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  • 8/19/2019 Problem Set 9 Solutions.pdf

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    MH1100/MTH112: Calculus I.

    Problem set #11.Solutions

    Problem 1: (#6.1.9 to #6.1.14 from [Stewart].)

    Which of the following functions are 1-1. Briefly justify.

    (i)   f (x) = x2 − 2x.(ii)   f (x) = 10 − 3x.

    (iii)   g(x) = 1/x.

    (iv)   g(x) = |x|.

    (v)   h(x) = 1 + cos x.

    (vi)   h(x) = 1 + cos x, for 0 ≤ x ≤ π.Solution

    We’ll go through these one by one.

    (i) The graph of this function is a shifted parabola, so it is clearly not1-1. Another way to see this is to observe that  f (0) = 0 = f (2).

    (ii) The graph of this function is a non-horizontal straight line, so it willclearly pass the horizontal line test. Another way to see that f (x) is1-1 is to observe that  f ′(x) = −3, so this is a decreasing function.

    (iii) The graph of  g(x) = 1/x is a standard hyperbola, which is clearly 1-1.Another way to see that this function is 1-1 is to deduce that if   x1and  x2  are two points where  g(x1) = g(x2), then 1/x1  = 1/x2, whichimplies that x1 =  x2.

    (iv) The function g(x) = |x| is not 1-1, because  g(−x) = g(x) for all x ∈ R.(v) The function   h(x) = 1 + cos x   is not 1-1 because for every   x ∈   R,

    h(x + 2π) = h(x).

    (vi) This   h(x) is 1-1. A transparent way of seeing this is to note thath′(x) =  − sin x, which is   <   0 when 0   < x <   π. Thus   h(x) is adecreasing function on this interval.

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    Problem 2: (Based on #6.1.19 from [Stewart].)

    If  h(x) = x +√ 

    x, then what is  h−1(6)? What about h

    h−1(π)

    ?

    Solution

    To begin, note that both  x  and √ x are increasing functions, so this functionh(x) is certainly 1-1, and so the inverse function  h−1(x) is logically defined.It also follows from the IVT that 6 lies in the range of  h(x).

    To determine h−1(6) we just have to answer the question: Which numberx̃ gives  h(x̃) = 6?

    Substituting in guesses, we can quickly find that  h(4) = 6. A more sys-tematic approach would compute as follows. (Below, note that the domainof  h  is [0, ∞), so we can assume that x̃ ≥ 0.)

    h(x̃) = 6

    ⇒  x̃ +

    √ x̃ = 6

    ⇒√ ̃x2 + 2 ∗ 1

    2√ ̃x

     + 1

    4 − 25

    4  = 0

    ⇒√ 

    x̃ + 1

    2

    2=

     25

    4

    ⇒√ 

    x̃ = 2

    ⇒   x̃ = 4.

    The second question is what is  h

    h−1(π)

    . Without a single calculationwe can say that the answer is   π. This is because this number is the answerto the question: “What is the value in  h(x) of the number whose value in

    h(x) is   π?”

    Problem 3:

    Consider f (x) = x + cos x. Explain why  f (x) is 1-1. What is  f −1(1)?

    Solution:

    The function is 1-1 because   f ′(x) = 1 − sin x   is always greater than zero(except at isolated points) so  f (x) is an increasing function.

    The value   f −1

    (1) is just the number x̃   which answers the question:Which number x̃  gives  f (x̃) = 1? If we just think about this function (per-haps substituting in a few test values) we’ll quickly realize that  f (0) = 1.Thus:   f −1(1) = 0.

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    Problem 4: (#6.1.22 from [Stewart].)

    In the theory of relativity, the mass of a particle with speed  v   is

    m =  f (v) =  m0 

    1 −   v2c2

    .

    where   m0   is the “rest mass” of the particle and   c   is the speed of lightin a vacuum. What is the inverse function   f −1(m) of   f   and what is itsinterpretation?

