Download Practice Problems for Exam 2: Calculus with Trigonometric Functions and Logarithms and more Exams Mathematics in PDF only on Docsity! M151B, Fall 2008, Practice Problems for Exam 2 Calculators will not be allowed on the exam. 1. Let f(x) = cos x − sin x, −π 4 ≤ x ≤ 3π 4 , and compute df −1 dx (1). 2. Show that d dx cos−1 x = − 1√ 1 − x2 , −1 < x < +1. 3. Let y = xtan x, 0 ≤ x < π 2 , and compute dy dx . 4. Use a linear approximation to estimate a value for ln(.99). 5. Consider a right triangle with hypotenuse length l and sidelengths 3 and x. Suppose x is measured as x = 4± .05, and use linear approximation to approximate the associated range of error on l. 6. Suppose that f(x) is continuous on the interval [2, 5] and differentiable on the interval (2, 5). Show that if 1 ≤ f ′(x) ≤ 4 for all x ∈ [2, 5], then 3 ≤ f(5) − f(2) ≤ 12. 7. Suppose that f(x) is twice differentiable in an open interval containing the point x = c and has a local minimum at the same point. Show that the function g(x) = ef(x) has a local minimum at x = c. 8. Let f(x) = e−x x − 1 , x 6= 1. 8a. Locate the critical points of f and determine the intervals on which f is increasing and the intervals on which f is decreasing. 8b. Locate the possible inflection points for f and determine the intervals on which f is concave up and the intervals on which it is concave down. 8c. Determine the boundary behavior of f by computing limits as x → ±∞. 8d. Use your information from Parts a-c to sketch a graph of this function. 9. A long rectangular sheet of metal, 12 inches wide, is to be made into a rain gutter by turning up two sides at right angles to the sheet. How many inches should be turned up to give the gutter its greatest capacity? 10. A piece of wire 10 meters long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is minimized? How should the wire be cut so that the total area is maximized? 1 11. Compute the following limits. 11a. lim x→0 ex − 1 x . 11b. lim x→∞ (x − √ x2 − 1). 11c. lim x→∞ ( x x + 1 )x. Solutions 1. First, f ′(x) = − sin x − cos x. Also, f(0) = 1 ⇒ f−1(1) = 0. We have, then, df−1 dx (1) = 1 f ′(f−1(1)) = 1 f ′(0) = 1 −1 = −1. 2. Set f(x) = cos x and use the formula df−1 dx (x) = 1 f ′(f−1(x)) = 1 − sin(cos−1 x) . In order to evaluate cos−1 x, set θ = cos−1 x (we use θ because this is an angle) and note that consequently cos θ = x ⇒ sin θ = √ 1 − cos2 θ = √ 1 − x2. Notice here that since the range of cos−1 x is [0, π], we know that θ ∈ [0, π], and so we know sin θ ≥ 0. This chooses the sign in front of √ 1 − cos2 θ. We finally have 1 − sin(cos−1 x) = − 1√ 1 − x2 . 3. If we take the natural logarithm of both sides, we have ln y = lnxtan x = (tanx)(ln x). Now differentiate each side with respect to x to obtain 1 y dy dx = (sec2 x)(ln x) + tan x x . Multiplying this last expression by y = xtan x, we conclude dy dx = xtan x((sec2 x)(ln x) + tan x x ). 2 The boundary behavior was obtained in Part c. Here’s a sketch: 9. Let x denote the width of sheet to be turned up on one side and let y denote the width of sheet left flat. If the length of the sheet is L then the volume is V = xyL, where y can be eliminated by the relation y = 12 − 2x. (For this problem it seems fairly natural to avoid bringing up the variable y, but I’ve used it here for consistency with our standard process.) In this way the function we would like to maximize is V (x) = x(12 − 2x)L, 0 ≤ x ≤ 6. We find the critical points by computing dV dx = (12 − 4x)L = 0 ⇒ x = 3. Evaluating V (0) = 0 V (3) = 18 V (6) = 0, we conclude that the maximum capacity occurs when x = 3 inches are turned up on either side. 10. Let x be the length of each side of the square, and let y be the length of each side of the equilateral triangle. Then the total length of wire is 10 = 4x + 3y, while the total area is A = area of square + area of triangle = x2 + √ 3 4 y2. 5 (You can derive the area formula for an equilateral triangle from the formula 1 2 bh and either the sidelengths for a 30-60-90 triangle or the Pythagorean theorem.) Solving our constraint for y, we have y = 10 3 − 4 3 x, so that A(x) = x2 + √ 3 4 ( 10 3 − 4 3 x)2, 0 ≤ x ≤ 10 4 . Proceeding as usual, we compute A′(x) = 2x + √ 3 2 ( 10 3 − 4 3 x)(−4 3 ) ⇒ x(2 + 8 √ 3 9 ) = 20 √ 3 9 . We conclude that x = 20 √ 3 18 + 8 √ 3 = 10 √ 3 9 + 4 √ 3 . From our expression for A′(x) we see that A′(x) < 0 for x < 10 √ 3 9+4 √ 3 , while A′(x) > 0 for x < 10 √ 3 9+4 √ 3 . We conclude that A(x) decreases for all x to the left of this point and increases for all x to the right of it, and is consequently a global minimum. This says that the area is minimized if the length of wire taken for the square is 4x = 40 √ 3 9 + 4 √ 3 . In order to find the global maximum, we must check A(x) at the two endpoints. We have A(0) = 100 √ 3 36 A( 10 4 ) = 100 16 . Clearly, A(10 4 ) is larger, and this corresponds with putting all of the wire into the square. 11a. lim x→0 ex − 1 x = lim x→0 ex 1 = 1. 11b. lim x→∞ (x − √ x2 − 1) = lim x→∞ x(1 − √ 1 − 1 x2 ) = lim x→∞ 1 − √ 1 − 1 x2 1 x . We can now apply l’Hospital’s rule to find that this limit is lim x→∞ −1 2 (1 − 1 x2 )− 1 2 ( 2 x3 ) − 1 x2 = 0. 11c. lim x→∞ ( x x + 1 )x = lim x→∞ eln( x x+1 )x = lim x→∞ ex ln( x x+1 ) = elimx→∞ x ln( x x+1 ). 6 We compute the limit in the exponent as lim x→∞ ln( x x+1 ) 1 x = lim x→∞ ln x − ln(x + 1) 1 x = lim x→∞ 1 x − 1 x+1 − 1 x2 = lim x→∞ 1 x(x+1) − 1 x2 = lim x→∞ − x 2 x2 + x = −1. We conclude lim x→∞ ( x x + 1 )x = e−1. Note: It’s slicker—but for our purposes less instructive—to simply notice that this is the inverse of the limit e = lim x→∞ (1 + 1 x )x. 7