Calculus I HW 9/13 Solutions
1. A Weird Function
{
Dene the function
f (x) =
Note that
f
is continuous at
x2 sin x1
0
c ̸= 0
since
x ̸= 0
.
x=0
if
if
x2 sin x1
is a combination of functions
known to be continuous where dened. Also note that
lim f (x) = lim x2 sin
x→0
x→0
1
=0
x
as shown in Example 1 of section 3.4 of your text using the Sandwich Theorem.
Since the limit not only exists, but also
f
is continuous at
0,
f (0) = 0 = limx→0 f (x),
we have that
too.
(a)
Use the limit denition of the derivative to nd
f ′ (0).
(You may cite an example
limit from the text, if you wish.)
Solution
By denition,
f ′ (0) = lim
h→0
f (0 + h) − f (0)
h
h2 sin h1 − 0
1
= lim h sin = 0
h→0
h→0
h
h
= lim
The nal limit above is zero as shown in the Sandwich Theorem section of your
text. (If you didn't cite that, you should have used the Sandwich Theorem to
prove it.)
(b)
Calculate
f ′ (x)
at all points
x ̸= 0.
Solution
For
x ̸= 0, f (x) = x2 sin (1/x),
so we can nd
f ′ (x)
through normal derivative
calculations:
)(
(
(
) (
)
)
1
d
1
d 2
1
2
2
sin
x sin
=
x
+x
sin
by the product
x
dx
x
dx
x
(
)(
)
1
1
d 1
′
2
= 2x sin + x cos
by the chain rule and sin = cos
x
x
dx x
d
f (x) =
dx
′
1
rule
= 2x sin
(
)(
)
1
1
1
+ x2 cos
− 2
x
x
x
= 2x sin
by the power rule since
1
1
− cos
x
x
1
= x−1
x
by algebra
(c)
Using the formula you found in part (b), calculate
limx→0 f ′ (x).
Solution
lim f ′ (x) = lim 2x sin
x→0
x→0
1
1
− cos
x
x
since the limit does not depend on
f ′ (0)
Now,
lim 2x sin
x→0
1
1
= 2 lim x sin = 2 ∗ 0 = 0
x→0
x
x
where the middle equality comes from the same limit we cited for part (a).
′
Since that part is 0, the original limit limx→0 f (x) does the same thing that
limx→0 (− cos (1/x)) does. However, this limit diverges due to oscillation on ei′
+
ther side because as x → 0 , 1/x → ∞, etc. Therefore, lim f (x) diverges due to
x→0
Note
′
This function is very weird: it is the only function we've discussed where f
′
exists everywhere but is not continuous everywhere. f is not continuous at 0
′
′
since f (0) = 0 (by part (a)) but limx→0 f (x) d.n.e. (by part (c)).
2. A Function with a Parameter
{
ax2 + 3x + a2 + 2a + 2
Let fa (x) =
eax
if
if
x<0
.
x≥0
(a)
For which values of
a
is
fa
continuous on all of
R?
Solution
To discuss continuity on all of
places. For any
a , fa
R,
it's easiest to look at continuity in three
is continuous at
c
for
c<0
since it's a polynomial there,
and polynomials are continuous everywhere. For any
a, fa
at
c
for
c>0
since
it's an exponential function there, and those are continuous everywhere too.
Finally, we must examine which values of
2
a
make
fa
continuous at
0.
oscillation.
Since exponential functions are continuous
a. Therefore, we merely
fa (0) = ea∗0 = 1.
fa (0) = limx→0+ fa (x)
for any
need to check when the limit from the left is equal to
lim fa (x) = lim− ax2 + 3x + a2 + 2a + 2
x→0−
x→0
= a ∗ 02 + 3 ∗ 0 + a2 + 2a + 2 = a2 + 2a + 2
a2 + 2a + 2 = fa (0) = 1 ⇒ a2 + 2a + 1 = 0
2
⇒ (a + 1) = 0 ⇒ a = −1.
Therefore,
fa
is only continuous on all of
R
if
a = −1.
(b)
For which values of
a
is
fa
dierentiable on all of
R?
Solution
c (for any a) for c ̸= 0 since it's a polynomial
If fa is dierentiable at 0, it's
continuous at 0. Therefore, the only a that could possibly work is the a from
′
part (a): −1. The real question is: Does f−1 (0) exist? To answer this, we
fa
is automatically dierentiable at
