Fall M151B, 2008, Practice Problems for Exam 2
Calculators will not be allowed on the exam. 1. Let f (x) = cos x - sin x, and compute 2. Show that
df -1 (1). dx
-
3 x , 4 4
1 d cos-1 x = - , dx 1 - x2 y = xtan x ,
-1 < x < +1. , 2
3. Let 0x< 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. e-x , x = 1. x-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. f (x) = 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 8. Let
11. Compute the following limits. 11a. ex - 1 . x0 x lim lim (x - lim ( x2 - 1).
11b.
x
11c.
x
x x ) . x+1
Solutions
1. First, Also, f (0) = 1 f -1 (1) = 0. We have, then, f (x) = - sin x - cos x.
1 1 1 df -1 (1) = -1 = = = -1. dx f (f (1)) f (0) -1 2. Set f (x) = cos x and use the formula df -1 1 1 (x) = -1 = . dx f (f (x)) - sin(cos-1 x) In order to evaluate cos-1 x, set = cos-1 x (we use because this is an angle) and note that consequently cos = x sin = 1 - cos2 = 1 - x2 . Notice here that since the range of cos-1 x is [0, ], we know that [0, ], and so we know sin 0. This chooses the sign in front of 1 - cos2 . We finally have 1 1 = - . -1 x) - sin(cos 1 - x2 3. If we take the natural logarithm of both sides, we have ln y = ln xtan x = (tan x)(ln x). Now differentiate each side with respect to x to obtain 1 dy tan x = (sec2 x)(ln x) + . y dx x Multiplying this last expression by y = xtan x , we conclude tan x dy = xtan x ((sec2 x)(ln x) + ). dx x 2
4. We start with f (x) = ln x and use the linear approximation f (x) = f (a) + f (x)(x - a), where it is reasonable here to take a = 1. We find f (x) ln 1 + 1(x - 1) = x - 1. We can now compute f (.99) .99 - 1 = -.01. (The exact value, to four decimal places, is -.0101.) 5. First, the length l is given by the Pythagorean Theorem, l = 32 + x2 . By linear approximation, we have l(x + x) - l(x) l (x)x, where |l(x + x) - l(x)| is the absolute error, x = 4, x = .05 and l (x) = We have, then, x . x2 + 9
4 l (4)(.05) = (.05) = .04. 5 l(4 .05) = 5 .04.
We conclude 6. By the Mean Value Theorem, we know that there exists some value c (2, 5) so that f (c) = f (5) - f (2) . 3
Since the largest possible value for f (c) on this interval is 4 and since the smallest possible value for f (c) on this interval is 1, we have the inequality 1 f (5) - f (2) 4. 3
Multiplying this last inequality by 3, find we 3 f (5) - f (2) 12. 7. By our assumptions on f , we know f (c) = 0 and f (c) > 0. We need to show that precisely the same two conditions hold for g(x) = ef (x) . We have, first g (c) = ef (c) f (c) = 0, 3
and second g (c) = ef (c) f (c)2 + ef (c) f (c) = ef (c) f (c) > 0, where in arriving at this last inequality we have observed that ef (x) > 0 for any finite value of f (x). 8a. Compute f (x) = - xe-x , (x - 1)2
and observe that the critical points are x = 0, 1. We see that f is increasing on (-, 0] and decreasing on [0, 1) (1, ). (We omit x = 1 because f isn't defined there.) 8b. Compute f (x) = e-x (x2 + 1) , (x - 1)3
and observe that the only possible inflection point is x = 1. We see that f (x) is concave down on (-, 1) and concave up on (1, ). 8c. For the boundary behavior, we have e-x = - x- x - 1 e-x lim = 0. x+ x - 1 lim 8d. In order to anchor the plot, we evaluate f at the critical points, the possible inflection points and the endpoints. We have f (0) = - 1 e-x = - lim x1- x - 1 e-x = + . lim+ x1 x - 1
4
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, We find the critical points by computing dV = (12 - 4x)L = 0 x = 3. dx 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 3 2 y . A = area of square + area of triangle = x2 + 4 5 0 x 6.
(You can derive the area formula for an equilateral triangle from the formula 1 bh and either 2 the sidelengths for a 30-60-90 triangle or the Pythagorean theorem.) Solving our constraint for y, we have 10 4 - x, y= 3 3 so ...