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  • Math1013 Calculus I

    Midterm Exam Solution (White-Green Version)

    Part I: Multiple Choice Questions

    MC Answers: White-Green Version

    Question 1 2 3 4 5 6 7 8 9 10 Total

    Answer c d c e a b b d e e

    MC Answers: Orange-Yellow Version

    Question 1 2 3 4 5 6 7 8 9 10 Total

    Answer d c d c d e b a c b

    Solution (White-Green Version.)

    1. Which of the following intervals is the domain of the function defined by the formula f(x) =x2 1

    x3lnx+ 1

    ?

    (a) (0,) (b) (e,) (c) (1e ,) (d) (1e ,) (e) (0, e)

    Solution

    The answer is (c), since lnx+ 1 > 0 x > e1.

    2. Across its whole domain, which of the functions given by the following graphs has an inversefunction?

    (a)

    x

    y

    (b)

    x

    y

    (c)

    x

    y

    (d)

    x

    y

    (e)

    x

    y

    Solution

    The answer is (d): the graph passes the horizontal line test, or the function is one-to-one on itsdomain.

  • 1

    3. Find the limit limx 1

    2

    4x2 + 2x4x2 8x+ 3

    (a) 12

    (b) 0 (c) 12

    (d) 1 (e) Does not exist

    Solution

    The answer is (c): limx 1

    2

    4x2 + 2x4x2 8x+ 3 = limx 1

    2

    2x(2x 1)(2x 1)(x 3) = limx 1

    2

    2x2x 3 =

    1

    2

    4. The temperature function of an object at t 0 hours is given by T (t) = 250t + 80t

    25 + 5t, in C. As

    t , approximately to what steady temperature will the object approach?

    (a) 0 C (b) 10 C (c) 16 C (d) 40 C (e) 50 C

    Solution

    The answer is (e): limt

    250t + 80t

    25 + 5t= lim

    t

    250 + 80 1t

    25t + 5

    = 50

    5. Using the graph of f given below, find limx1+

    f(g(x)) where g(x) = 1 2x.

    -4 4

    3

    -3

    x

    y

    y = f(x)

    (a) -2 (b) 0 (c) 1 (d) 3 (e) does not exist

    Solution

    The answer is (a): As x 1+, g(x) 1, hence f(g(x)) 2.

  • 2

    6. Find limx

    (

    x2 + x

    x+

    x

    )

    .

    (a) (b) (c) -1 (d) 1 (e) 0

    Solution

    The answer is (b): limx

    (

    x2 + x

    x+

    x

    )

    = limx

    (

    1 +1

    x+

    x

    )

    = .

    7. Find the equation of the tangent line to the graph of the function y = f(x) = 6x2 + 3x at thepoint (2, f(2)).

    (a) y = 15x+ 12 (b) y = 21x+ 24 (c) y = 21x 60

    (d) y = 12x 6 (e) y = 15x 48

    Solution

    The answer is (b): f (x) = 12x+ 3, f (2) = 21, f(2) = 18.Tangent line equation y = 21(x 2) 18 = 21x+ 24.

    8. Compute g(1) for g(x) = esinx.

    (a) e (b) (c) e (d) (e) e

    Solution

    The answer is (d): g(x) = esinx cos x, g(1) = esin cos = .

    9. If f(x) = g(x2 + 7) and g(x) = x2/3, compute the derivative f (1).

    (a) 23 (b) 12 (c) 12 (d) 13 (e) 23Solution

    The answer is (e): f (x) = g(x2 + 7) 2x, g(x) = 23x1/3, hence f (1) = g(8) 2 = 2381/3 2 =23 .

  • 3

    10. The number of gallons of water in a tank t minutes after the tank started to drain is V (t) =25(30 2t)2. How fast is the water running out at the end of 8 minutes?

    (a) 250 gal/min (b) 500 gal/min (c) 700 gal/min (d) 1000 gal/min (e) 1400 gal/min

    Solution

    The answer is (e): V (t) = 100(30 2t), V (8) = 1400 gal/min.

