INSTRUCTIONS SHOW ALL WORK in problems 2 3 4 ?· INSTRUCTIONS SHOW ALL WORK in problems 2, 3, ... Show…

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  • 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) limx3

    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

  • MIDTERM 1Form A, Page 3

    1. ( CONTINUED)

    1. ( CONTINUED)

    (d) (1 pt) limx0-

    f (x) = 0

    (e) (1 pt) limx0+

    f (x) = -

    (f) (1 pt) limx0

    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, +)

    4 SOLUTIONS-MIDTERM 1- AU 15.nb

  • MIDTERM 1Form A, Page 4

    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= t

    2 - 2 - (-1)t - 1

    = t2 - 1t - 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 - t

    2

    1 - t=

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

    = 1 + t , t < 1

    OR

    SOLUTIONS-MIDTERM 1- AU 15.nb 5

  • OR

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

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

    t - 1= t

    2 - 2 - (-1)t - 1

    =

    = t2 - 1t - 1

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

    = t + 1, t < 1

    6 SOLUTIONS-MIDTERM 1- AU 15.nb

  • MIDTERM 1Form A, Page 5

    2. (CONTINUED)

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

    vinst = limt1

    vAV (t) =limt1

    (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

    SOLUTIONS-MIDTERM 1- AU 15.nb 7

  • MIDTERM 1Form A, Page 6

    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)

    limx3

    x2 - 2 x - 3

    x + 1 - 2=

    = limx3

    x2 - 2 x - 3

    x + 1 - 2

    x + 1 + 2

    x + 1 + 2=

    = limx3

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

    x + 1 - 4=

    = limx3

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

    x - 3=

    = limx3

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

    =4 4 +2 =16

    8 SOLUTIONS-MIDTERM 1- AU 15.nb

  • (b) (7 pts) limx0+

    x sin (ln x)

    NOTE:We cannot apply the Product Law,since lim

    x0+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 limx0+

    x = limx0+

    -x) = 0,the inequality above implies that

    limx0+

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

    EXPLANATION:Since lim

    x0+ln x=-,

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

    limx0+

    sin (ln x) DOES NOT EXIST.

    (c) (7 pts) limx2-

    (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

    limx2-

    ln xx - 2

    = -

    SOLUTIONS-MIDTERM 1- AU 15.nb 9

  • MIDTERM 1Form A, Page 7

    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 limx0

    g (x) = g (0).

    g (0) =0 + 16

    0 - 16= -1;

    limx0-

    g (x) =limx0-

    2 x + b

    x - 5=0 + b

    0 - 5=

    -b

    5

    limx0+

    g (x) =limx0+

    x + 16

    x2 - 16=0 + 16

    0 - 16= -1

    Since limx0

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

    it has to be true that

    limx0-

    g (x) = limx0+

    g (x) = limx0

    g (x).

    Therefore,

    -b5

    = -1, and , so b = 5.

    So, if b = 5 , limx0

    g (x) = g (0),

    and therefore, g is continuous at x = 0.

    10 SOLUTIONS-MIDTERM 1- AU 15.nb

  • 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 +

    bx

    limx- 1 -5x

    = 2;

    (b) limx+

    g (x) = limx+

    x + 16

    x2 - 16= lim

    x+

    1 + 16x1

    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)

    SOLUTIONS-MIDTERM 1- AU 15.nb 11

  • (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 limx5

    g (x) =5 + 16

    52 -16;

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

    g (x) =-8 + b

    -4 - 5.

    But,

    limx4+

    g (x) =limx4+

    x + 16

    x2 - 16=limx4+

    (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.

    MIDTERM 1Form A, Page 8

    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 () = 0 True False

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

    12 SOLUTIONS-MIDTERM 1- AU 15.nb

  • (c)

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

    limx0

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

    limx0

    [ 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

    x0g (x) = lim

    x0(x) = 0, but

    limx0

    [ f (x) g (x)] = limx0

    1x x = 1 0.

    (d)

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

    x0g (x) = 2,

    then limx0

    f (x)

    g (x)= 5. True False

    Because limx0

    f (x)

    g (x)=

    limx0 f (x)

    limx0 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) =

    has a solution in (1, 3).

    True False

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

    (f)

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

    limx5

    f (x) = +.

    True False

    SOLUTIONS-MIDTERM 1- AU 15.nb 13

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

    5 - x, if x > 5.

    Then limx5+

    f (x) = +, and therefore, x = 5 is a vertical asymptote.

    So, f is a positive function with vertical asymptote x = 5 butlimx5

    f (x) +, since limx5+

    f (x) = + limx5-

    f (x) = 1,

    which means that limx5

    f (x) does not exist.

    14 SOLUTIONS-MIDTERM 1- AU 15.nb

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