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

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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 x 4 (c) f 0 (2) = 16 f 0 (3) = 26 3. (a) y x =5 ; a constant function. (b) No; dy/dx . 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 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 a 3. (a) 5 (b) 5 4. If we choose a very small neighborhood of the point L + a 2 ,wecannot f 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 ,wegettheso lut ion 3 / 4 <x . 32

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(b) The solution is x< 9 . (c) The solution is 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 +1 < 6 . Subtracting 1 from all sides, we get 7 5 as the solution. (b) The solution is 2 / 3 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 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 =l im 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
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 f ned at x =4 ; i.e., x is not in the domain of the function. (c) for x 6 , the function reduces to y = x 5 ,so lim x 4 y = 1 . 5. No, because q = v +1 , as such, is de f ned at every value of v, whereas the given rational function is not de f ned at v =2 and v = 2 . The only permissible way to rewrite is to qualify the equation q = v by the restrictions v 6 and v 6 = 2 . 6. Yes; each function is not only continuous but also smooth. 34

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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 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 f 0 (1) = 3 c and f 0 (2) = 12 c . (c) f 0 ( x )=10 x 3 f 0 (1) = 10 and f 0 (2) = 10 8 =1 1 4 (d) f 0 ( x )= x 1 / 3 = 3 x f 0 (1) = 1 and f 0 (2) = 3 2 (e) f 0 ( w )=2 w 2 / 3 f 0 (1) = 2 and f 0 (2) = 2 · 2 2 / 3 =2 1 / 3 (f) f 0 ( w 1 2 w 7 / 6 f 0 (1) = 1 2 and f 0 (2) = 1 2 (2 7 / 6 1 · 2 7 / 6 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 is the MC function.
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## 6-10 - CHAPTER 6 Exercise 6.2 1. (a) y 4(x + x)2 + 9 (4x2 +...

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