    Solution:

    First of all note that the domain of the function is [0, c]. (This is not surpris-ing! Nothing travels faster than light, right?) Then note that the functionis increasing. (This is easy to check formally. Or we can intuitively reason:

    “As   v   increases from 0 to   c,   v2

    c2  gets closer to 1, so the bottom line gets

    smaller, so the whole function increases.”) So it will be 1-1 and there willbe no problems discussing an inverse function.

    To actually build the inverse function: The value of  f −1(m) will be thev  which solves the equation:

    f (v) = m

    ⇒   m =  m0 

    1 −   v2c2

    ⇒   1 −m0

    m

    2=

      v2

    c2

    ⇒   v =  c 

    1 −m0

    m

    2.

    Thus:

    f −1(m) = c

     1 −

    m0m

    2.

    The “interpretation” of this function, is that given a body with rest mass

    m0, the speed it has to be going at to acquire a mass  m  is  c 

    1 − m0m 2.

    3

    !"#$%&

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    Problem 5: (#6.1.27 from [Stewart].)

    Find a formula for the inverse function  f −1(x) to the function

    f (x) = 1 − √ x1 +

    √ x

    .

    Solution:

    If we write this function

    f (x) =  1

    1 +√ 

    x −   1

    1√ x

     + 1

    we see that  f (x) is the sum of two decreasing functions, so it is decreasing.Thus f (x) is 1-1, and has a logically defined inverse. Furthermore, note thatlimx→∞ f (x) = −1. Thus we deduce that the range of  f (x) is [1, −1), whichwill be the domain of the inverse function.

    To determine the rule for the inverse, let  x ∈ [1, −1). Then x̃ =  f −1(x)satisfies:

    x =  1 −

    √ x̃

    1 +√ ̃

    x

    ⇒   x + x√ 

    x̃ = 1 −√ 

    ⇒   (x + 1) √ ̃x = 1 − x⇒

    √ x̃ =

      1 − x1 + x

    ⇒   x̃ =

    1 − x1 + x

    2.

    Thus:

    f −1(x) =

    1 − x1 + x

    2.

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    Problem 6: (#6.1.42 from [Stewart].)

    Why is

    f (x) = 

    x3 + x2 + x + 1

    1-1? Determine (f −1)′(2).

    Solution

    Why is f (x) a 1-1 function? First, note that x3+x2+x+1 is an increasingfunction because its derivative is 3x2 + 2x + 1 = 3(x + 1/3)2 + 2/3, which isalways  >  0. Thus  f (x) is the composition of two increasing functions, so itis increasing as well, and so it is 1-1.

    To use the formula for the derivative of an inverse, we’ll need to knowf −1(2), so must find an x̃  such that

    4 = x̃3 + x̃2 + x̃ + 1 ⇒  x̃3 + x̃2 + x̃ − 3 = 0.

    A bit of trial and error leads to the solution x̃ = 1, so:

    f −1(2) = 1.

    Now we computed the requested derivative using the standard formulafor the derivative of an inverse:

    (f −1)′(2) =   1f ′(f −1(2))

    =  1

    f ′(1)

    =  1

      12√ x3+x2+x+1

     ∗ (3x2 + 2x + 1)

    x=1

    =  1

    12√ 4 ∗ (3 ∗ 1 + 2 + 1)

    =  11

    4 ∗6

    =  2

    3.

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    Problem 7: (#6.1.46 from [Stewart].)

    Suppose that  g(x) is the inverse function of a diff erentiable function  f (x),and let  G(x) =   1g(x) . If  f (3) = 2 and  f 

    ′(3) =   19 , find  G′(2).

