Calculus I Practice Final Exam A

Transcription

1 Calculus I Practice Final Exam A 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. The velocity of a particle moving along a horizontal line is given by the following graph. Find the distance travelled by the particle from time t= to t=. A. B. 2 C. 3 D. 26 R. Boerner, School of Mathematical & Statistical Sciences Arizona State University

2 2. Determine whether the given statements are true or false. Statement I: If f is differentiable at x = c, then f is continuous at x = c. Statement II: The derivative of a function f at x is equal to f(x+h) f(x) if h is small. h Statement III: If f is a differentiable function, and a is a point in its domain, then f(x) = f(a) + f (a)(x a) for all x in the domain. A. All three statements are true. B. I is true, the other two are false. C. II is true, the other two are false. D. III is true, the other two are false. x 5 3. Algebraically calculate the exact limit: lim x 5 x 5 A. / B. 2 5 C..224 D. x 5 4. Algebraically calculate the exact limit: lim x x 5 A. B. / C. D. 5. Differentiate f(x) = e x + x e E. Does not exist. E. Does not exist. A. f (x) = e x + x e B. f (x) = xe x + ex e C. f (x) = e x + ex e D. f (x) = xe x + x e x 6. Algebraically calculate the exact limit: lim 2 +x x +x A. B. C. D. E.None of the above. 7. Differentiate f(x) = sin x sin x A. f (x) = B. f (x) = C. f (x) = sin 2 x cos 2 x D. f (x) = E. f (x) = sin x x 2 + cot x 8. Differentiate f(x) = ln(ln x). cos x x 2 sin x x 2 + sin x cos x A. f (x) = x B. f (x) = ln x C. f (x) = x ln x D. f (x) = ln ( x ) x 2 9. Differentiate F(x) = t 4 dt. 4t 3 B. 4x 3 C. t 8 D. x 8 E. 2x 9 26 R. Boerner, School of Mathematical & Statistical Sciences Arizona State University

3 . Given the following graph of a function, select which graph represents the derivative. A. B. C. D. 26 R. Boerner, School of Mathematical & Statistical Sciences Arizona State University

4 2. Use the left Riemann sum with n = 2 rectangles to approximate ln x dx digits. and round to 3 A..23 B..386 C..45 D When you increase the radius of a sphere by 3%, by what percentage does the volume approximately increase? A. % B. 3% C. 9% D. 27% 3. Given a positive, differentiable function f with f() = and f() = e, evaluate f (x) f(x) dx. A. B. C. e D. e E. None of the above/can t be determined based on the given information. Free Response:. Find the intervals of concavity and the location of the inflection point(s) for the function f(x) = x 4 6x 3 + 2x 2. Justify your answer with the concavity test. 2. A box is to be made out of a cm by 2 cm piece of cardboard. Squares of side length x cm will be cut out of each corner, and then the ends and sides will be folded up to form a box with an open top. What is the maximum volume of such a box in cubic centimeters? Round your answer to two decimals places. 3. Find the slope of the tangent line to the curve x 3 + y 3 = x + y at the point (,). 4A. Evaluate the definite integral exactly and simplify as much as possible. Do not give a decimal approximation. /2 x 2 dx 4B. A stone is dropped into a well with initial velocity zero. The acceleration caused by Earth s gravity is approximately m/s 2. The stone hits bottom after 2 seconds. How deep is the well? 5. Evaluate the definite integral sin(x 3 ) dx by exploiting the symmetry of the function sin(x 3 ). Explain your reasoning in one sentence. 26 R. Boerner, School of Mathematical & Statistical Sciences Arizona State University

5 Answers: Multiple Choice: B 2D 3D 4C 5C 6D 7D 8C 9E D A 2C 3B. First we compute the second derivative of f(x) = x 4 6x 3 + 2x 2 and factor it: f (x) = 2x 2 36x + 24 = 2(x )(x 2). Since f (x) is continuous, it can only change sign where it is zero, which is at x =, 2. We now carry out the concavity test: Intervals: (, ) (,2) (2, ) x value for testing:.5 3 f (x) is on this interval positive negative positive f is on this interval concave up concave down concave up Therefore, the inflection points are x=,2. The function f is concave up on (, ) (2, ) and concave down on (,2). 2. If x is the side of the square we cut out from each corner, the volume of the box is V(x) = (2 2x)( 2x)x = 2x 6x 2 + 4x 3. The domain of V(x) is [,5]. To find the maximum volume, we compute the derivative: V (x) = 2 2x + 2x 2 = 4(3x 2 3x + 5). 26 R. Boerner, School of Mathematical & Statistical Sciences Arizona State University

6 Since V (x) is defined for all x, the only critical points are points where V (x) =. Using the quadratic formula, we find the zeros of V (x) to be x = 5 ± 5 3. Only the smaller of these two numbers is in the domain. Since V(x) is zero at the domain end points, the maximum must occur at the critical point inside the domain, x = The corresponding value of V(x), rounded to two decimal places, is cubic centimeters. 3. We perform implicit differentiation on x 3 + y 3 = x + y: and plug in x =, y = : 3x 2 + 3y 2 y = + y Solving this for y, we get y = y = + y 4A. Here we use the fact that the antiderivative of theorem of calculs: /2 x 2 dx x 2 is arcsin x and apply the fundamental = [arcsin x] 2 = arcsin 2 arcsin = π 6. (Decimal approximations from calculators, no matter how many digits they show, are not the exact answer.) 4B. Let us call the stone s acceleration function a(t). It is given that a(t) = (in units of m/s 2, using an upwards pointing axis.) By integrating with respect to time once, we find the velocity function to be v(t) = t + v, with the integration constant v, which represents initial velocity, given as v =. Therefore, v(t) = t. We now integrate again to find the stone s position function s(t): s(t) = 5t 2 + s. 26 R. Boerner, School of Mathematical & Statistical Sciences Arizona State University

7 The integration constant s represents initial position, which is determined solely by our choice of coordinate system. For convenience, let us agree that the initial position marks the zero point of our vertical axis. Then s =. Therefore, s(t) = 5t 2. By evaluating s(2) = 2, we find the depth of the well: 2 meters. 5. The integral is zero, because the integrand f(x) = sin(x 3 ) is an odd function. You can verify this directly by checking that f( x) = f(x) and using that sin itself is an odd 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, School of Mathematical & Statistical Sciences Arizona State University