Calculus: Early Transcendentals (6th) -- Student Solutions by James Stewart

By James Stewart

Offers thoroughly worked-out options to all odd-numbered routines in the textual content, giving scholars how to payment their solutions and make sure that they took the right kind steps to reach at a solution.

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Since f 0 (1) = 0, we will round off our figure so that there is a horizontal tangent directly over x = 1. Last, we make sure that the curve has a slope of −1 as we pass over x = 2. Two of the many possibilities are shown. 21. Using Definition 2 with f (x) = 3x2 − 5x and the point (2, 2), we have 3(2 + h)2 − 5(2 + h) − 2 f(2 + h) − f(2) (12 + 12h + 3h2 − 10 − 5h) − 2 = lim = lim h→0 h→0 h→0 h h h 3h2 + 7h = lim = lim (3h + 7) = 7 h→0 h→0 h f 0 (2) = lim So an equation of the tangent line at (2, 2) is y − 2 = 7(x − 2) or y = 7x − 12.

This enables us to sketch the graph for x ≥ 0. Then we use the fact that f is an even function to reflect this part of the graph about the y-axis to obtain the entire graph. Or, we could consider also the cases x < −3, −3 ≤ x < −1, and −1 ≤ x < 0. 7. Remember that |a| = a if a ≥ 0 and that |a| = −a if a < 0. Thus, x + |x| = + 2x if x ≥ 0 0 if x < 0 and y + |y| = + 2y 0 if y ≥ 0 if y < 0 We will consider the equation x + |x| = y + |y| in four cases. (1) x ≥ 0, y ≥ 0 2x = 2y (2) x ≥ 0, y < 0 2x = 0 x=y x=0 (3) x < 0, y ≥ 0 0 = 2y (4) x < 0, y < 0 0=0 0=y Case 1 gives us the line y = x with nonnegative x and y.

7 DERIVATIVES AND RATES OF CHANGE 23. (a) Using Definition 2 with F (x) = 5x/(1 + x2 ) and the point (2, 2), we have ¤ (b) 5(2 + h) −2 F (2 + h) − F (2) 1 + (2 + h)2 F (2) = lim = lim h→0 h→0 h h 0 = lim 5h + 10 5h + 10 − 2(h2 + 4h + 5) −2 + 4h + 5 h2 + 4h + 5 = lim h→0 h h h2 h→0 = lim h→0 −2h2 − 3h h(−2h − 3) −2h − 3 −3 = lim = lim = h(h2 + 4h + 5) h→0 h(h2 + 4h + 5) h→0 h2 + 4h + 5 5 So an equation of the tangent line at (2, 2) is y − 2 = − 35 (x − 2) or y = − 35 x + 16 . 5 25. Use Definition 2 with f (x) = 3 − 2x + 4x2 .

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