Download pdf - INSTRUCTIONS SHOW ALL WORK in problems 2 3 4 · INSTRUCTIONS SHOW ALL WORK in problems 2, 3, ... Show your work. You may NOT use a table of values, ... g (x) = 2 x +b x-5 if x < 0

Math 1151 NAME : ___ ___ ___ ___ ___ ___ ___ ___MIDTERM 1September 15, 2015 OSU Name.# : ___ ___ ___ ___ ___ ___ __Form APage 1 of 8 Lecturer::___ ___ ___ ___ ___ ___ ___

Recitation Instructor : ___ ___ ___ ___

Recitation Time : ___ ___ ___ ___ ___ __

INSTRUCTIONS

● SHOW ALL WORK in problems 2, 3, and 4 .Incorrect answers with work shown may receive partial credit,but unsubstantiated correct answers may receive NO credit.

You don' t have to show work in problems 1 and 5.

● Give EXACT answers unless asked to do otherwise.

● Calculators are NOT permitted!PDA' s , laptops, and cell phones are prohibited.Do not have these devices out!

● The exam duration is 55 minutes.

● The exam consists of 5 problems starting on page 2 and ending on page 8.Make sure your exam is not missing any pages before you start.

PROBLEMNUMBER

SCORE

1 (20)2 (20)3 (21)4 (21)5 (18)

TOTAL (100)

MIDTERM 1Form A, Page 2

1. (20 pts)The graph of a function f is given in the figure below.

Use the graph of f to answer the questions below.

y = f ( x )

-2 -1 1 2 3 4 5 6x

-6

-4

-2

2

4

6

y

(I) (1 pt) Find the domain of f.

Domain of f = -∞, 3) ⋃ (3, +∞)

(II) (1 pt) Find the range of f.

Range of f = (-∞, +∞)

2 SOLUTIONS-MIDTERM 1- AU 15.nb

(III) Find the following values.

Note : Possible answers include + ∞ , -∞, or "does not exist".

(a) (1 pt) limx→3

f (x) = 0

(b) (1 pt) f (3) = DNE (does not exist)

(c) (1 pt) f (5) = 2

SOLUTIONS-MIDTERM 1- AU 15.nb 3


1. ( CONTINUED)

1. ( CONTINUED)

(d) (1 pt) limx→0-

f (x) = 0

(e) (1 pt) limx→0+

f (x) = -∞

(f) (1 pt) limx→0

f (x) = DNE

(g) (1 pt) limx→-∞

f (x) = +∞

(h) (1 pt) limx→+∞

f (x) = -6

(IV) (2 pts) Find all vertical asymptotes.

x = 0

(V) (2 pts) Find all horizontal asymptotes.

y = -6

(VI) (6 pts) Determine the intervals of continuity for f .

(-∞, 0], (0, 3), (3, 5], (5, +∞)



2. (20 pts) SHOW YOUR WORK!0-4 -3 -2 -1 0 1 2 3 4

s

The position, s (t), of an object moving along ahorizontal line (see the figure above) is given by

s (t) = t2 - 2.

(I) Mark the position of the object at the time t = 1 on the line above.

s (1) = 12 - 2 = -1

(II) Find the average velocity, vAV , of the objectduring the time interval [1, 3].

vAV =s (3) - s (1)

3 - 1=32 - 2 -12 - 2

3 - 1=82

= 4

(III) Compute the average velocity, vAV (t), of the objectduring the time interval

(a) [1, t], for t > 1;

vAV (t) =s (t) - s (1)

t - 1= t2 - 2 - (-1)

t - 1= t2 - 1

t - 1=

= (t - 1) (t + 1)t - 1

= t + 1 , t > 1

(b) [t, 1], for 0 < t < 1.

vAV (t) =s (1) - s (t)

1 - t=

-1 - t2 - 2

1 - t= 1 - t2

1 - t=

= (1 - t) (1 + t)1 - t

= 1 + t , t < 1

OR


OR

vAV (t) =s (1) - s (t)

1 - t= s (t) - s (1)

t - 1= t2 - 2 - (-1)

t - 1=

= t2 - 1t - 1

= (t - 1) (t + 1)t - 1

= t + 1, t < 1



2. (CONTINUED)

(IV) Find the instantaneous velocity, vinst, of the object at t = 1.Justify your answer.

vinst = limt→1

vAV (t) =limt→1

(t + 1) = 1 + 1 = 2

(V) Let s (t) = t2 - 2.The graph of of the function s is given in the figure below.

s = s ( t )

1 2 3t

-2

-1

1

2

3

4

5

6

7

s

(a) Assume P is a point on the graph of s. Fill in the blank.

P = (1, s (1)) = (1, -1)

(b) Plot the point P and draw the tangent line at this pointin the figure above.

(c) Find the slope, mtan, of the tangent line in part (b). Explain.

mtan = vinst = 2



3. (21 pts) Evaluate the limit or say that the limit does not exist .If a limit does not exist, explain why. Show your work.

You may NOT use a table of values, a graph,or L' Hospitals' s Rule to justify your answer.

Note : Possible answers include + ∞ or - ∞.

(a) (7 pts)

limx→3

x2 - 2 x - 3

x + 1 - 2=

= limx→3

x2 - 2 x - 3

x + 1 - 2

x + 1 + 2

x + 1 + 2=

= limx→3

(x - 3) (x + 1) x + 1 + 2

x + 1 - 4=

= limx→3

(x - 3) (x + 1) x + 1 + 2

x - 3=

= limx→3

(x + 1) x + 1 +2 = (3 + 1) 3 + 1 +2 =

=4 4 +2 =16


(b) (7 pts) limx→0+

x sin (ln x)

NOTE:We cannot apply the Product Law,since lim

x→0+sin (ln x) DOES NOT EXIST.

