Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 2.8 THEDERIVATIVEASAFUNCTION ¤ 71
(b) After 800 ounces of gold have been produced, the rate at which the production cost is increasing is $17/ounce. So the cost
of producing the 800th (or 801st) ounce is about $17.
(c) In the short term, the values of f 0 (x) will decrease because more efficient use is made of start-up costs as x increases. But
eventually f 0 (x) might increase due to large-scale operations.
47. T 0 (10) is the rate at which the temperature is changing at 10:00 AM.ToestimatethevalueofT 0 (10), we will average the
difference quotients obtained using the times t =8and t =12.LetA =
B =
T (12) − T (10)
12 − 10
=
88 − 81
2
T (8) − T (10)
8 − 10
=
72 − 81
−2
=4.5 and
=3.5. ThenT 0 T (t) − T (10)
(10) = lim
≈ A + B = 4.5+3.5 =4 ◦ F/h.
t→10 t − 10 2 2
49. (a) S 0 (T ) is the rate at which the oxygen solubility changes with respect to the water temperature. Its units are (mg/L)/ ◦ C.
(b) For T =16 ◦ C, it appears that the tangent line to the curve goes through the points (0, 14) and (32, 6). So
S 0 (16) ≈ 6 − 14
32 − 0 = − 8
32 = −0.25 (mg/L)/◦ C. This means that as the temperature increases past 16 ◦ C, the oxygen
solubility is decreasing at a rate of 0.25 (mg/L)/ ◦ C.
51. Since f(x) =x sin(1/x) when x 6= 0and f(0) = 0, wehave
f 0 f(0 + h) − f(0) h sin(1/h) − 0
(0) = lim
=lim
=limsin(1/h). This limit does not exist since sin(1/h) takes the
h→0 h
h→0 h
h→0
values −1 and 1 on any interval containing 0. (Compare with Example 4 in Section 2.2.)
2.8 The Derivative as a Function
1. It appears that f is an odd function, so f 0 will be an even
function—that is, f 0 (−a) =f 0 (a).
(a) f 0 (−3) ≈ 1.5
(b) f 0 (−2) ≈ 1
(c) f 0 (−1) ≈ 0
(d) f 0 (0) ≈−4
(e) f 0 (1) ≈ 0
(f ) f 0 (2) ≈ 1
(g) f 0 (3) ≈ 1.5
3. (a) 0 = II, since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0,then
negative again. The actual function values in graph II follow the same pattern.
(b) 0 = IV, since from left to right, the slopes of the tangents to graph (b) start out at a fixed positive quantity, then suddenly
become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents.
(c) 0 = I, since the slopes of the tangents to graph (c) are negative for x0,asarethefunctionvaluesof
graph I.
(d) 0 = III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then 0, then
positive, then 0, then negative again, and the function values in graph III follow the same pattern.