or an exponential function in those regions.
examine the two one-sided version of the limit denition of the derivative.
lim
h→0+
f−1 (0 + h) − f−1 (0)
e(−1)h − e(−1)0
= lim
h
h
h→0+
d −x =
e by denition
dx
x=0
= −e−x x=0 = −e−0 = −1.
2
lim−
h→0
f−1 (0 + h) − f−1 (0)
(−1) h2 + 3h + (−1) + 2 (−1) + 2 − e(−1)0
= lim−
h
h
h→0
= lim−
h→0
−h2 + 3h
= lim− −h + 3 = 3.
h
h→0
Since the limit on one side is
−1 and the limit on the other side is 3, the
′
f−1
(0) doesn't either. No value of a would
two sided limit does not exist, and
make
fa
dierentiable everywhere.
(c)
For which values of
a
is
fa
twice dierentiable on all of
3
R?
Solution
Since no value of
a
makes
continuous, no value of
a
fa
dierentiable at
could make
fa
0,
and
fa′′
existing implies
twice dierentiable on all of
R.
fa′
is
3. Proving Inequalities
Let
f (x) = 2x − x ln x
for
x > 0.
(a)
Find
f ′ (x)
x > 0.
for
Solution
d
d
d
f ′ (x) =
(2x − x ln x) = 2
(x) −
(x ln x)
dx
dx
dx
((
)
)
d
d
=2∗1−
(x) ln x + x
ln x by product rule
dx
dx
(
)
1
= 2 ∗ 1 − 1 ∗ ln x + x ∗
= 2 − (ln x + 1) = 2 − ln x − 1 = 1 − ln x
x
(b)
Show that for
x ∈ [1, e], 0 ≤ f ′ (x) ≤ 1.
Solution
f ′ (e) = 1 − ln e = 1 − 1 = 0.
With this information alone, it might be that for some x in between 1 and e,
f ′ (x) ∈
/ [0, 1]. However, that can't happen because f ′′ (x) = 0 − 1/x < 0 for
x ∈ [1, e], so f ′ is only allowed to decrease from 1 to 0 on [1, e].
Note that
f ′ (1) = 1 − ln 1 = 1 − 0 = 1
and
(c)
Use part (b) and a fact from the text/lecture to nd linear functions
h (x)
such that
g (x) ≤ f (x) ≤ h (x)
for
g (x)
and
f.
The
x ∈ [1, e].
Solution
Part (b) is an inequality involving
f ′,
and this asks for one involving
only time we've seen something like that in this course is a corollary to the
Mean Value Theorem.
Specically, the one that states: If
(a, b) and continuous on [a, b] and m ≤ f ′ (x) ≤ M
m (b − a) ≤ f (b) − f (a) ≤ M (b − a).
on
4
f
for
is dierentiable
x ∈ (a, b),
then
Setting
a=1
and
b=x
means that part (b) yields
formula for
f
(for some x ∈ [1, e]) in the statement
0 (x − 1) ≤ f (x) − f (1) ≤ 1 (x − 1).
above, this
Using the
and simplifying, we have
0 ≤ f (x) − (2 (1) − 1 ln 1) ≤ x − 1
⇔ 0 ≤ f (x) − 2 ≤ x − 1
(since
ln 1 = 0)
⇔ 2 ≤ f (x) ≤ x + 1
g (x) = 2
(So
and
h (x) = x + 1
will work.)
(d)
Use the inequality you found in part (c) to show that
2−
x+1
≤ ln x ≤ 2
x
(
x−1
x
)
for
x ∈ [1, e] .
Solution
2 ≤ f (x) ≤ x + 1 ⇒ 2 ≤ 2x − x ln x ≤ x + 1
⇒ 2 − 2x ≤ −x ln x ≤ 1 − x
⇒ 2x − 2 ≥ x ln x ≥ x − 1
since mult. by a neg. switches ineq. direction
2x − 2
x−1
≥ ln x ≥
since x ∈ [1, e] so x > 0
x
x
(
)
x−1
2x − (x + 1)
x+1
⇒2
≥ ln x ≥
=2−
as desired.
x
x
x
⇒
4. Graphing a Function
Let
x+1
f (x) = √
.
|x2 − 1|
(a)
Determine where
f
is increasing and where it is decreasing.
5
Solution
There are two ways to approach this problem.
√
absolute value as a single function (like
(d/dx) |x| = x/ |x| where dened).
|x| =
One way is to work with the
x2
or merely using the fact that
However, that would lead to some extremely
messy expressions (but it would work). Another way is to simply break things
up into cases, so that we don't have to worry about the absolute value.
2
To use this cases approach, we need to know when x − 1 < 0.
2
2
happens exactly for x ∈ (−1, 1) (so that x ∈ [0, 1)). Analogously, x −
when
x>1
or
x < −1
Therefore, we can write the following:
x+1
 x+1