On the other hand,-1 ≤ sin (ln x) ≤ 1

and , therefore,

-x ≤ x sin (ln x) ≤ x , (since x > 0).

Since limx→0+

x = limx→0+

-x) = 0,

the inequality above implies that

limx→0+

x sin (ln x) = 0, by the Squeeze Theorem .

EXPLANATION:Since lim

x→0+ln x=-∞,

sin (ln x) oscillates between 1 and - 1 as x → 0+,and, therefore

limx→0+

sin (ln x) DOES NOT EXIST.

(c) (7 pts) limx→2-

(ln x)→ln 2>0

(x - 2)→0-= -∞

`or

Since ln x approaches ln 2 > 0,and the denominator (x - 2) approaches 0,and is negative as x → 2-, it follows that

limx→2-

ln xx - 2

= -∞



4. (21 pts) (I) (6 pts) Determine the value of a constant b for whichthe function g is continuous at 0.Show your work. Justify your answer!

g (x) =

2 x +bx-5 if x < 0 ,

x +16x2-16

if x ≥ 0, x ≠ 4.

We have to show that limx→0

g (x) = g (0).

g (0) =0 + 16

0 - 16= -1;

limx→0-

g (x) =limx→0-

2 x + b

x - 5=0 + b

0 - 5=

-b

5

limx→0+

g (x) =limx→0+

x + 16

x2 - 16=0 + 16

0 - 16= -1

Since limx→0

g (x) has to exist in order for g to be continuous at x = 0,

it has to be true that

limx→0-

g (x) = limx→0+

g (x) = limx→0

g (x).

Therefore,

-b5

= -1, and , so b = 5.

So, if b = 5 , limx→0

g (x) = g (0),

and therefore, g is continuous at x = 0.


In the questions below the function g is given in part (I).

(II) (4 pts) Evaluate the limit. Show your work!

(a) limx→-∞

g (x) = limx→-∞

2 x + b

x - 5= lim

x→-∞

2 + bx

1 - 5x

=limx→-∞ 2 + b

x

limx→-∞ 1 - 5x

= 2;

(b) limx→+∞

g (x) = limx→+∞

x + 16

x2 - 16= lim

x→+∞

1 + 16x→1

x - 16x→∞

= 0.

OR

(a) limx→-∞

g (x) = limx→-∞

2 x + b

x - 5= 2;

(b) limx→+∞

g (x) = limx→+∞

x + 16

x2 - 16= 0.

(III) (4 pts) Determine whether g has any horizontal asymptotes. If so,write the equation (or equations) of all horizontal asymptotes.

y = 2, by part (II) (a)

y = 0, by part (II) (b)


(IV) (7 pts) Determine whether g has any vertical asymptotes. If so,write the equation (or equations) of all vertical asymptotes.Show your work. Justify your answer!

There is no vertical asymptote at x = 5, because limx→5

g (x) =5 + 16

52 -16;

there is no vertical asymptote at x = -4, because limx→-4

g (x) =-8 + b

-4 - 5.

But,

limx→4+

g (x) =limx→4+

x + 16

x2 - 16=limx→4+

(x + 16)→20

(x - 4)→0+ (x + 4)→8=

(or the numerator goes to 20, and the denominatoris positive and goes to 0, as x → 4+)

= +∞

So, the function g has only one vertical asymptote,x = 4.


5. (18 pts) Circle True if the statement is ALWAYS true;circle False otherwise. No explanations are required.

(a)

Let f be a one - to - one function and f-1 its inverse.

If the point (2, 5) lies on the graph of f,then the point (5, 2) lies on the graph of f-1 .

True False

f (2) = 5 ⟺ f-1 (5) = 2

(b) sin-1 (π) = 0 True False

Since π > 1 , π is not in the domain of sin-1;orsin-1 (π) ≠ 0, since sin (0) = 0 ≠ π


(c)

If f and g are two functions defined on (-1, 1), and if

limx→0

g (x) = 0, then it must be true that

limx→0

[ f (x) g (x)] = 0. True False

Let f (x) = 1/x, if x ≠ 0, and f (0) = 4 and g (x) = x.Then lim

x→0g (x) = lim

x→0(x) = 0, but

limx→0

[ f (x) g (x)] = limx→0

1x x = 1≠ 0.

(d)

If f is continuous on (-1, 1),and if f (0) = 10 and lim

x→0g (x) = 2,

then limx→0

f (x)

g (x)= 5. True False

Because limx→0

f (x)

g (x)=

limx→0 f (x)

limx→0 g (x)=f (0)

2=10

2= 5

(e)

If f is continuous on [1, 3], and if

f (1) = 0 and f (3) = 4, then the equation f (x) = π

has a solution in (1, 3).

True False

Since f continuous on [1, 3], f (1) = 0 < π < 4 = f (3),by the Intermidiate Value Theorem there exists ac in (1, 3) such that f (c) = π .

(f)

Let f be a positive function with verticalasymptote x = 5. Then

limx→5

f (x) = +∞.

True False


Let f (x) = 1, if x < 5, and f (x) =1

5 - x, if x > 5.

Then limx→5+ f (x) = +∞, and therefore, x = 5 is a vertical asymptote.

So, f is a positive function with vertical asymptote x = 5 butlimx→5

f (x) ≠ +∞, since limx→5+ f (x) = +∞ ≠ lim

x→5- f (x) = 1,

which means that limx→5

f (x) does not exist.
