MATH 1A WORKSHEET 24

MATH 1A WORKSHEET 24

THU, APR 14, 2016

(1)

Find

all

slant

asymptotes

of

f (x) =

3x2 +4x-2 x-1

.

We

use

polynomial

long

division

to

get

f (x)

=

3x

+

7

+

5 x-1

,

so

y

=

3x

+

7

is

the

unique

slant

asymptote.

(2) Let 1 = {x + y = 1} and 2 = {x = 2} and 3 = {x + y = 4} be three lines. Points P1, P2, P3 lie on 1, 2, 3

respectively, and P1 is fixed at (1, 0) but P2 and P3 are free to move about the lines 2 and 3 respectively. Where

are the location of P2, P3 that minimize the combined sum of the distances P1P2 + P2P3?

The point P2 is of the form (2, y2) for some y2 R, and the point P3 is of the form (x3, 4 - x3) for some x3 R.

Set

(y2, x3) = P1P2 + P2P3 = (1 - 2)2 + (0 - y2)2 + (2 - x3)2 + (y2 - (4 - x3))2

for all y2, x3. The idea is to fix P2 (i.e. fix y2), find the point P3 (i.e. find x3) which minimizes (y2, x3) and call

it f (y2), then minimize f (y2) while letting y2 vary.1 To minimize (y2, x3) for fixed y2 and varying x3, it suffices to

minimize the second term (2 - x3)2 + (y2 - (4 - x3))2 since the first term (1 - 2)2 + (0 - y2)2 is constant (since

y2 is fixed); for this, it suffices to minimize the square (2 - x3)2 + (y2 - (4 - x3))2 of the expression. This expression

is differentiable everywhere and has derivative 2(2 - x3)(-1) + 2(y2 - (4 - x3)) = 4x3 + 2y2 - 12; the only critical

point

is

at x3

=

6-y2 2

which

is

an

absolute minimum

since

4x3 + 2y2 - 12 < 0

if

x3

<

6-y2 2

and

4x3 + 2y2 - 12 > 0

if

x3

>

6-y2 2

.

The above shows that, for a given point P2 = (2, y2), the point P3 which minimizes P1P2 + P2P3 is P3 =

(

6-y2 2

,

y2 -2 2

).

Set

f (y2) = =

(1 - 2)2 + (0 - y2)2 +

1

+

y22

+

|

y2 -2 2

|

(2

-

6-y2 2

)2

+

(y2

-

(4

-

6-y2 2

))2

which is differentiable everywhere except y2 = 2. Thus to find the absolute minimum of f (y2) for y2 R, it suffices

to find the absolute minimum of f (y2) for y2 (-, 2), find the absolute minimum of f (y2) for y2 (2, ), and

compare it to f (2).

Assume y2 (-, 2). Then f (y2) =

1 + y22 +

2-y2 2

so f

(y2) =

2y2

2 1+y22

-

1 2

which is

0 when

y2

= ? 1 ,

3

both

of which lie in the interval (-, 2).

Sign

analysis

shows

that

f

(y2)

<

0

for

each

of

the

intervals

(-, - 1 )

3

and

(- 1 ,

3

1 ),

3

and

f

(y2)

>

0

on

the

interval

( 1 , 2),

3

thus

f (y2)

is

decreasing

on

the

entire

interval

(-,

1 )

3

and

increasing

on

the

interval

( 1 ,

3

2);

this

means

that

f ( 1 )

3

f (y2)

for

any

y2

(-,

2).

Assume y2 (2, ). Then f (y2) =

1 + y22 +

y2 -2 2

so

f (y2)

=

2y2

2 1+y22

+

1 2

which

is

always

positive

when

y2 (2, ). This means f (y2) is increasing on the entire interval (2, ); in particular f (2) < f (y2) for any

y2 (2, ).

The

above

arguments

show

that

f (y2)

is

minimized

when

y2

=

1 ,

3

in

which

case

P2

=

(2, 1 )

3

and

P3

=

(3 - 1 , 1 - 1).

2323

(3)

Find

the

function

which

satisfies

f

(x)

=

-x2

+

6e

1 2

x

and

f (0) = 7.

The

general

formula

for

the

antiderivative

of

f

(x)

is

f (x)

=

-

1 3

x3

+

12e

1 2

x

+

C.

We can use our given initial

condition

f (0)

=

7,

which

means

0 + 12e0

+C

=

7,

or

C

=

-5.

Thus

f (x)

=

-

1 3

x3

+

12e

1 2

x

-

5.

(4) Estimate the area under the graph of f (x) = 2x + 1 from x = 0 to x = 5 using 10 approximating rectangles and right

endpoints. Is this an over/underestimate?

1The idea is similar in spirit to the following. (1) "If I want to find the cheapest house in the U.S., I can find the cheapest house in each of the 50 states then find the cheapest among the 50 that I found." (2) "If I have 60 soldiers standing in a formation of 6 rows and 10 columns, to find the tallest soldier it suffices to find the tallest soldier in each of the 6 rows, then find the tallest soldier among the 6 that are the tallest in their rows."

The

10

right

endpoints

of

the

rectangles

are

1 2

,

2 2

,

3 2

,

.

.

.

,

10 2

,

so

the

heights

of

the

rectangles

are

f

(

1 2

),

f

(

2 2

),

f

(

3 2

),

.

..

,

f

(

10 2

)

or in other words 2, 3, 4, . . . , 11.

Thus the total area of the 10 rectangles is

1 2

?

2

+

1 2

?

3

+

1 2

?

4

+

???

+

1 2

?

11

=

1 2

(2

+

3+4

+???

+ 11)

=

1 2

(

11(12) 2

-

1)

=

65 2

.

Here

we've

used

the

formula

n(n + 1)

1+2+???+n =

.

2

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