 √x2 −1
f (x) = √
=
|x2 − 1| 
if
√x+1
 1−x2
if
undened
To analyze the monotonicity of
f,
if
x>1
x=1
if
√ 2
√
x −1−(x+1)(x/ x2 −1)


√
x2 −1
√
1−x2 −(x+1)(−x/ 1−x2 )
=
2

1−x


undened

undened

−1−x


 (x2 −1)3/2
=
1+x
 (1−x2 )3/2

undened
if
if
if
if
if
if
6
or
f ′.
x < −1
−1<x<1
if
if
.
x = −1
x>1
x=1

(
)
√
d
d

√1
(x+1)) x2 −1−(x+1)
( dx
(x2 −1))
(

dx
2

2 x −1

√

2

( x2 −1

)
()
√
d
d
2
= ( dx (x+1)) 1−x −(x+1) √ 1 2 ( dx
(1−x2 ))
2 1−x


√
2


( 1−x2 )


undened
=
or
we need to know about
undened
1−x2 +x2 +x
 (1−x2 )3/2
x < −1
−1<x<1
 d (x+1) √x2 −1−(x+1) d √x2 −1
)
( dx
dx

√

2

x2 −1)
(

√
√
d
2 −(x+1) d
1−x2
dx
f ′ (x) = ( dx (x+1)) 1−x
√
2

2
1−x

(
)


 2
x −1−x2 −x


 (x2 −1)3/2
or
if
if
or
x>1
x = −1
or
−1<x<1
if
x=1
or
x < −1
or
if
x>1
if
−1<x<1
x = 1 or x = −1
if
x>1
or
x < −1
−1<x<1
x=1
x>1
or
or
x = −1
x < −1
−1<x<1
x=1
or
x < −1
x = −1
x = −1
This
1>0
Note that if x > 1 or x < 1,
2
then 1 − x is positive. Thus,
f ′ (x)
x2 − 1
has the same sign as
and is positive. Similarly, if


−1 − x
1+x


if
if
undened
if
−1 < x < 1,
x > 1 or x < −1
.
−1<x<1
x = 1 or x = −1
f ′ (x) > 0 for x ∈ (−∞, −1), and f ′ (x) < 0 for x ∈ (1, ∞) and f ′ (x) >
0 for x ∈ (−1, 1). Since positive derivative implies increasing, and negative
derivative implies decreasing, f is increasing on (−∞, −1) and (−1, 1) and f is
decreasing on (1, ∞).
Hence,
(b)
Where is the function concave up and where is it concave down?
inection points of
Find all
f.
Solution
Concavity depends on the second derivative, so we should dierentiate the expression we had from part (a):

−1−x


 (x2 −1)3/2
d
1+x
f (x) =
(1−x2 )3/2
dx 

undened
′′
if
x>1
x=1
(
)
 d
3/2
3/2
d
( dx (−1−x))(x2 −1) −(−1−x) dx
(x2 −1)




(x2 −1)3 (
)
3/2
2 3/2
d
d
(1+x)
1−x
−(1+x) dx
(
)(
)
(1−x2 )
=
dx


(1−x2 )3


undened
)
(

1/2 d
3/2
2
−(x2 −1)
−(−1−x) 32 (x2 −1)
dx (x −1)



3
2

((x −1) 1/2
)
2
2 3/2
3
d
1−x
1−x
−(1+x)
(
) dx
(1−x2 )
(
)
=
2


(1−x2 )3


undened
(
)

3/2
1/2
−(x2 −1)
−(−1−x) 3x(x2 −1)



3

(x2 −1)
(
)
1/2
2 3/2
1−x
−(1+x)
−3x
)
(1−x2 )
= (


(1−x2 )3


undened
7
x < −1
−1<x<1
if
if
or
if
if
if
or
x = −1
if
if
if
if
if
if
x>1
or
x < −1
−1<x<1
x=1
x>1
or
or
x = −1
x < −1
−1<x<1
x=1
x>1
or
or
x = −1
x < −1
−1<x<1
x=1
or
x = −1

−(x2 −1)−(−1−x)(3x)