    Solution

    We’ll compute using the standard formula for the derivative of an inverse,and other standard formulae. The computation will need g(2). Because g(x)is the inverse of  f (x),  g(2) is the number x̃   such that  f  (x̃) = 2. From thegiven datum that  f (3) = 2, we deduce  g (2) = 3. Now:

    G′(2) =   d

    dx   1

    g(x)

    x=2

    =

    −   1

    (g(x))2 ∗   d

    dx [g(x)]

    x=2

    =

    −   1

    (g(x))2 ∗   1

    f ′(g(x))

    x=2

    =   −   1(g(2))2

     ∗   1f ′(g(2))

    =

      −

     1

    32

     ∗

      1

    f ′(3)

    =   −19 ∗   1

    19

    =   −1.

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    Problem 8: (#6.1.49 from [Stewart].)

    1. Find an expression for the inverse of a function g(x) = f (x + c), wheref (x) is a one-to-one function. Interpret your result geometrically interms of the graph of a function and it’s inverse.

    2. Find an expression for the inverse of  h(x) = f (cx), where  c̸ = 0.

    Solution to (i)

    To begin, note that if  f (x) is 1-1, then  g(x) =  f (x + c) is 1-1 as well.(What would a proof look like? Assume that  x1   and   x2  are two numberswhere g(x1) = g(x2). This implies that  f (x1 + c) = f (x2 + c). Because f (x)is 1-1 this implies that  x1 + c =  x2 + c, which implies that  x1 =  x2.)

    To determine a formula for   g−1(x): note that   g−1(x) is the unique x̃such that  g (x̃) = x. Thus:

    g(x̃) = x

    ⇒   f (x̃ + c) = x⇒   x̃ + c =  f −1(x)⇒   x̃ =  f −1(x) − c.

    So:g−1(x) = f −1(x) − c.

    This is exactly what our geometric understanding of inverse functionspredicts: The graph of  g(x) is obtained from the graph of  f (x) by shifting itc units to the left. So when we reflect this arrangement over the line  y  =  x,the graph of  g(x) is reflected to the graph of  g−1(x), the graph of  f (x) isreflected to the graph of  f −1(x), and we expect that the graph of  g−1(x) isobtained from the graph of  f −1(x) shifting it  down  c  units.

    Solution to (ii).

    We argue in the same way:

    h(x̃) = x

    ⇒   f (cx̃) = x⇒   cx̃ =  f −1(x)⇒   x̃ =  1

    cf −1(x).

    Thus:   h−1(x) =   1cf −1(x).

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    Problem 9: (#6.2.17 and #6.2.18 from [Stewart].)

    Solution to (i).

    The two points we are given determine two equations for the constants  C and  a:

    1. 6 = C a1.

    2. 24 = C a3.

    If we divide the second equation by the first we deduce:   a2 = 4. Thusa = 2. (We do not discuss exponential functions with negative base.)

    Then substituting  a  = 2 into the first equation gives  C  = 3.Thus:   f (x) = 3 ∗ 2x.

    Solution to (ii).

    In this case the two equations we get are:

    1. 2 = C a0.

    2. 2/9 = Ca2.

    These can be solved to deduce:   f (x) = 2 13

    x, which can be rewritten

    f (x) = 2 ∗ 3−x

    .

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    Problem 10⋆: (#6.1.50 from [Stewart].)

    Let  f (x) be a 1-1, twice diff erentiable function with inverse function  g(x).Show that:

    g′′(x) = −   f ′′(g(x))

    [f ′(g(x))]3.

    Solution

    According to the standard formula, the derivative of  g(x) is:

    g′(x) =  1

    f ′(g(x)).

    If we diff erentiate this formula we get:

    g′′(x) =   −   1(f ′(g(x)))2

     ∗ f ′′(g(x)) ∗ g′(x)

    =   −   1(f ′(g(x)))2

     ∗ f ′′(g(x)) ∗   1f ′(g(x))

    =   −   1(f ′(g(x)))3

     ∗ f ′′(g(x))

    which is the formula we had to show.

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    Problem 11:

    Use geometric transformations to sketch the graph of the function

    f (x) = 2 + 5(1 − e−x).

    Solution

    We seek a sequence of elementary transformations which will take the stan-dard exponential function f (x) = ex into this function.

