MA 3972-MA-Book April 11, 2018 15:14
226 STEP 4. Review the Knowledge You Need to Score High
dy
dx
∣∣
∣∣
y= 2
=
1
3(2)^2 −6(2)+ 3
=
1
3
- Substitutingx=100 into the expression
x− 100
√
x− 10
would lead to
0
0
. Apply
L’Hôpital’sRule, and you have limx→ 100
1
1
2 x
−^12
or
1
1
2 (100)
−^12 =^20 .Another approach to
solve the problem is as follows. Multiply
both numerator and denominator by the
conjugate of the denominator (
√
x+10):
xlim→ 100
(x−100)
(√
x− 10
)·
(√
x+ 10
)
(√
x+ 10
)=
xlim→ 100
(x−100)
(√
x+ 10
)
(x−100)
xlim→ 100 (
√
x+10)= 10 + 10 = 20.
An alternative solution is to factor the
numerator:
xlim→ 10
(√
x− 10
)(√
x+ 10
)
(√
x− 10
) = 20.
- (a) f′>0on(−1, 2),f is increasing on
(−1, 2),f′<0 on (2, 8), fis
decreasing on (2, 8).
(b) Atx=2,f′=0 and f′′<0, thus at
x=2,f has a relative maximum.
Since it is the only relative extremum
on the interval, it is an absolute
maximum. Since fis a continuous
function on a closed interval and at its
endpointsf(−1)<0 and f(8)= 1 /2,
f has an absolute minimum atx=−1.
(c) Atx=5,fhas a change of concavity
and f′exists atx=5.
(d) f′′<0on(−1, 5), fis concave
downward on (−1, 5).
f′′>0 on (5, 8), f is concave
upward on (5, 8).
(e) A possible graph of fis given in
Figure 10.8-1.
- 123456780 1
(2, 3)
(8, 1⁄ 2)
(^3) f
y
x
Figure 10.8-1
- (a) v(t)=3 ft/sec att=6. The tangent
line to the graph ofv(t)att=6 has a
slope of approximatelym=1. (The
tangent line passes through the points
(8, 5) and (6, 3); thusm=1.)
Therefore, the acceleration is 1 ft/sec^2.
(b) The acceleration is a minimum
att=0, since the slope of the tangent to
the curve ofv(t) is the smallest att=0.