(x2 −1)5/2
2
1−x
(
)−(1+x)(−3x)
=

(1−x2 )5/2



undened
 2
2x +3x+1


 (x2 −1)5/2
=
if
2
2x +3x+1
(1−x2 )5/2


undened

(2x+1)(x+1)


 (x2 −1)5/2
=
(2x+1)(x+1)
 (1−x2 )5/2

undened
if
if
if
if
if
if
if
if
x>1
or
x < −1
−1<x<1
x=1
x>1
or
or
x = −1
x < −1
−1<x<1
x=1
x>1
or
x = −1
or
x < −1
−1<x<1
x=1
or
x = −1
By the same sort of reasoning used for part (a),
{
′′
f (x)
has the same sign as
(2x + 1) (x + 1)
if
undened
if
x ̸= 1
x=1
x ̸= −1
.
x = −1
and
or
x ∈ (−∞, −1), 2x + 1, x + 1 < 0, so the product above is positive.
x ∈ (−1, −1/2), 2x + 1 < 0 and x + 1 > 0, so the product above is negative.
If x ∈ (−1/2, ∞), 2x + 1, x + 1 > 0, so the product above is positive. Since
′′
′′
positive f corresponds to concave up, and negative f corresponds to concave
down, f is concave up on (−∞, −1), (−1/2, 1), and (1, ∞), and f is concave
down on (−1, −1/2).
Note that if
If
Furthermore, inection points are points on the graph where concavity changes.
f
is not dened at
(
f
Therefore
(
−
1
2
)
−1,
but it
is dened at −1/2.
−1/2 + 1
1/2
1
1/2
= √
=√
=√ .
=√
3
|1/4 − 1|
3/4
2
(−1/2) − 1
√ )
−1/2, 1/ 3
is an inection point.
(c)
Find all horizontal asymptotes, vertical asymptotes, and holes in the graph of
f.
Solution
To nd the horizontal asymptotes, we merely need to examine the limits at
x+1
x+1
lim f (x) = lim √
= lim √
x→∞
x→∞
|x2 − 1| x→∞ x2 − 1
8
since for
x > 1,
±∞.
that's equal to
f (x)
This limit has the form
∞/∞,
but L'Hospital's rule doesn't lead anywhere
helpful with this limit. Instead, write
x−1
1+0
x+1
x+1
1 + 1/x
lim √
= lim √
∗ −1 = lim √
=√
= 1.
2
2
2
x→∞
x→∞
1−0
x − 1 x→∞ x − 1 x
1 − 1/x
Therefore,
y=1
is a horizontal asymptote. Similarly,
x+1
x+1
lim f (x) = lim √
= lim √
2
x→−∞
x→−∞
x2 − 1
|x − 1|
x→−∞
since for
x > 1,
that's equal to
−1
x+1
(−x)
−1 − 1/x
−1 − 0
= lim √
∗ √
= lim √
=√
= −1.
2
2
−2
x→−∞
x→−∞
1−0
x −1
1 − 1/x
x
Therefore,
y = −1
is a horizontal asymptote as well.
To nd the vertical asymptotes and holes, we must examine limits at places
where
f
is undened.
x+1
x+1
lim+ √
= lim+ √
2
x→1
x2 − 1
|x − 1| x→1
This has the form
x=1
since
x2 − 1 > 0
for
x>1
2/0, and since square roots are positive, this is ∞.
Therefore,
is a vertical asymptote.
lim
x→−1+
x+1
x+1
√
= lim + √
2
x→−1
1 − x2
|x − 1|
since
1 − x2 > 0
for
x ∈ (−1, 1)
√
x+1
1+x
0
√
= lim √
= lim √
= √ = 0 ̸= ±∞
+
+
x→−1
1 + x 1 − x x→−1
1−x
2
x+1
x+1
2
lim √
= lim √
since x − 1 > 0 for x < −1
x→−1−
|x2 − 1| x→−1− x2 − 1
√
x+1
− −x − 1
√
√
= lim − √
= lim
= 0 ̸= ±∞
x→−1
1 − x −x − 1 x→−1−
x−1
x approaches −1 are ±∞,
(−1, 0) .
Since neither of the limits as
0,
there's just
a hole at
and in fact they're both
(d)
Sketch a possible graph of
f
that incorporates all of the information gathered
in the previous parts.
9
f (x)
Solution
To collect it all in one place, the known information is:
• f
is increasing on
(−∞, −1)
• f
is concave up on
and
(−1, 1)
and
(−∞, −1), (−1/2, 1),
(−1, −1/2).
√ )
(
• −1/2, 1/ 3 is an inection point.
f
and
is decreasing on
(1, ∞),
and
f
(1, ∞).
is concave
down on
• y=1
(from
x → ∞)
and
y = −1
(from
x → −∞)
are horizontal asymp-
totes.
• x=1
is a vertical asymptote.
• (−1, 0)