    The sequence of transformations, and the corresponding geometric trans-formations, are:

    !5 0 5!1

    0

    1

    2

    3

    4

    5

    6

    7

    8

    9

    10

    f (x) = ex

    f (x)→f (−x)−→

    !5 0 5!1

    0

    1

    2

    3

    4

    5

    6

    7

    8

    9

    10

    f (x) = e−x

    f (x)→−f (x)−→

    !5 0 5!10

    !9

    !8

    !7

    !6

    !5

    !4

    !3

    !2

    !1

    0

    1

    f (x) = −e−x

    10

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    f (x)→f (x)+1−→

    !5 0 5!10

    !9

    !8

    !7

    !6

    !5

    !4

    !3

    !2

    !1

    0

    1

    2

    3

    f (x) = 1 − e−x

    f (x)→5f (x)−→

    !5 0 5!20

    !15

    !10

    !5

    0

    5

    10

    f (x) = 5(1

    −e−x)

    f (x)→f (x)+2−→

    !5 0 5!20

    !18

    !16

    !14

    !12

    !10

    !8

    !6

    !4

    !2

    0

    2

    4

    6

    8

    10

    f (x) = 2 + 5(1 − e−x

    )

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    Problem 12: (#6.2.25 from [Stewart].)

    Determine the limit

    limx→∞

    e3x − e−3xe3x + e−3x

    .

    Solution

    This limit can be reduced to the standard limit limx→∞ e−x = 0 by thededuction:

    limx→∞

    e3x − e−3xe3x + e−3x

      = limx→∞

    e3x

    e3x ∗  1 − e

    −6x

    1 + e−6x

    =   1 − limx→∞ e−6x1 + limx→∞ e−6x

    =  1

    1

    = 1.

    Problem 13: (#6.2.29 from [Stewart].)

    Use a version of the squeeze theorem to determine the limit

    limx→∞

    e−2x cos x

    .

    Solution

    We have−1 ≤ cos x ≤ 1.

    Because   e−2x >  0 for all  x, we can deduce from the above inequalitiesthat

    −e−2x ≤ e−2x cos x ≤ e−2x.Now because

    limx→∞

    e−2x = limx→∞

    (−e−2x) = 0we deduce form the natural version of the squeeze theorem as  x → ∞  thatthe given limit limx→∞

    e−2x cos x

     exists, and equals 0.

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    Problem 14: (#6.2.67 from [Stewart].)

    Find the absolute maximum and absolute minimum of 

    f (x) = xe−x2/8

    on the interval [−1, 4].Solution

    We are asked for the extreme values of a continuous function f (x) = xe−x2/8

    on the closed interval [−1, 4]. We can obtain them from an application of the closed interval method.

    The first thing we need to do is find the critical numbers of  f (x). Thefunction is diff erentiable at every point, so the critical numbers are the setof numbers at which the derivative is zero:   f ′(c) = 0. The derivative is:

    f ′(x) = e−x2/8 −  2x

    8  ∗ x ∗ e−x2/8 =

    1 −  x

    2

    4

    e−x

    2/8.

    The only point inside the given interval [−1, 4] at which this derivative iszero is at  x  = 2.

    To finish we need to compare the values at the critical number   x   = 2with the values at the boundaries of the interval, at  x =

     −1 and at  x = 4.

    The three values are:

    •   f (−1) = −e−1/8 ≈ −0.88•   f (2) = 2e−1/2 ≈ 1.21•   f (4) = 4e−2 ≈ 0.54We conclude that on the interval [−1, 4], the given function   f (x) =

    xe−x2/8 achieves an absolute maximum of  ≈ 1.21 at  x  = 2, and an absolute

    minimum of  ≈ −0.88 at x  = −1.

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    Problem 15: (#6.2.99 from [Stewart].)

    Given  f (x) = 3 + x + ex, calculate

    f −1′

    (4).

