256 STEP 4. Review the Knowledge You Need to Score High
x
y
10
6
x^2 – 36
Figure 11.9-4
Step 2: Find the minimum value ofP.
Enter
yt=x+ 10 ∗x/
(√
(x∧ 2 − 36 )
)
.
Use the [Minimum] function of
the calculator and obtain the
minimum point (9.306, 22.388).
Step 3: Verify with the First Derivative
Test.
Entery 2 =(y1(x),x) and
observe. (See Figure 11.9-5.)
9.306
rel. min.
decr. incr.
y 1 = f
y 2 = f′
Figure 11.9-5
Step 4: Check endpoints.
The domain ofxis (6,∞).
Sincex= 9 .306 is the only
relative extremum, it is the
absolute minimum.
Thus, the maximum length of the
pipe is 22.388 feet.
limx→− 1
1 +cosπx
x^2 − 1
=limx→− 1
−πsinπx
2 x
0
− 2
= 0
32.
dx
dt
=− 4 t
dy
dt
= 2 t. The speed of the object
is
√
(− 4 t)^2 +(2t)^2 =
√
16 t^2 + 4 t^2 =
√
20 t^2 = 2 t
√
- Evaluated
att=
1
2
, the speed is 2
(
1
2
)√
5 =
√
5.
- Integrate
∫
2 x
√
x+ 3 dxby parts with
u= 2 x,du= 2 dx,dv=
√
x+ 4 dx, and
v=
2
3
(x+4)^3 /^2. Then
∫
2 x
√
x+ 3 dx=
4
3
x(x+3)^3 /^2 −
4
3
∫
(x+3)^3 /^2 dx=
4
3
x(x+3)^3 /^2 −
4
3
(
2
5
(x+3)^5 /^2
)
. Simplifying this
expression, we get
∫
2 x
√
x+ 3 dx
=
4
5
(x+3)^3 /^2 (x−2).