is a hole in the graph.
To combine the information from the rst two bullet points, note that on
(−∞, −1), f is increasing and concave up (so it looks like the right half of
`). On (−1, −1/2), f is increasing and concave down (so it looks like the left
half of a). On (−1/2, 1), f is increasing and concave up (so it looks like the
right half of `). On (1, ∞), f is decreasing and concave up (so it looks like the
left half of `).
Putting all of this information together produces something like the following (with a lled in circle representing the inection point, an open circle
representing the hole, and dashed lines representing asymptotes):
5. Optimizing a Function
A rectangle has its base on the x-axis, and its other two vertices on the graph
y = 1 + 2x2 − x4 , as in this example diagram:
of
10
What is the length of the base of such a rectangle that would maximize
its area?
Note:
You may assume that an area-maximizing rectangle will be
symmetric about the
y -axis.
(Hint:
If you get stuck on a dicult algebra
x2 .)
problem, try substituting a new variable for
Solution
Let
(x, y)
be the coordinates of the top-right corner of the rectangle. Then the
2x and the height is y , so the the area we need to maximize is
2xy . Since (x,(y) lies on the)curve y = 1+2x2 −x4 , the area we need to maximize
2
4
is f (x) = 2x 1 + 2x − x . The exact domain for a problem like this doesn't
length of the base is
really matter since the endpoints of the domain would yield rectangles with
0
area, but note that
x≥0
since anything less would make
√
1 + 2x2 − x4 = 0 at x = ±
[ √
√ ]
1 + 2 2 .)
happens to be 0,
right vertex. (Also,
so the domain
−2+
√
(x, y)
22 −4(−1)1
2(−1)
not the top
√
√
=± 1+2 2
′
The non-endpoints-of-the-domain candidates for extrema are where f does
′
not exist and where f is zero. Since f is a polynomial, its derivative exists
′
everywhere, and we merely have to solve f (x) = 0.
f ′ (x) = 0 ⇒
⇒
))
d ( (
2x 1 + 2x2 − x4 = 0
dx
)
d (
2x + 4x3 − 2x5 = 0 ⇒ 2 + 4 ∗ 3x2 − 2 ∗ 5x4 = 0
dx
⇒ 2 + 12x2 − 10x4 = 0 ⇒ 5x4 − 6x2 − 1 = 0
√
√
√
2
−
(−6)
±
(−6) − 4 (5) (−1)
6 ± 56
3 ± 14
2
⇒x =
=
=
2 (5)
10
5
√
√
3 ± 14
⇒x=±
5
√
√
√
√
3 + 14
⇒x=±
since 3 −
14 < 3 − 9 = 0
5
11
√
⇒x=
3+
√
14
5
since
x≥0
for
x
to be in the domain
Since this is the only critical point, and the area (which is terrible and unnecessary to calculate exactly) for this
x-value.
x is positive, this must be the area-maximizing
So the area-maximizing base-length is
√
3+
2x = 2
√
14
5
.
Notes
Incidentally, this
x-value
hump of the graph is.
is slightly more than
1,
which is where the top of the
Thus, this area-maximizing rectangle is slightly wider
and slightly less tall than the rectangle of maximum height.
6. Optimizing a Distance
Find the point on the graph of
y=
(
)
(√ )
2
3 x2 − √
x
3
(
closest to the point
7 7
, √
9 6 3
)
.
Straightforward Solution
the curve. Then the distance to the given
√ (x, y) is a( point on
( √ ))2
2
(x − 7/9) + y − 7/ 6 3 . Since for every point on the curve,
point is
(√ ) 2 ( √ )
y=
3 x − 2/ 3 x, we may write this distance as
√(
)2 ((( )
(
) )
)2
√
7
2
7
2
√
√
x−
+
3 x −
x −
.