    Solution

    To begin, notice that this function is the sum of increasing functions, so itis increasing, so it is 1-1 and has a well-defined inverse. The calculation of the derivative of the inverse will be a standard application of the formulafor the derivative of an inverse function. The formula will need  f −1(4). If we notice that  f (0) = 3 + 0 + e0 = 4 we conclude that f −1(4) = 0.

    Then:

    d

    dxf −1(x)

    x=4

    =  1

    f ′(f −1(4))

    =  1

    (1 + ex)|x=0

    =  1

    1 + 1

    =  1

    2.

    Problem 16: (#6.2.100 from [Stewart].)

    Evaluate the limitlimx→π

    esinx − 1x − π .

    Solution

    The trick here is to recognize that this is a limit defining a certain derivative.Set

    g(x) = esinx.

    Then:

    limx→π

    esinx − 1x − π = limx→π

    esinx − esin0x − π

    =   g′(π

    )=

    cos x esinx

    x=π

    = (−1)esinπ=   −1.

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    Problem 17: (#6.3.35 and #6.3.36 from [Stewart].)

    Solve the following equations:

    (a)   e2x − ex − 6 = 0.(b) ln(2x + 1) = 2 − ln x.

    Solution to (a)

    The equation is equivalent to

    (ex

    −3)(ex + 2) = 0.

    An  x  solves this equation if and only if (ex − 3) = 0 or (ex + 2) = 0. Theonly solution is  x  = ln3.

    Solution to (b)

    Because ln (2x + 1) + ln x  = ln (x(2x + 1)), we can rewrite the given equa-tion:

    ln (x(2x + 1)) = 2.

    Because ex is 1-1, an  x  solves this equation if and only if 

    x(2x + 1) = e2

    ⇔   2x2 + x − e2 = 0⇔   2 x +   142 −   18 − e2 = 0⇔   x = −14 ±   14

    √ 8e2 + 1.

    Because x  must be >  0 for ln x to make sense, we take the positive solutiononly:

    x = −14

     + 1

    4

     8e2 + 1.

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    Problem 18:

    Solve the following inequalities:

    (a) 2 <  ln x  4.

    Solution to (a).

    Because ex and its inverse ln x are increasing functions, a < b  if and only if ea < eb. Thus:

    2 <  ln x  4 if and only if 2 − 3x > ln 4. Theset of solutions to this inequality is

    x <  1

    3(2 − ln4).

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    Problem 19: (#6.3.55 from [Stewart].)

    If  f (x) =√ 

    3 − e2x, then:(i) What is the domain of  f (x)?

    (ii) What is  f −1, stated with its domain?

    Solution

    An  x   lies in the domain of  f   if and only if 3 − e2x ≥ 0. So:

    dom(f ) =−∞,

     1

    2 ln 3

    .

    The domain of  f −1(x) is the range of  f (x), which is [0,√ 

    3).

    To determine the rule for the inverse function: let   x ∈   [0, √ 3) and letx̃ ∈ −∞,  12 ln 3  such that  f (x̃) = x. Then:

    √ 3 − e2x̃ = x

    ⇔   3 − e2x̃ = x2⇔   e2x̃ = 3 − x2⇔   2x̃ = ln(3 − x2)

    ⇔  x̃ =   12 ln(3

    −x2).

    (Note that this deduction confirms our “guess” for the range of  f (x).)Thus the rule for  f −1 : [0,

    √ 3) → R is:

    f −1(x) = 1

    2 ln(3 − x2).

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    Problem 20: (#6.3.56 from [Stewart].)

    If  f (x) = ln(2 + ln x), then:

    (i) What is the domain of  f (x)?

    (ii) What is  f −1, stated with its domain?

    Solution

    The domain of ln x   is (0, ∞), so the domain of ln(2 + ln x) is the set of   xsuch that 2 + ln x > 0. Thus:

    dom(f ) =

    e−2, ∞ .The domain of the inverse function  f −1(x) is the range of  f , which is all

    of  R.

