Math 1151 Midterm 1 Name:
Solutions
Form A OSU name.#:
January 30, 2024 Lecturer:
Duration: 45 Minutes Recitation Instructor:
Page 1 of 5 Recitation Time:
Instructions:
These questions are here for you to demonstrate your understanding of the course material.
To do that, show all relevant supporting work to receive full credit on all Problems unless explicitly stated not to.
Incorrect answers with substantially correct work may receive partial credit.
Answers without justification will receive no credit.
Using a table of values or L’Hôpitals Rule are not justification for computing limits.
Using graphs not given in this assignment are not justification.
Give exact answers unless instructed to do otherwise.
Intervals should be expressed simplified in interval notation.
If a requested value does not exist, write “DNE”. Write that limits “DNE” only if they do not exist and are neither 1 nor
1.
You are expected to evaluate any famous functions at standard values.
This is a closed book/closed note exam.
Calculators, tablets, and other devices are not permitted.
You are responsible for any instructions in Carmen pertaining to this exam.
Your solutions are to be uploaded to Gradescope as a single pdf file.
Failure to follow the Gradescope formatting guildelines will result in substantial penalties.
Check that the file you upload is the file you intend to turn in. No credit will be given if the wrong file is uploaded or the file
is posted to the wrong assignment.
SIGN YOUR NAME:
The following is the statement of the Intermediate Value Theorem:
Theorem. If f is a continuous function on the closed interval [a, b ], and d is any value between f (a) and f (b ), then
there is a number c in (a, b ) such that f (c ) = d .
Question: 1 2 3 4 Total
Points: 15 12 19 14 60
Score:
Math 1151 Midterm 1 Form A
0
5858
1) a) (8 points) Find the form of the following limit and evaluate it.
p p
3x 1 8
lim
x !3 x (x 2 x 6)
x3 2 0 F ORM :
VALUE : Is
adf.IT
tIsxIKEioy sEIro
IF
8 1
Cos49
b) (7 points) Find the form of the following limit and evaluate it.
2x + cos (⇡ x )
8
lim
x !4 x 2 16
o
F ORM : f
one sided limit with form Check
sign
N
2x Cos Ix is close to 9 so positive VALUE :
since 44 X 4 is negative is closeto 4 so 4 is positive
so X 16 x 4 1 4 is negative
2
is is negative
Eg
Math 1151 Midterm 1 Form A
2) A function f has graph given in the figure below.
4
3
2
1
4 3 2 1 1 2 3 4 5
1
2
3
4
a) (2 points) Find the Intervals of Continuity of f. (You do not need to show work for this question.)
Intervals of Continuity: 4 1 10 2 2 3 3 5
b) (2 points) Evaluate the following limit. (You do not need to show work for this question.)
x !2
lim f (x ) : 1
c) (2 points) Evaluate the following limit. (You do not need to show work for this question.)
p 14
16 x 2
lim+ :
x !0 f (x )
d) (6 points) Suppose g is any function satisfying f (x ) g (x ) x 2 + 6x + 11 for all x in the interval ( 4, 1). Find
the value of lim g (x ). EXPLAIN.
x! 3
2 by squeeze them
ftp.sfx 2 fromgraph
58 2
ftp.sfa 1 6 11 2
46 11 g
since
im
VALUE : 2
Math 1151 Midterm 1 Form A
3) Let g be a piecewise-defined function given by:
8 2
> 3 + 4x 5x ,
>
> if x 1
>
< 4 x
g (x ) =
>
> p
>
> 2x 2 + 7 4
: , if x > 1
x +1
a) (3 points) Evaluate lim g (x )
x ! 1+
into.EE lim g (x ) =
x ! 1+
no
b) (2 points) Evaluate lim g (x )
x !4
lim g (x ) =
5
x !4
Continuous
c) (2 points) Evaluate
atx
lim g (x )
4
x! 1
in lim g (x ) =
x! 1
a
d) (4 points) Evaluate lim g (x )
x !1
t.im
at E
1 II
lim g (x ) =
x !1
e) (4 points) Find the vertical asymptotes of g . EXPLAIN by citing appropriate limits. How do you know that there are
no others?
VA at 1 because N from a
fim gx
is continuous else so no other possibilities
g everywhere
f) (4 points) Find the horizontal asymptotes of g . EXPLAIN by citing appropriate limits. How do you know that there
are no others?
HA at y f because F No others because
fiz.ge
0
11,941
Math 1151 Midterm 1 Form A
4) In this problem, we will continue to use the piecewise-defined function from Problem 3, given by:
8
>
> 3 + 4x 5x 2
>
> , if x 1
< 4 x
g (x ) =
>
> p
>
> 2x 2 + 7 4
: , if x > 1
x +1
a) (3 points) Is g (x ) continuous at x = 1? EXPLAIN. You may cite your answers from Problem 3.
(Note: For full credit, you must use the definition of continuity).
No For uuityweneed gx g but from prob 3 a
does not exist
11 go
b) (3 points) Is g (x ) continuous at x = 4? EXPLAIN. You may cite your answers from Problem 3.
(Note: For full credit, you must use the definition of continuity).
914 21651 4 5 From prob 3b
tiny 9
Yes is continuous at x 4 because fins 90 94
g
c) The function r is continuous on the interval ( 2, 8). Some values of the function r are given in the table below.
x 1 0 1 2 3 4 5 6 7
r(x) 6 2 1 0 2 7 5 3 4
6
Show that the equation r (x ) = has a solution on the interval (1, 4) using the Intermediate Value Theorem (IVT) by
7
completing the steps below.
(The statement of the Intermediate Value Theorem is on the front page of this exam.)
i) (1 point) IVT requires the choice of a number, d . What is your choice for d ?
d=
ii) (2 points) Fill in the blanks with 1 and 4 appropriately:
l
rat
1 4
r 147 7
so
Ira r( )<d <r( )
iii) (5 points) Based on your answers above, EXPLAIN how to use IVT to determine that the equation has a solution
in the interval (1, 4).
r is continuous on 1,4since it is inside 1 2 8 with
ra 2514 By IVT there is C in 114 with r c
C is a solution of ra