9
3
6 3
Suppose that
By a theorem from a problem in the text/Paul's math notes, or just thinking about the fact that
r (x) =
minimized at the same value of
√
x
(
x
is an increasing function, this distance is
that minimizes its square.
(( )
(
)
)2
√
2
7
2
Let f (x) = square of dist. =
+
3 x − √
x− √
3
6 3
( )2
7
7
= x2 − 2 ∗ x +
9
9
(
)( )
)(
)
(
((√ ) )2 ( ( 2 ) )2 ( 7 )2 (√ ) ( 2 )
√
7
7
2
√
√
x + √
x−2
3 x2 + − √
−2
3 x2 √
3 x2 +2 √ x
+
3
6 3
3
6 3
3
6 3
= x2 −
7
x−
9
)2
14
49
4
49
7
14
x+
+ 3x4 + x2 +
− 4x3 − x2 + x
9
81
3
36 ∗ 3
3
9
12
= 3x4 − 4x3 +
49
49
+
.
81 36 ∗ 3
This is the function we wish to minimize. Since the equation of the curve is
just a polynomial, the domain here is all of
∞
or
−∞
R = (−∞, ∞).
Since
x
approaching
certainly wouldn't minimize the distance, there really is a minimum,
′
and we merely need to look at candidates coming from f .
′
3
2
Note that f (x) = 12x − 12x . Since this is a polynomial, it's dened
′
′
everywhere, so there are no f undened candidates, only f = 0 candidates.
)
(
f ′ (x) = 0 ⇒ 12 x3 − x2 = 0 ⇒ x2 (x − 1) = 0 ⇒ x = 1
or
0.
To nd out which of these is the distance-minimizing x-coordinate, we can just
49
49
49
49
plug into f and compare. f (1) = 3 − 4 +
81 + 36∗3 = −1 + 81 + 36∗3 , but
49
49
f (0) = 81 + 36∗3 > f (1). Therefore, the distance is minimized at x = 1. Since
the problem asked for the point, we need to calculate the y -coordinate, too. The
point is
( √
( (√ )
√ )
( √ ) )
2
1, 3 (1) − 2/ 3 ∗ 1 = 1, 3 − 2/ 3
.
Normal Line Solution (More annoying arithmetic)
There is a geometric fact that if you draw a line segment from a point to the
closest point on some (dierentiable) curve, the segment makes a right angle
with the curve; i.e.
the normal line to the curve at the closest point passes
through that other point.
To make use of this fact here, we must nd all normal lines of the curve and
(
( √ ))
(√ ) 2 ( √ )
7/9, 7/ 6 3 . Set f (x) =
3 x − 2/ 3 x;
see which, if any, pass through
then the equation for the normal line to the curve
y = f (x)
at
x=a
is
−1
(x − a)
f ′ (a)
(
) )
(( )
√
2
−1
2
√
√ (x − a)
3 a −
a = (√ )
⇔y−
3
2 3 a − 2/ 3
(
)
(√ )
2
a−x
2
√
⇔y−
3 a + √
a = (√ )
3
2 3 a − 2/ 3
(
( √ ))
Since the point 7/9, 7/ 6 3
is supposed to lie on the normal line,
plug those numbers in for x and y and get a true equation:
(
)
(√ )
2
a − 7/9
7
√ −
√
3 a2 + √
a = (√ )
6 3
3
2 3 a − 2/ 3
y − f (a) =
⇒
√
7
a − 7/9
√
− 3a2 + 2a = 3 ∗ (√ )
6
2 3 a − 2/ 3
13
(mult. thru by
√
3)
we can
(
⇒
)( ( )
)
(
)
√
√
7
2
7
− 3a2 + 2a
2
3 a− √
= 3∗ a −
6
9
3
(
⇒
)
(
)
7
7
2
− 3a + 2a (2 (3) a − 2) = 3 ∗ a −
6
9
(mult. thru by
(mult. thru by
2
(√ )
2
3 a− √ )
3
√
3)
7
7
+ 6a2 − 4a = 3a −
3
3
3
2
3
2
⇒ 12a − 18a + 6a = 0 ⇒ 2a − 3a + a = 0
⇒ 7a − 18a3 + 12a −
(
)
⇒ a 2 + a − 3a2 = 0 ⇒ a (1 − a) (2 + 3a) = 0 ⇒ a = 0, 1, −3/2
x-coordinate, we can
y -coordinates
and compare distances (or better yet, their
(
( √ ))
(√ ) 2 ( √ )
squares) to the given point 7/9, 7/ 6 3 . f (1) =
3 1 − 2/ 3 1 =
√
( √ )
(√ ) 2 ( √ )
(√ )
2
3 − 2/ 3 , f (0) =
3 0 − 2/ 3 0 = 0, and f (−3/2) =
3 (−3/2) −
√
√
( √ )
2/ 3 (−3/2) = 9 3/4 + 3.
(
)
(
)
√
7 7
2
