426 STEP 5. Build Your Test-Taking Confidence
- The correct answer is (D).
y
x
0 0.5 1.0
(0.5, 1.5)
y = f(x)
(1, 2.324)
1.0
Sincey(0)=1, the graph ofypasses through
the point (0, 1). The slope of the tangent line
at (0, 1) is
dy
dx
∣∣
∣∣
x= 0
=e^0 =1. The equation of the
tangent isy− 1 =1(x−0) ory=x+1. At
x= 0 .5,y= 0. 5 + 1 = 1 .5. Thus, the point
(0.5, 1.5) is your next starting point, and
dy
dx
∣∣
∣
∣x= 0. 5 =e
- (^5) =√e. The equation of the next
line isy− 1. 5 =
√
e(x− 0 .5) or
y=
√
e(x− 0 .5)+ 1 .5. Atx=1,
y=
√
e(1− 0 .5)+ 1. 5 ≈ 2 .324.
- The correct answer is (B).
The property
∫ 0
−k
f(x)dx=−
∫k
0
f(x)dx
implies that the regions bounded by the graph
offand thex-axis are such that one region is
above thex-axis and the other region is below.
The property also implies thatf is an odd
function, which means thatf(−x)=−f(x)or
that the graph off is symmetrical with respect
to the origin. The only graph that satisfies
those conditions is choice (B).
- The correct answer is (C).
If 2x
dy
dx
− 7 = 1 ⇒
dy
dx
=
8
2 x
=
4
x
.
Integratedy=
4 dx
x
to get
y=4ln|x|+C. Sincey= 4 .5 whenx=3,
4. 5 =4ln3+C⇒ 0. 10555 =C. Thus,
y=4ln|x|+ 0 .10555. Atx= 3 .1,y= 4 .631.
- The correct answer is (A)
Step 1. Begin by finding the first non-negative
value oftsuch thatv(t)=0. To accomplish
this, use your graphing calculator, set
y 1 =sin(x^2 +1), and graph.
MAIN RAD EXACT FUNC
F1-
Tools
F2-
Zoom
F3
Trace
F4
Regraph
F5-
Math
F6-
Draw
F7-
Pen
Use the [Zero] function and find the first
non-negative value ofxsuch thaty 1 =0. Note
thatx= 1 .46342.
Step 2. The position function of the particle is
s(t)=
∫
v(t)dt.Sinces(1)=5, we have
∫ 1. 46342
1
v(t)=s(1.46342)−s(1).Using your
calculator, evaluate
∫ 1. 46342
1
sin (x^2 +1)dxand
obtain 0.250325. Therefore,
. 250325 =s(1.46342)−s(1) or
. 250325 =s(1.46342)− 5 .Thus,s(1.46342)
= 5. 250325 ≈ 5 .250.