Math 120 A - Autumn 2016Midterm Exam Number Two

November 17th, 2016

Name: Student ID no. :

Signature: Section:

1 142 153 154 16

Total 60

• This exam consists of FOUR problems on FIVE pages, including this cover sheet.

• Show all work for full credit.

• You may use a TI-30X IIS calculator during this exam. Other calculators and electronicdevice are not permitted.

• You do not need to simplify your answers.

• If you use a trial-and-error or guess-and-check method when a more rigorous method isavailable, you will not receive full credit.

• If you write on the back of the page, please indicate that you have done so!

• Draw a box around your final answer to each problem.

• You may use one hand-written double-sided 8.5” by 11” page of notes.

• You have 50 minutes to complete the exam.

–

1. Happy Thursday! I bought you this graph.

-3 -2 -1 0 1 2 3 4 5

-2

-1

0

1

2

3

4

5

x

f(x)

(a) [4 points] Compute f(f(f(4))).

(b) [5 points] Find the domain and range of f�1(x).

(c) [5 points] Let g(x) = f(2x+ 1) + 1. Find the domain and range of g(x).

fff -H4D)=f(ffiD=f( 3) =D

Domain of f"

= range of f

=[-2,l]U(2§Range off "=domain of f =f3,DU[3,§

Domain Range

Original ( -3 ,DU[ 3.5 ] [ '

2.1 ]U( 2,5 )

Shift up1 ( '

3,1 )U[ 3,5 ] [ 1,2 ]u( 3,6 )

Shift left 1 ( ' 4,0) 42,4 ] [ 1,2 ]U( 3,6 )

Scatgyhorijs. -2,4€Et2]u@

2. [15 points] Gomba is on a diet. His weight is a linear-to-linear rational function of time.

Right now, Gomba weighs 21 pounds.

In 1 month, he will weigh 20.5 pounds.

In 7 months, he will weigh 18.75 pounds.

In the long run, what will Gomba’s weight approach?

(Assume Gomba will live forever.)

*- a××+thd f(o)=hf=21 - b=

21dg%omba's weight f( D= 9+4=20.5 - atb

¥20.5+20.5A

lwetixjnjil,

" * :# .pt?IIiIiI:P7 (20.5-0.51)+211=131.25+18.75115.6€

-12.25=1.251

D= 9.8

b=21( 9.8 )b=205.8

a=20.5--5/9.8)a=

15.6

3. [5 points per part] The rent for a one-bedroom apartment in Beattle is growing exponen-tially. (Even though the city is filled with bees.)

(a) In the year 2000, the rent in Beattle was $1020, and it increases by 2.3% per year.

Write a function f(t) for the rent in Beattle t years after 2000.

(b) The average monthly rent in Tickoma is also growing exponentially.

In the year 2007, the rent in Tickoma was $500 less than the rent in Beattle.

In the year 2016, the rent in Tickoma is $1000.

Write a function g(t) for the rent in Tickoma t years after 2000.

(c) When will the rents in Beattle and Tickoma be equal?

(Round your answer to the nearest year.)

⇐=A.btb.lt#

-10204.0235¥= Aobt

g(7)=A .

b7= fl 7) - 500=696

g( 16)=Aob' 6=1000 edit

r%÷¥17;¥:] ...

.gs#5mmb

1020 ( 1.0235=525 ( 1.04109 )t

In ( 1020 ) + In ( 1.025) = In ( 525 ) th ( 1.04109T)

In ( 1020 ) t £ In ( 1.023 )=h( 525 ) t th ( 1.04109)

+ ( In ( 1.023) - In ( 1.04109))= In ( 525 ) - In ( 1020 )

t.int?oIYHYoI?m=3s#

4. A polar rectangle is the region bounded by two concentric circular arcs and two raysthrough the center of those arcs. Okay, fine, here’s a picture:

y

x

✓

(a) [4 points] Write a formula for the area of this polar rectangle.(Your answer will involve x, y, and ✓. Let ✓ be measured in radians.)

(b) [4 points] Write a formula for the perimeter of this polar rectangle.

(c) [8 points] Suppose you have 24 meters of fencing, and you want to construct a fencein the shape of a polar rectangle with central angle ✓ = 1.2 radians. What is themaximum possible area of your fence?

),

A=tsox2-£oy÷

Inkiest. ¥iII

24=(2+1.2)×+4+1.2)y

t

24=3.2×-0 .8y

y=4¥30 A= 's(1.3×3 - £4.2)( 4×-393

A = 0.6×3 - 0.646×2-240×+900)

A =- 9×2+144×-540

fsfaph !

Area ,

maxareayy.gg-540 .

YEH -36cg

>x