    Part II: Answer each of the following questions.

    11. [10 pts] The graphs of y = f(x), y = f (x), and y = f (x) are shown below.

    x

    y

    Graph A

    Graph B

    Graph C

    Graph C

    Graph A

    Graph B

    (a) Match the following functions with their graphs: [4 pts]

    The graph of y = f (x) is : Graph B

    The graph of y = f (x) is : Graph A

    (b) Give brief reason to support your answers above. [6 pts]

    Solution

    Derivative is the slope of the graph, or the rate of change of the function.

    When x > 0, the positive slope of Graph C gradually increases and then decreases to 0, andturns into negative slope afterward. Graph B matches the changing slope of Graph C; i.e.C = B.

    Similarly, Graph A matches the changing slope of Graph B; i.e., B = A.

  • 4

    12. [10 pts] A function f is defined and continuous on the interval 0 < x < 5. It is known that

    f(x) =

    2x+ 5 if 0 < x < 2

    ax2 + bx+ c if 2 < x < 5

    and the function has the following right-hand derivative and derivative at x = 2 and x = 3 respec-tively:

    f +(2) = 0 and f(3) = 1.

    (a) Find the function value f(2). Justify your answer for full credit. [3 pts]

    Solution

    Since f is continuous at x = 2,

    f(2) = limx2

    f(x) = limx2

    (2x+ 5) = 2 2 + 5 = 9

    (b) Express the right hand derivative f +(2) as the limit of an appropriate expression. [3 pts]

    Solution

    f +(2) = limh0+

    [a(2 + h)2 + b(2 + h) + c] (4a+ 2b+ c)h

    or= lim

    h0+

    [a(2 + h)2 + b(2 + h) + c] 9h

    or= lim

    h0+

    4ah+ ah2 + bh

    hor= lim

    h0+(4a + b+ ah)

    or= lim

    x2+

    (ax2 + bx+ c) (4a+ 2b+ c)x 2 = . . .

    Or f +(2) = limx2+

    (2ax+ b).

    (c) Find the coefficients a, b, and c. [4 pts]

    Solution

    f(2) = 4a+ 2b+ c = 9 (1)

    f +(2) = 4a+ b = 0 (2)

    f (3) = 6a+ b = 1 (3)

    (3) (2): 2a = 1, a = 12.

    Hence by (2), b = 4a = 2, and by (1), c = 9 4a 2b = 9 2 + 4 = 11.

  • 5

    13. [15 pts] The functions f, g are differentiable everywhere. Using the given table of function values,

    x -2 -1 0 1 2

    f(x) 1 -4 -6 1 -2

    f (x) 2 -5 1 3 5

    g(x) -1 1 3 6 10

    g(x) -5 -2 1 3 5

    compute the following derivatives:

    (a)d

    dx

    (

    x2f(x)

    g(x)

    )

    x=2[5 pts]

    Solution

    d

    dx

    (

    x2f(x)

    g(x)

    )

    x=2=

    g(x)(x2f(x)) x2f(x)g(x)[g(x)]2

    x=2

    =g(2)[2(2)f(2) + (2)2f (2)] (2)2f(2)g(2)

    [g(2)]2

    =(1)[2(2)(1) + 4(2)] 4(5)

    (1)2 = 16

    (b)d

    dxln([f(x)]2 + 2)

    x=1

    [5 pts]

    Solution

    d

    dxln([f(x)]2 + 2)

    x=1

    =1

    [f(x)]2 + 2

    d

    dx([f(x)]2 + 2)

    x=1

    =1

    [f(1)]2 + 2 2f(1)f (1) = 2 1 3

    12 + 2= 2

    (c) (f g1) (1), assuming that g1 exists and is also differentiable. [5 pts]

    Solution

    (f g1) (1) = f (g1(1))(g1)(1) = f (g1(1)) 1g(g1(1))

    = f (1) 1g(1) =

    52 =

    5

    2

    Note that g1(1) = 1 by the given table.

  • 6

    14. [15 pts] Answer the following true-false questions. Give brief reason to justify each answer.

