MA 3972-MA-Book April 11, 2018 16:32
366 STEP 5. Build Your Test-Taking Confidence
- The correct answer is (D).
0
(–1.9, 5.3)
(–1.9, f(1.9))
(–2, 5)
y f
x
The tangent line for local linearization
approximation passes through the point(−2, 5)
and has a slope ofm=3. The equation of the
tangent isy− 5 = 3 (x+ 2 ).Atx=− 1 .9,
y− 5 = 3 (− 1. 9 + 2 ),ory= 5 .3. Therefore,
f(− 1. 9 )≈ 5 .3.
- The correct answer is (D).
ToolsF1 TraceF3
MAIN RAD EXACT FUNC
ZoomF2 ReGraphF4 MathF5 DrawF6 F7Pen
[−2, 2] by [−5, 10]
Using your graphing calculator, examine the
graph off′(x). Note that the graph off′(x)
crosses thex-axis atx=− 1 .407 andx= 0 .345.
The graph also shows the following:
+
f
f ́
Incr decreasing incr
–2
0
–1.407 0.345 2
–+ 0
Therefore, the functionfincreases on the
intervals (−2,− 1 .407) and (0.345, 2).
- The correct answer is (D).
The total distance traveled is∫
3
0
∣∣
v(t)
∣∣
dt=
∫ 3
0
∣∣
2 (−t)cos( 2 t)
∣∣
dt. Using the
TI-89 graphing calculator, enter
∫
(abs(2∧(−t)∗cos(2t)),t, 0, 3) and obtain
0. 812884 ≈ 0 .813.
- The correct answer is (D).
ToolsF1 TraceF3
MAIN RAD EXACT FUNC
ZoomF2 ReGraphF4 MathF5 DrawF6 F7Pen
[0, 2 π] by [−2, 4]
Using the TI-89 graphing calculator, enter the
second derivative
d^2 y
dx^2
intoy 1 as follows:
y 1 =d
(
5 ∗e∧
(
−x
2
)
∗sin
(
3 x
2
)
,x,2
)
Inspecting the graph ofy 1 , you see that the
curve crosses thex-axis three times, which
means the second derivative changes sign three
times. Since the functionyis a twice
differentiable function, it has three points of
inflection.
- The correct answer is (B).
y
x
f
0 1 6 11 16
Note that the length of each rectangle is 5.
Therefore:
∫ 16
1
f(x)dx≈5(f(1)+f(6)+f(11))
≈5(1. 1 + 2. 1 + 2 .6)≈ 29
