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Answers To Early Transcendentals Calculus

Thomas Calculus Early Transcendentals 14th Edition Hass SOLUTIONS

Manual

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Thomas Calculus Early Transcendentals 14th Edition Hass TEST Banking concern

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CHAPTER 2 LIMITS AND CONTINUITY

2.one RATE S OF CHANGE AND TANGENTDue south TO CURVES

one. (a) f f (3)  f (2) 289 f f (1)  f ( ane) two0

x 3two ane nineteen (b) 10 1( 1) 2 1

2. (a) thousand g (3)  g (1) three (

1 ) g chiliad (4) g ( 2) 8  8

10 three ane 2 2 (b) 10 4 ( 2) half-dozen 0

h



3

h







3. (a) h iv 4 1 1 iv h h ii h six 0 3 three 3

t

3

 (b) t



iv. (a)

4 iv 2

m g (



) g (0) (2 ane) (2 1) 2

2 6 3

g m (



) g (



) (2 one)(two 1)

t

0

0 (b) t

(

) ii 0

5. R R (2) R(0)

8 1 one 3 1

2 0 two 2 ane

half-dozen. P

P (2)

P

(1) (8 16 ten) (1 4 5)

2 1 1 2 two 0

y ((2 h )two

5) (22 five) 44h h2

5 1 ivh h2

7. (a) x h h h 4 h. As h 0, 4 h 4 at P (2, one) the gradient is four.

(b) y ( one)

4(

10 two) y

one

4

10

8 y

4

x 9

y (7 (2 h )2 )(vii ii2 ) 7iv fourh h2 iii ivh h 2

viii. (a) x h

is 4.

h h 4 h. As h 0, 4 h four at P (two, 3) the slope

(b) y three

(

4)(

x 2) y 3

4

x 8  y

4

x 11

y ((2h ) 2 2(2 h ) 3) (2 2 2(2) 3) 4 4h h2 iv2h three ( 3) 2 h h ii

nine. (a) ten h h h two h. As h 0, 2 h 2 at

(b)

P(2, 3) the sl ope is ii.

y ( 3)

two(

x 2) y iii

2

x 4 y 2 ten 7.

10. (a) y

x

((ih ) 2 4(1 h)) (1 2 4(1))

h

1 twoh hii 4 4h ( three)

h hii twoh

h

h 2. Equally h 0, h 2 ii at P (1, iii) the

(b)

gradient is 2.

y ( iii)

(

ii)(

ten

one) y 3

2

x 2 y 2 x 1.

Answers To Early Transcendentals Calculus,

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