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Problem 1.1.12 Find the equation of the line: ( 12 , 1) and (1, 2) on line. Solution: The slope is or y = 2x. 2−1 1− 12 = 2. The equation of the line is y − 2 = 2(x − 1) Problem 1.1.24 Find the equation of the line: parallel to y − x = 13; y-intercept is − 12 . Solution: The slope is 1. So y = x − 12 . Problem 1.1.36 How many units must you move in the y-direction is you start on a line of slope .2 and move 5 units in the x-direction? Solution: You must move (.2)(5) = 1 unit in the y direction. Problem 1.1.52 A restaurant paying 5.75 had a quit ratio of .2. When the pay was raised to 6, the quit ratio was .18. a) Assuming a linear relationship between wagex and Q(x) the quit ratio, find Q(x). = −.08. Solution: The change in quit ratio over the change in wage is − .02 .25 So Q(x) − .18 = −.08(x − 6) or Q(x) = −.08x + .66. b) When is Q(x) = .1? Solution Q(x) = −0.8x + .66 = .1 ⇒ −0.8x = −.56. So x = 7. Problem 1.1.68 Prove Slope Property 5 of straight lines. Solution: Consider the picture in the book. We have two perpendicular lines l1 and l2 intersecting at the origin with slopes m1 > m2 respectively. By slope property 1, we know that along the x = 1 line, the distance from 1 l1 to the x-axis is m1 and the distance from l2 to the x-axis is m2 . So the picture is correct. The triangle bounded by the x-axis, the x = 1 line, and l1 is right, and so by Pythagoras’ Theorem a2 = 1 + m21 . Similarly, we have b2 = 1 + m22 for the bottom right triangle. The last right triangle is the one formed by x = 1, l1 and l2 , and the Pythagorean relationship is a2 + b2 = (m1 − m2 )2 . Substituting into the last equation the other two equations, we have 1 + m21 + 1 + m22 = (m1 − m2 )2 = m21 − 2m1 m2 + m22 Simplifying, we get m1 m2 = −1. Problem 1.2.22 Find the slope of the tangent line to the curve y = x2 at the point where x = − 12 . Solution: The slope at x is 2x, thus the slope of the tangent line at x = − 12 is -1. Problem 1.2.28 Find the point on the graph of y = x2 where the tangent line is parallel to the line 3x − y = 2. Solution: The slope of that line is 3, so we must find the point on the curve whose slope is 3 as well. Since the slope at x of the curve x2 is 2x, thus at ( 32 , 49 ) the tangent line is parallel to the given line. Problem 1.2.30 Find the slope of the curve y = x3 at the point ( 23 , 27 ). 8 Solution: We simply plug 3 2 into 3x2 and get 27 4 as our answer. Problem 1.2.36 Find a, f (a), and the slope of the curve at (a, f (a)). Solution: (a, f (a)) lies at the intersection of the two lines y = 2 − x and 2 y = − 15 x + 65 . Substituting for y, we have the following equation 1 6 2−x=− x+ 5 5 Collecting like terms, we deduce that x = 1. Thus a = 1. Now f (a) = 2 − a = 1. The slope of the curve at that point is simply the slope of the tangent line, which is the line y = 2 − x. Since this line has slope −1, we know that the slope of the curve at (a, f (a)) is -1 as well. 3 Problem Use the Power Rule to find the derivative of (x− 5 )2 . 3 6 Solution: Using the exponent rule we know that (x− 5 )2 = x− 5 . Now ap11 plying the Power Rule, the derivative is − 56 x− 5 . Problem Let L be the tangent line to the curve f (x) = x4 − 6 at (2, 10). Let P be the line perpendicular to L and also passing through (2, 10). Find the equation for P . Solution: An application of the Power Rule gives us f 0 (x) = 4x3 . Thus at 2, the derivative is 32. So the slope of L is 32. This implies that the slope of 1 1 1 1 . Thus the equation for P is y −10 = − 32 (x−2) or y = − 32 x+10 16 . P is − 32 Problem 1.3.6 Find The derivative of f (x) = Solution: √1 x √1 . x 1 3 = x− 2 . So by the Power Rule f 0 (x) = − 12 x− 2 = − 2x1√x . Problem 1.3.30 If f (x) = x12 , find f (5) and f 0 (5). Solution: Using the Power Rule, we get that f 0 (x) = −2 and f 0 (5) = 125 . −2 . x3 So f (5) = 1 25 Problem 1.3.44 The line y = ax+b is tangent to y = x3 at the point (−3, −27). Find a and b. Solution: The derivative of x3 is 3x2 . Thus at x = −3 the slope of the tangent line is 27. So the equation of the tangent line is y + 27 = 27(x + 3) 3 or y = 27x + 54. Therefore a = 27 and b = 54. Problem 1.3.54 dy if y = x−4 dx Solution: dy dx = −4x−5 . Problem 1.3.62 Estimate f 0 (1). Solution: Evaluating the slopes of the chords from (1, .8) to the the other points, we get the following values: 1.3 − .8 1.1 − .8 1.4 − .8 = 1, = 1.25, = 1.5 1.6 − 1 1.4 − 1 1.2 − 1 This sequence of numbers is increasing, and one would guess that f 0 (1) > 1.5, say for example 1.75. Problem 1.3.68 √ Compute the difference quotient when f (x) = x3 − 2x2 + 5. Solution: f (x + h) − f (x) (x + h)3 − 2(x + h)2 + = h h = √ 5 − x3 + 2x2 − √ 5 x3 + 3x2 h + 3xh2 + 3h3 − 2x2 − 4xh − 2h2 − x3 + 2x2 h 2 = 3x + 3xh + 3h2 − 4x − 2h Problem 1.3.76 Apply the three step method to compute the derivative of f (x) = 2x2 + x. Solution: Step 1. f (x + h) − f (x) 2(x + h)2 + x + h − 2x2 − x = h h 4 = 2x2 + 4xh + 2h2 + x + h − 2x2 − x = 4x + 2h + 1 h Step 2. Letting h → 0, this expression becomes 4x + 1. Step 3. Thus the derivative f 0 (x) = 4x + 1. Problem 1.3.88 Show that d d f (x) = (f (x) + c) dx dx Solution: If one were to solve it pictorially, simply draw a differentiable function f , and then pick a c and draw out f + c on the same plane. Then select a point on f and draw the tangent line. Then select the point on f + c directly above or below the previous point and draw its tangent line as well. You will notice that these two lines are parallel, telling us that f and f + c have the same slope at all x. To solve it algebraically, simply note the following: f (x + h) + c − (f (x) + c) f (x + h) − f (x) = h h Thus their difference quotients are the same, and so therefore their derivatives are the same as well. 5