MA 3972-MA-Book May 9, 2018 10:9
130 STEP 4. Review the Knowledge You Need to Score High
- Findf′(x)iff(x)=ln(3x).
Answer: f′(x)=
1
3 x
(3)=
1
x
.
- Find the approximate value off′(3). (See Figure 7.8-1.)
y
x
f
(4, 3)
(2, 1)
0
Figure 7.8-1
Answer: Using the slope of the line segment joining (2, 1) and (4, 3),f′(3)=
3 − 1
4 − 2
=1.
- Find
dy
dx
ifxy= 5 x^2.
Answer: Using implicit differentiation, 1y+x
dy
dx
= 10 x. Thus,
dy
dx
=
10 x−y
x
.
Or simply solve foryleading toy= 5 xand thus,
dy
dx
= 5.
- Ify=
5
x^2
, find
d^2 y
dx^2
.
Answer: Rewritey= 5 x−^2. Then,
dy
dx
=− 10 x−^3 and
d^2 y
dx^2
= 30 x−^4 =
30
x^4
.
- Using a calculator, write an equation of the line tangent to the graphf(x)=− 2 x^4 at
the point wheref′(x)=−1.
Answer: f′(x)=− 8 x^3. Using a calculator, enter [Solve][− 8 x∧ 3 =−1,x] and
obtainx=
1
2
⇒ f′
(
1
2
)
=−1. Using the calculatorf
(
1
2
)
=−
1
8
. Thus,
tangent isy+
1
8
=− 1
(
x−
1
2
)
.
- limx→ 2
x^2 +x− 6
x^2 − 4
Answer: Since
x^2 +x− 6
x^2 − 4
→
0
0
, consider limx→ 2
2 x+ 1
2 x
=
5
4
.
- limx→∞
lnx
x
Answer: Since
lnx
x
→
∞
∞
, consider limx→∞
1 /x
1
=xlim→∞
1
x