Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 2.7 DERIVATIVES AND RATES OF CHANGE ¤ 67
(c)
The graph of y =2x +1is tangent to the graph of y =4x − x 2 at the
point (1, 3). Now zoom in toward the point (1, 3) until the parabola and
the tangent line are indistiguishable.
5. Using (1) with f(x) = x − 1 and P (3, 2),
x − 2
x − 1
f(x) − f(a)
m = lim
=lim
x − 2 − 2
x→a x − a x→3 x − 3
=lim
x→3
3 − x
(x − 2)(x − 3) =lim
x→3
x − 1 − 2(x − 2)
=lim
x − 2
x→3 x − 3
−1
x − 2 = −1
1 = −1
Tangent line: y − 2=−1(x − 3) ⇔ y − 2=−x +3 ⇔ y = −x +5
√ √
x − 1 ( √ x − 1)( √ x +1)
7. Using (1), m =lim =lim
x→1 x − 1 x→1 (x − 1)( √ x − 1
= lim
x +1) x→1 (x − 1)( √ x +1) =lim 1
√ = 1
x→1 x +1 2 .
Tangent line: y − 1= 1 2 (x − 1) ⇔ y = 1 2 x + 1 2
9. (a) Using (2) with y = f(x) =3+4x 2 − 2x 3 ,
f(a + h) − f(a) 3+4(a + h) 2 − 2(a + h) 3 − (3 + 4a 2 − 2a 3 )
m =lim
= lim
h→0 h
h→0 h
=lim
h→0
3+4(a 2 +2ah + h 2 ) − 2(a 3 +3a 2 h +3ah 2 + h 3 ) − 3 − 4a 2 +2a 3
h
=lim
h→0
3+4a 2 +8ah +4h 2 − 2a 3 − 6a 2 h − 6ah 2 − 2h 3 − 3 − 4a 2 +2a 3
h
8ah +4h 2 − 6a 2 h − 6ah 2 − 2h 3 h(8a +4h − 6a 2 − 6ah − 2h 2 )
=lim
=lim
h→0 h
h→0 h
=lim
h→0
(8a +4h − 6a 2 − 6ah − 2h 2 )=8a − 6a 2
(b) At (1, 5): m =8(1)− 6(1) 2 =2,soanequationofthetangentline
is y − 5=2(x − 1) ⇔ y =2x +3.
(c)
At (2, 3): m =8(2)− 6(2) 2 = −8, so an equation of the tangent
line is y − 3=−8(x − 2) ⇔ y = −8x +19.
11. (a) The particle is moving to the right when s is increasing; that is, on the intervals (0, 1) and (4, 6). The particle is moving to
the left when s is decreasing; that is, on the interval (2, 3). The particle is standing still when s is constant; that is, on the
intervals (1, 2) and (3, 4).