square of dist. between
, √
and
1, 3 − √
9 6 3
3
(
)2
)2
)2 (
(
√
√
7
2
7
4
19
= 1−
3− √ − √
=
3− √
+
+
9
81
3 6 3
6 3
4
19
192
3
18
182
=
+3−
+
≈
+3−
+
81
3
36 ∗ 3
81
3
36 ∗ 3
3
3
=
+3−6+3=
.
81
81
(
)
7 7
square of dist. between
, √
and (0, 0)
9 6 3
)2 (
(
)2
7
49
3
7
49
+ 0− √
+
≫
.
= 0−
=
9
81 36 ∗ 3
81
6 3
(
)
√
(
)
7 7
3 9 3 √
square of dist. between
, √
and
− ,
+ 3
9 6 3
2 4
)2 (
)2
)2 ( √
)2 ( √
(
7
3 7
9 3 √
7
41
13 3
+
+ 3− √
+
− √
= − −
= −
2 9
4
18
4
6 3
6 3
To nd out which of these is the distance-minimizing
just plug into
f
to get
132 ∗ 3
49
13 ∗ 7
412
+
+
−
182
16
36 ∗ 3
12
162 ∗ 3
45
12 ∗ 7
25
5
3
452
+
−
=
+ 48 +
−7≫
.
≈ 2+
18
16
36 ∗ 3
12
81
12
81
( √
√ )
Therefore, the closest point on the curve is
1, 3 − 2/ 3 .
=
14
7. Optimization Application
10 feet of wire is to be cut into two pieces.
One piece will be bent into the shape
of a square, and the other will be bent into a circle. How much wire should go
to the square to maximize the sum of the areas of the two shapes? How much
to minimize the sum of the areas?
Solution
r
Firstly, we must pick variables. Let
side length of the square. Then since
be the radius of the circle, and
10ft
s
be the
of wire is used to make both shapes,
by the formulas for circumference of a circle and perimeter of a square, we have
10ft = 2πr + 4s.
Solving the previous equation for
2
2
By area formulas, the sum of the areas is πr + s
this last expression
To optimize
f,
f (r);
s, we have s = (5ft − πr) /2.
2
= πr2 + (5ft − πr) /4. Call
it's what we must optimize.
r can be is 0, when
r can be is when all of
r = 5ft/π . Therefore, if
we must nd the domain. The smallest
all of the wire goes to making the square. The largest
10ft = 2πr
the wire goes to making the circle, so
and
[0, 5/π].
we work in feet, the domain is
The three types of candidates for extrema are endpoints of the interval,
f ′ is undened, and places where f ′ = 0. We must calculate f ′ :
places where
d
f (r) =
dr
(
(5ft − πr)
πr +
4
′
= 2πr +
f′
2 (5ft − πr) (−π)
4
2
)
2
(by chain/product rule)
= 2πr +
π 2 r − 5πft
.
2
is a polynomial, so it's dened everywhere.
f ′ (r) = 0 ⇔ 2πr +
⇔r=
π 2 r − 5πft
4π + π 2
5π
=0⇔
r=
ft
2
2
2
5πft/2
5
=
ft.
2
(4π + π ) /2
4+π
Therefore, our three candidates are
just need to compare the values of
f
r = 0, r = 5ft/π ,
and
r = 5ft/ (4 + π).
and which is smallest.
2
f (0) = π02 +
(
f
(
f
5
ft
π
)
(
=π
)
5
ft = π
4+π
(5ft − π0)
25 2
=
ft
4
4
)2
2
(5ft − π (5ft/π))
25 2
=
ft
4
π
)2
2
5
(5ft − π (5ft/ (4 + π)))
ft +
4+π
4
5
ft
π
(
We
at these three points to see which is largest
+
15
2
25ft2 (1 − π/ (4 + π))
4
(4 + π)
(
)2
25
25ft2
4
25π + 25 ∗ 4 2
2
=π
=
2 ft +
2 ft
4
4
+
π
(4 + π)
(8 + π)
=π
<
25
2
2 ft +
25 ∗ 4 + 25 ∗ 4
(8 + 3)
2
ft2 =
200 2
25 2
ft < 2ft2 <
ft .
121
4
π < 4, 25/π > 25/4, so that the maximum area sum is achieved
r = 5ft/π , and the minimum area sum is achieved at r = 5ft/ (4 + π).
Since
at
Since the problem asks how much wire should the square get, we need to
nd
4s = 10ft − 2πr
square gets
none
4s = 40/ (4 + π) ft
in each of these cases. If
r = 5ft/π , then 4s = 0, and the
If r = 5ft/ (4 + π), then
