# Calculus: One and Several Variables

This preview shows pages 1–3. Sign up to view the full content.

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57 118 SECTION 4.1 CHAPTER 4 SECTION 4.1 1. f is diﬀerentiable on (0 , 1) , continuous on [0,1]; and f (0) = f (1) = 0 . f ± ( c )=3 c 2 1; 3 c 2 1=0= c = 3 3 ± 3 3 / (0 , 1) ² 2. f is diﬀerentiable on ( 2 , 2) , continuous on [ 2 , 2]; and f ( 2) = f (2) = 0 . f ± ( c )=4 c 3 4 c ;4 c ( c 2 1)=0= c =0 , ² 1 3. f is diﬀerentiable on (0 , 2 π ) , continuous on [0 , 2 π ]; and f (0) = f (2 π )=0 . f ± ( c ) = 2cos2 c ; 2cos2 c =0= 2 c = π 2 + nπ, and c = π 4 + 2 ,n , ² 1 , ² 2 ... Thus, c = π 4 , 3 π 4 , 5 π 4 , 7 π 4 4. f is diﬀerentiable on (0 , 8) , continuous on [0 , 8]; and f (0) = f (8) = 0 . f ± ( c )= 2 3 c 1 / 3 2 3 c 2 / 3 = 2 3 c 1 / 3 1 c 2 / 3 f ± ( c )=0 = c =1 . 5. f ± ( c )=2 c , f ( b ) f ( a ) b a = 4 1 2 1 =3; 2 c =3 = c =3 / 2 6. f ± ( c 3 2 c 4, f ( b ) f ( a ) b a = 10 ( 1) 4 1 = 3; 3 2 c 4= 3= c =9 / 4 7. f ± ( c c 2 , f ( b ) f ( a ) b a = 27 1 3 1 = 13; 3 c 2 =13 = c = 1 3 39 ³ 1 3 39 is not in [ a,b ] ´ 8. f ± ( c 2 3 c 1 / 3 , f ( b ) f ( a ) b a = 4 1 8 1 = 3 7 ; 2 3 c 1 / 3 = 3 7 = c = (14) 3 9 3 9. f ± ( c c 1 c 2 , f ( b ) f ( a ) b a = 0 1 1 0 = 1; c 1 c 2 = 1= c = 1 2 2 ( 1 2 2 is not in [ ]) 10. f ± ( c c 2 3, f ( b ) f ( a ) b a = 2 2 1 ( 1) = 2; 3 c 2 2= c = ² 3 3 11. f is continuous on [ 1 , 1], diﬀerentiable on ( 1 , 1) and f ( 1) = f (1) = 0. f ± ( x x (5 x 2 ) (3 + x 2 ) 2 1 x 2 , f ± ( c ) = 0 for c in ( 1 , 1) implies c =0. 12. (a) f ± ( x 2 3 x 1 / 3 = 2 3 x 1 / 3 ³ = 0 for all x ( 1 , 1) . (b) f ± (0) does not exist. Therefore, f is not diﬀerentiable on ( 1 , 1) .

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

View Full Document
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57 SECTION 4.1 119 13. No. By the mean-value theorem there exists at least one number c (0 , 2) such that f ± ( c )= f (2) f (0) 2 0 = 3 2 > 1 . 14. No, by Rolle’s theorem: f (2) = f (3) = 1 but there is no value c (2 , 3) such that f ± ( c )=0 . 15. By the mean-value theorem there is a number c (2 , 6) such that f (6) f (2) = f ± ( c )(6 2) = f ± ( c )4 . Since 1 f ± ( x ) 3 for all x (2 , 6), it follows that 4 f (6) f (2) 12 . 16. f ( x x 2 + x +3 ,f ± ( x )=2 x +1. The slope of the line through ( 1 , 3) and (2 , 9) is 2. Setting 2 x + 1 = 2, we get x = 1 2 . The point on the graph of f where the tangent line is parallel to the line through ( 1 , 3) and (2 , 9) is: (1 / 2 , 15 / 4). 17. f is everywhere continuous and everywhere diﬀerentiable except possibly at x = 1 . f is continuous at x = 1: as you can check, lim x →− 1 f ( x , lim x →− 1 + f ( x , and f ( 1) = 0 . f is diﬀerentiable at x = 1 and f ± ( 1) = 2: as you can check, lim h 0 f ( 1+ h ) f ( 1) h = 2 and lim h 0 + f ( h ) f ( 1) h =2 . Thus f satisFes the conditions of the mean-value theorem on every closed interval [ a,b ]. f ± ( x ± 2 ,x ≤− 1 3 x 2 1 ,x> 1 f (2) f ( 3) 2 ( 3) = 6 ( 4) 2 ( 3) .
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/30/2011 for the course MATH 1431 taught by Professor Any during the Spring '08 term at University of Houston.

### Page1 / 111

ch04[1] - P1: PBU/OVY JWDD027-04 P2: PBU/OVY...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online