More Applications of Derivatives 205
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
endpoints f(−1)<0 andf(8)= 1 /2,
fhas an absolute minimum at
x=−1.
(c)Atx=5,fhas a change of concavity
and f′exists atx=5.
(d) f′′<0on(−1, 5),f is concave
downward on (−1, 5).
f′′>0 on (5, 8), fis concave
upward on (5, 8).
(e)A possible graph off is given in
Figure 9.9-1.
–1 0 1 2 3 4 5 6 7 8
(2,3)
(8,^1 ⁄ 2 )
(^3) f
y
x
Figure 9.9-1
- (a) v(t)=3 ft/s att=6. The tangent line
to the graph ofv(t)att=6 has a slope
of approximatelym=1. (The tangent
passes through the points (8, 5) and
(6, 3); thusm=1.) Therefore the
acceleration is 1 ft/s^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. - To convertr=4 cosθto a Cartesian
representation, recall thatr=
√
x^2 +y^2
and tanθ=
y
x
.Then,
√
x^2 +y^2 =4 cos
(
tan−^1
y
x
)
. Since
cos
(
tan−^1
y
x
)
=
x
√
x^2 +y^2
, the equation
becomes
√
x^2 +y^2 =
4 x
√
x^2 +y^2
. Multiply
through by
√
x^2 +y^2 to produce
x^2 +y^2 = 4 x. Completing the square
produces (x−2)^2 +y^2 =4.