of the wire for area-maximizing.
and the square gets most of the wire for area-minimizing.
8. Two exponent limits
(a)
(
lim
Find
x→0+
1
1−
ln x
)2x
.
Solution
Note that
limx→0+ ln x = −∞,
so
limx→0+ 1/ ln x = 0
(as the denominator
0. Therefore,
limx→0+ (1 − 1/ ln x) = 1 − 0 = 1. Since limx→0+ 2x = 0, this limit has the form
10 , so the answer is simply 10 = 1 . (Note: 10 is not an indeterminate form,
grows in absolute value without bound, the fraction approaches
so we couldn't use L'Hospital's rule, although it wouldn't be invalid to convert
this limit into a fraction.)
(b)
Find
1− 1
lim (2x)( ln x ) .
x→0+
Solution
By identical reasoning to that in part (a), this has the form
indeterminate form, it's just
0
.
16
01 .
That's not an
9. Two trig limits
(a) limx→∞ (ex + sin x) /ex
Solution
ex + sin x
ex
sin x
sin x
=
lim
+ lim
= 1 + lim
.
x→∞
x→∞ ex
x→∞ ex
x→∞ ex
ex
x
x
For this last limit, since −1 ≤ sin x ≤ 1, and e is positive, we have −1/e ≤
x
x
x
(sin x) /e ≤ 1/e . Since limx→∞ ±1/e = 0 (it has the form ±1/∞), the
x
Sandwich Theorem tells us that limx→∞ (sin x) /e = 0, so that the limit in the
problem is just 1 .
lim
(If we had tried L'Hospital's rule, then after four applications we'd be right
back where we started, and no intermediate step would be any easier to handle
than the current limit.)
(b) limx→0 (1 − cos x) /x2
Solution
Since
1 − cos 0 = 1 − 1 = 0
and
02 = 0,
this limit has the form
0/0,
and we may
apply L'Hospital's rule:
1 − cos x
= lim
x→0
x→0
x2
lim
d
dx
(1 − cos x)
d 2
dx x
This last limit can either be recalled to be
sin x
1
sin x
= lim
.
x→0 2x
2 x→0 x
= lim
1 (as proven in the text), or L'Hospital's
0/0). Either way, the answer to the
rule can be applied to it (as it has the form
original limit is seen to be
1/2
.
10. Three more limits
(a)
Find
lim ln x + ln (1/x) .
x→∞
Solution
While this limit has the indeterminate form
∞ − ∞,
it's easy to evaluate if we
simplify it with log rules:
lim ln x + ln
x→∞
(
)
1
= lim ln x + ln x−1 = lim ln x − ln x = 0
x→∞
x x→∞
(b)
Find
lim ln x − log10 (x) .
x→∞
17
Solution
While this limit has the indeterminate form
∞ − ∞,
it's easy to evaluate if we
rewrite it with log rules:
ln x
=
lim ln x −
x→∞
ln 10
(
1−
1
ln 10
)
lim ln x = ∞
x→∞
(c)
Find
√
3
x2 − 1
lim
x→∞
(8x + 1)
2/3
since
1/ ln 10 < 1.
.
Solution Method 1
∞/∞,
While this limit has the form
L'Hospital's rule just gives a similar limit
x is going to innity, multiplying
1 in a special form to divide the numerator and the denominator by the same
power of x will be useful:
√
√
3
3
x2 − 1
x2 − 1
x−2/3
lim
=
lim
∗
x→∞ (8x + 1)2/3
x→∞ (8x + 1)2/3
x−2/3
back again, and so isn't very helpful here. Since
by
√
3
1 − 1/x2
= lim
x→∞
(8 + 1/x)
2/3
√
3
1
= 2/3
8
(since
lim
x→∞
1
1
= 0) =
x
4
Solution Method 2
Another useful thing we can do is rewrite the function and use continuity to get
the answer with our theorem about rational functions:
√
3
x2 − 1
lim
x→∞
(8x + 1)
2/3
√
√
3
x2 − 1
x2 − 1
= lim √
= lim 3
2
x→∞ 3
x→∞
2
(8x + 1)
(8x + 1)
√
=
√
=
3
1
1
=
82
4
3
lim
x→∞
x2 − 1
(8x + 1)
2 by continuity of
√
3
x
by the thm. about limits at innity of rational functions
18