs x A when x = y = A, i.e. when the rectangle is a square. a. We use the definition of derivative: f () = lim h f( + h) f() h We cannot evaluate this limit based on the given data,but we know that the difference quotient is approximately equal to the limit if h is small in absolute value. There is only one h value that lets us use the given data: h =.. Therefore, f () f(.) f(). = Therefore, we conclude that f () is approximately. =. b. We apply the linear approximation formula f(x) f(a) + f (a)(x a) with a = and x =.2: f(.2) f() + f ().2. We substitute the result from a. and the given value for f() and get: 26 R. Boerner, ASU School of Mathematical and Statistical Sciences

8 f(.2) = Here we use the fact that the antiderivative of sec x tan x is sec x and apply the fundamental theorem of calculs: π 4 sec x tan x dx = [sec x] π 4 = sec π 4 sec The question asks you to simplify as much as possible. From trigonometry, we know that the trig functions have known values for certain angles, including, π, π, π and Specifically, sec π = 2 and sec =. Therefore, 4 π 4 sec x tan x dx = We calculate these limits from the inside out : a. lim arctan e x = lim arctan e u = lim arctan w = arctan = π. The arctan limit can be x u w 4 found by evaluation because arctan is continuous at. b. lim x + arctan e x = lim arctan u eu = lim arctan w = arctan =. Again, the arctan w limit can be found by evaluation because arctan is continuous, this time at. c. lim x 5. The definite integral arctan e x = lim arctan e u = lim arctan w = π. u w 2 (x 7 + 2x 5 + 4x 3 + 6x)(3 x 6 + 5x 4 + 7x 2 + 9)dx is zero because it is the integral of an odd function over a symmetric interval [ a, a]. The integrand is odd because it is the product of an odd and an even function. Note: while this guide is being made freely available to ASU students and the general public for personal use, it is not to be uploaded to third-party websites, especially not ones that profit from such content. If you found this document on a third-party website such as Course Hero or Chegg, the document is being served to you in violation of copyright law. 26 R. Boerner, ASU School of Mathematical and Statistical Sciences