146 STEP 4. Review the Knowledge You Need to Score High
[–4,4] by [–1,4]
Figure 7.9-10- (See Figure 7.9-11.) Entery 1 =cos(x)∗
(sin(x))∧2. A fundamental domain ofy 1
is [0, 2π]. Using the functions of the
calculator, you will find:
[–1,9.4] by [–1,1]
Figure 7.9-11
(a)relative maximum points at (0.955,
0.385), (π, 0), and (5.328, 0.385),
and relative minimum points at
(2.186,− 0 .385) and (4.097,− 0 .385);
(b)(points of inflection at (0.491, 0.196),
π
2,0
)
, (2.651,− 0 .196),
(3.632,− 0 .196),
(
3 π
2,0
)
, and (5.792, 0.196);
(c)no asymptote;
(d) function is increasing on intervals
(0, 0.955), (2.186,π), and
(4.097, 5.328), and decreasing on
intervals (0.955, 2.186), (π, 4.097),
and (5.328, 2π);
(e)function is concave upward on
intervals (0, 0.491),(
π
2,2. 651
)
,
(
3 .632,
3 π
2)
, and (5.792, 2π), andconcave downward on the intervals(
0 .491,
π
2)
, (2.651, 3.632), and
(
3 π
2,5. 792
)
.- Solvex=
t
2
fort= 2 xand substitute into
y=t^2 − 4 t+1.y=(2x)^2 −4(2x)+ 1
= 4 x^2 − 8 x+1. - Sincex=rcosθandy=rsinθ,y= 3 x− 5
becomesrsinθ= 3 rcosθ−5. Solving for
rproducesr(sinθ−3 cosθ)=−5 and
r=
− 5
sinθ−3 cosθ.
- The equationr= 2 −sinθis of the form
r=a−bsinθwith
a
b
1, so the graph is
a limaçon with no inner loop. Since
r(−θ)= 2 +sinθ/=r(θ), the graph is not
symmetric about the polar axis. However,
r(π−θ)= 2 −sin(π−θ) is equal to
2 sinθ=r(θ), so the graph is symmetric
about the linex=
π
2
. Finally,
r(θ+π)= 2 −sin(θ+π)= 2 +sinθand so
the graph is not symmetric about the pole.
- The vectors
〈
3,− 2〉
and〈
1,k〉
will be
orthogonal if the dot product is equal to
zero.〈
3,− 2〉
·〈
1,k〉
= 3. 1 − 2 kwill be
equal to zero when 2k=3sok=3
2
.
- The dot product of
〈
5,− 3〉
and〈
5, 3〉
is
5 · 5 +− 3 · 3 = 25 − 9 =16, so the vectors
are not orthogonal. To find the angle
between the vectors, begin by dividing the
dot product by the product of the
magnitudes of the two vectors. Both
vectors have a magnitude of√
34, so the
quotient becomes16
34
=
8
17
. The angle
between the vectors is
θ=cos−^1(
8
17)
≈ 1 .081 radians.