100 STEP 4. Review the Knowledge You Need to Score High
- Sincef′(4) is equivalent to the slope of the
tangent tof(x)atx=4, there are several
ways you can find its approximate value.
Method 1: Using the slope of the line
segment joining the points at
x=4 andx=5.
f(5)=1 and f(4)= 4
m=
f(5)− f(4)
5 − 4
=
1 − 4
1
=− 3
Method 2: Using the slope of the line
segment joining the points at
x=3 andx=4.
f(3)=5 and f(4)= 4
m=
f(4)− f(3)
4 − 3
=
4 − 5
4 − 3
=− 1
Method 3: Using the slope of the line
segment joining the points at
x=3 andx=5.
f(3)=5 and f(5)= 1
m=
f(5)− f(3)
5 − 3
=
1 − 5
5 − 3
=− 2
Note that−2 is the average of the results
from methods 1 and 2. Thus
f′(4)≈−3,−1, or−2 depending on
which line segment you use.
- You can use the difference quotient
f(a+h)−f(a)
h
to approximate f′(a).
Leth=1; f′(2)≈
f(3)−f(2)
3 − 2
≈
14 − 9
3 − 2
≈5.
Or, you can use the symmetric difference
quotient
f(a+h)−f(a−h)
2 h
to
approximate f′(a).
Leth=1; f′(2)≈
f(3)− f(1)
2 − 0
≈
14 − 6
2
≈4.
Thus,f′(2)≈4 or 5 depending on your
method.
- Entery 1 =x^5 + 3 x−8. The graph ofy1is
strictly increasing. Thusf(x) has an
inverse. Note that f(0)=−8. Thus the
point (0,−8) is on the graph off(x),
which implies that the point (−8, 0) is on
the graph off−^1 (x).
f′(x)= 5 x^4 +3 and f′(0)= 3.
Since (f−^1 )′(−8)=
1
f′(0)
, thus
(f−^1 )′(−8)=
1
3
.
17.
dy
dx
=
1
x
and
dy
dx
∣∣
∣∣
x=e
=
1
e
Thus, the slope of the tangent toy=lnx
atx=eis
1
e
.Atx=e,y=lnx=lne=1,
which means the point (e,1)isonthe
curve ofy=lnx. Therefore, an equation
of the tangent isy− 1 =
1
e
(x−e)ory=
x
e
.
(See Figure 6.10-1.)
[−1.8] by [−3,3]
Figure 6.10-1