    (a) The equationx4 2x 2

    x= |x| has at least two roots in the interval [-2,2]. [5 pts]

    True False

    Brief reason.

    Solution

    Let f(x) =x4 2x 2

    x |x|. Then

    f(2) =24 2 2 2

    2 |2| = 3 > 0, f(1) = 1

    4 2 1 21

    |1| = 4 < 0.

    Since f is continuous over the interval [1, 2], by the Intermediate Value Theorem, there is anx1 in the interval (1, 2) such that f(x1) = 0.

    Similarly, f is continuous over the interval [2,12 ], and

    f(2) = 16 + 4 22 2 = 11, f(1

    2) =

    116 + 1 2

    12 1

    2=

    11

    8> 0,

    by the Intermediate Value Theorem, there must be another root x2 of f in (2,12 ) such thatf(x2) = 0.

    Consequently, f has at least two roots in the interval [2, 2].(b) The function f(x) = |x| sin(2x) is not differentiable at x = 0. [5 pts]

    True False

    Brief reason.

    Solution

    The left-hand derivative and right-hand derivative are:

    limh0+

    f(h) f(0)h

    = limh0+

    h sin(2h)

    h= lim

    h0+

    h sin(2h)

    h= lim

    h0+sin(2h) = 0

    limh0

    f(h) f(0)h

    = limh0

    |h| sin(2h)h

    = limh0

    h sin(2h)h

    = limh0+

    ( sin(2h)) = 0

    Thus f (0) = 0 exists, and f is differentiable at x = 0. Or,

    f (0) = limh0

    f(h) f(0)h

    = limh0

    |h| sin(2h)h

    = limh0

    2|h| limh0

    sin(2h)

    2h= 0 1 = 0

    (c) If f (a) and g(a) do not exist, then the quotient functionf

    gcan not be differentiable at x = a.

    True False [5 pts]

    Brief reason.

    Solution

    For example f(x) = |x|+1 and g(x) = |x|+1 are both not differentiable at x = 0, but f(x)g(x)

    = 1

    is clearly differentiable at x = 0.

  • 7

    15. [10 pts] Consider the curve defined by the equation x4 xy2 + y3 = 8.

    (a) Find the equation of the tangent line to the curve at the point where the curve intersects they-axis. [6 pts]

    Solution

    Differentiating both sides of the equation, we have

    4x3 y2 2xy dydx

    + 3y2dy

    dx= 0

    dy

    dx=

    y2 4x33y2 2xy

    When x = 0, y = 2, and the slope of the tangent tothe curve at (0, 2) is

    dy

    dx

    (0,2)

    =22

    3 22 =1

    3

    The equation of the tangent line to the curve at (0, 2)is

    y =1

    3x+ 2

    3 3

    3

    3

    x

    y

    (b) Is it true that the curve has at least four vertical tangent lines? Given reason for you answer.[4 pts]

    Solution

    True.

    There is a vertical tangent line at (x, y) whendy

    dx , or dx

    dy= 0; in particular,

    3y2 2xy = y(3y 2x) = 0, i.e., y = 0 or 3y 2x = 0..

    Putting y = 0 into the equation of the given curve, we have x = 48. Hence there are two

    vertical tangent lines at the points ( 48, 0) and ( 4

    8, 0) respectively.

    Putting y = 23x into the equation, we have

    x4 49x3 +

    8

    27x3 = 8 x4 4

    27x3 8 = 0

    Consider the continuous function f(x) = x4 427x3 8. Note that f(3) = 81 + 4 8 > 0,f(0) = 8 < 0, and f(3) = 81 4 8 > 0. f(x) must have at least two roots in the interval(3, 3) by the Intermediate Value Theorem.Graphically speaking, the line y = 23x intersects the given curve at least twice.

    Thus there are at least four points on the curve with vertical tangent line.

    Exercise Consider the interval of increase/decrease of f(x) = x4 427x3 8 and show thatthe equation x4 427x3 8 = 0 has exactly two real roots.

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