{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

6-10 - CHAPTER 6 Exercise 6.2 1(a y 4(x x)2 9(4x2 9 = = 8x...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 6 Exercise 6.2 1. (a) y x = 4( x + x ) 2 + 9 (4 x 2 + 9) x = 8 x + 4 x (b) dy/dx = f 0 ( x ) = 8 x (c) f 0 (3) = 24 and f 0 (4) = 32 2. (a) y x = 10 x + 5 x 4 (b) dy/dx = 10 x 4 (c) f 0 (2) = 16 f 0 (3) = 26 3. (a) y x = 5 ; a constant function. (b) No; dy/dx = 5 . Exercise 6.4 1. Left-side limit = right-side limit = 15. Yes, the limit is 15. 2. The function can be rewritten as q = ( v 3 + 6 v 2 + 12) /v = v 2 + 6 v + 12 ( v 6 = 0 ). Thus (a) lim v 0 q = 12 (b) lim v 2 q = 28 (c) lim v a q = a 2 + 6 a + 12 3. (a) 5 (b) 5 4. If we choose a very small neighborhood of the point L + a 2 , we cannot fi nd a neighborhood of N such that for every value of v in the N-neighborhood, q will be in the ( L + a 2 )-neighborhood. Exercise 6.5 1. (a) Adding 3 x 2 to both sides, we get 3 < 4 x . Multiplying both sides of the latter by 1 / 4 , we get the solution 3 / 4 < x . 32
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(b) The solution is x < 9 . (c) The solution is x < 1 / 2 (d) The solution is 3 / 2 < x . 2. The continued inequality is 8 x 3 < 0 < 8 x . Adding 8 x to all sides, and then multiplying by 1 / 8 (thereby reversing the sense of inequality), we get the solution 0 < x < 3 / 8 . 3. (a) By (6.9), we can write 6 < x + 1 < 6 . Subtracting 1 from all sides, we get 7 < x < 5 as the solution. (b) The solution is 2 / 3 < x < 2 . (c) The solution is 4 x 1 . Exercise 6.6 1. (a) lim v 0 q = 7 0 + 0 = 7 (b) lim v 3 q = 7 27 + 9 = 11 (c) lim v →− 1 q = 7 + 9 + 1 = 17 2. (a) lim v →− 1 q = lim v →− 1 ( v + 2) · lim v →− 1 ( v 3) = 1( 4) = 4 (b) lim v 0 q = 2( 3) = 6 (c) lim v 5 q = 7(2) = 14 3. (a) lim q v 0 = lim v 0 (3 v + 5) / lim v 0 ( v + 2) = 5 / 2 = 2 1 2 (b) lim v 5 q = (15 + 5) / (5 + 2) = 20 / 7 = 2 6 7 (c) lim v →− 1 q = ( 3 + 5) / ( 1 + 2) = 2 / 1 = 2 33
Background image of page 2
Exercise 6.7 1. For example, 2. (a) lim v N q = N 2 5 N 2 = g ( N ) (b) Yes. (c) Yes. 3. (a) lim v N q = ( N + 2) / ( N 2 + 2) = g ( N ) (b) Yes. (c) The function is continuous in the domain 4. (a) No. (b) No, because f ( x ) is not de fi ned at x = 4 ; i.e., x = 4 is not in the domain of the function. (c) for x 6 = 4 , the function reduces to y = x 5 , so lim x 4 y = 1 . 5. No, because q = v + 1 , as such, is de fi ned at every value of v, whereas the given rational function is not de fi ned at v = 2 and v = 2 . The only permissible way to rewrite is to qualify the equation q = v + 1 by the restrictions v 6 = 2 and v 6 = 2 . 6. Yes; each function is not only continuous but also smooth. 34
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
CHAPTER 7 Exercise 7.1 1. (a) dy/dx = 12 x 11 (b) dy/dx = 0 (c) dy/dx = 35 x 4 (d) dw/du = 3 u 2 (e) dw/du = 2 u 1 / 2 (f) dw/du = u 3 / 4 2. (a) 4 x 5 (b) 3 x 2 / 3 (c) 20 w 3 (d) 2 cx (e) abu b 1 (f) abu b 1 3. (a) f 0 ( x ) = 18 ; thus f 0 (1) = 18 and f 0 (2) = 18 . (b) f 0 ( x ) = 3 cx 2 ; thus f 0 (1) = 3 c and f 0 (2) = 12 c . (c) f 0 ( x ) = 10 x 3 ; thus f 0 (1) = 10 and f 0 (2) = 10 8 = 1 1 4 (d) f 0 ( x ) = x 1 / 3 = 3 x ; thus f 0 (1) = 1 and f 0 (2) = 3 2 (e) f 0 ( w ) = 2 w 2 / 3 ; thus f 0 (1) = 2 and f 0 (2) = 2 · 2 2 / 3 = 2 1 / 3 (f) f 0 ( w ) = 1 2 w 7 / 6 ; thus f 0 (1) = 1 2 and f 0 (2) = 1 2 (2 7 / 6 ) = 2 1 · 2 7 / 6 = 2 13 / 6 4. Refer to the following two graphs Exercise 7.2 1. VC = Q 3 5 Q 2 + 12 Q . The derivative d dQ VC = 3 Q 2 10 Q + 12 is the MC function.
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}