    To determine the rule for the inverse function, let   x ∈   R   and let x̃ ∈e−2, ∞  such that  f  (x̃) = x. Then:

    ln(2 + ln x̃) = x⇔   2 + ln x̃ =  ex⇔   x̃ =  e(ex−2).

    Thus the rule for the inverse function  f −1

    : R → R isf −1(x) = e(e

    x−2).

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    Problem 21: (#6.3.61 from [Stewart].)

    If  f (x) = e(x3), then what is  f −1(x), the corresponding inverse function?

    Solution

    The domain of   f   is all of   R. The domain of the inverse function   f −1(x)

    is the range of   f , which is (0, ∞). To determine the rule for the inversefunction, let  x ∈ (0, ∞), and let x̃ ∈ R such that  f  (x̃) = x. Then:

    x =  e(x̃)3

    ⇔  (x̃)3 = ln x

    ⇔   x̃ = (ln x) 13 .

    Thus the rule for the inverse function  f −1 : (0, ∞) → R is:

    f −1(x) = (ln x)1/3 .

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    Problem 22: (#6.3.64 from [Stewart].)

    If  f (x) =   ex

    1+2ex , then what is  f −1(x), the corresponding inverse function?

    Solution

    To begin, note that this function does not obviously appear to be 1-1. Tounderstand it better, let’s calculate the derivative, which is

    f ′(x) =  ex(1 + 2ex) − ex ∗ 2ex

    (1 + 2ex)2  =

      ex

    (1 + 2ex)2.

    This is always   >   0, so the function is increasing, and so is 1-1 and has a

    well-defined inverse.The domain of  f (x) is all of  R. To understand the range, note that

    limx→−∞

    ex

    1 + 2ex  = 0

    while

    limx→+∞

    ex

    1 + 2ex  = lim

    x→∞1

    e−x + 2  =

     1

    2.

    Thus the range of  f (x) is

    0,  12

    .

    The domain of the inverse function is the range of  f , so is 0, 12.

    To determine the rule for the inverse function, let   x ∈ 0,  12, and letx̃ ∈ R such that  f  (x̃) = x. Then:

    x =   ex̃

    1+2ex̃

    ⇔   x(1 + 2ex̃) = ex̃⇔   ex̃(1 − 2x) = x⇔   x̃ = ln

      x1−2x

    .

    Thus the inverse function  f −1 :

    0,  12

    → R is given by the rule

    f −1(x) = ln

      x1 − x

    .

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    Problem 23⋆: (#6.3.67 from [Stewart].)

    Consider the function

    f (x) = ln

    x + 

    x2 + 1

    .

    Show that f (x) is an odd function of  x. Then find the corresponding inversefunction.

    Solution

    To begin: note that  f (x) is constructed from increasing functions, so it is

    itself increasing, and 1-1.The first thing we are asked to show is that f (x) is odd:   f (−x) = −f (x).

    Observe:

    f (−x) = ln−x +

     (−x)2 + 1

    = ln

    −x +

     x2 + 1 ∗  x  +

    √ x2 + 1

    x +√ 

    x2 + 1

    = ln−x2 + (x2 + 1)

    x +√ 

    x2 + 1

    = ln

      1

    x + √ x2 + 1=   − ln

    x +

     x2 + 1

    =   −f (x).

    What is the domain of  f (x)? The issue is that ln x is only defined whenx > 0. Because  x  +

    √ x2 + 1 ≥ 1 for all  x ≥ 0, we deduce that [0, ∞) lies in

    the domain of  f (x). Also: we just showed that f (−x) =  f (x), so (−∞, 0]must lie in the domain of  f (x) of  f  as well. Thus: the domain of  f (x) is thewhole of  R.

    What is the range of  f (x)? Note that  f (0) = 0,  f (x) is increasing, and

    limx→∞ f (x) = ∞, which implies that [0, ∞) lies in the range of  f (x). Also:f (x) is odd, as we just showed, so (−∞, 0] must lie in the range of  f (x) aswell. So the range of  f   is all of  R.

    Thus the inverse function is a function with domain  R and range  R. To

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    determine the rule, let  x  and x̃

    ∈R  be reals such that  f  (x̃) = x. Then:

    ln

    x̃ +√ 

    x̃2 + 1

     =  x

    ⇔√ 

    x̃2 + 1 = ex − x̃⇔   x̃2 + 1 = (e2x − 2exx̃ + x̃2)⇔   2exx̃ =  e2x − 1⇔   x̃ =   12 (ex − e−x) .

    So:

    f −1(x) = 1

    2

    ex − e−x .

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    Problem 24: (#6.3.70 from [Stewart].)

    Analyse the given limits by using the equation g(x) = eln g(x) to push all thedependence on x  into the index.

    Solutions

    Solution to (a).

    limx→∞

    xlnx = limx→∞

    e(lnx)

    lnx= lim

    x→∞e(lnx)

    2

    = ∞.

    In this analysis we have used the fact that if limx→∞ f (x) =  ∞, and

    limx→∞ g(x) = ∞, then limx→∞ f (g(x)) = ∞   as well. We haven’t provedthis explicitly, although it is exactly what you would expect from such limits.

    Solution to (b).

    limx→0+

    x− lnx = limx→0+

    e−(lnx)2

    = 0.

    In this limit, the property we have used is that if limx→−∞ f (x) =  L, andlimx→0+ g(x) = −∞, then limx→0+ f (g(x)) = L. Again, this is an expectedproperty which we haven’t proved explicitly.

    Solution to (c).

    limx→0+

    x1/x = limx→0+

    e1

    x∗lnx = 0.

    The reasoning here is that as   x →   0+, 1/x →   +∞   and ln x → −∞, sotheir product tends to −∞. And because limx→−∞ ex = 0, the whole limitapproaches zero.

    Solution to (d).

    limx→∞

    (ln2x)− lnx = limx→∞

    e− lnx∗ln(ln2x) = 0.

    This is a variation on the logic in Part (c).

    Curious students are invited to construct proofs of the “intuitive” prop-erties we have used in the above discussions.

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    Problem 25: (#6.4.37 from [Stewart].)

    Find an equation of the tangent line to the curve

    y  = ln (x2 − 3x + 1)at the point (3, 0).

    Solution

    This will just be a simple calculation. We’ll need the derivative of thefunction at the point 3.

    dy

    dx  =

      1

    x2 − 3x + 1 ∗ (2x − 3).

    At x  = 3 this derivative is: 3.So the tangent line is given by:

    y = 0 + 3(x − 3) = 3x − 9.

    Problem 26: (#6.4.46 from [Stewart].)

    Use logarithmic diff erentiation to diff erentiate the function

    f (x) =√ 

    xex2−x(x + 1)2/3.

    Solution

    The first step in logarithmic diff erentiation is to take the logarithm of bothsides:

    ln f (x) = ln√ 

    xex2−x(x + 1)2/3

    .

    The second step is to use the properties of the logarithm to simplify theexpression:

    ln f (x) = 1

    2 ln x + (x2 − x) + 2

    3 ln (x + 1).

    The third step is to diff erentiate both sides, using the chain rule:

    f ′(x)f (x)

      =  1

    2x + 2x − 1 +   2

    3(x + 1).

    Finally, solve for the derivative you want, substituting back in  f (x):

    f ′(x) =

     1

    2x + 2x − 1 +   2

    3(x + 1)

    √ xex

    2−x(x + 1)2

    3 .

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    Problem 27: (#6.4.47 from [Stewart].)

    Use logarithmic diff erentiation to diff erentiate the function

    f (x) = xx.

    Solution

    We’ll do this one a bit more quickly:

    f (x) = xx

    ⇒   ln f (x) = ln (xx)

    ⇒   ln f (x) = x ln x⇒   f ′(x)

    f (x)   = ln x +  xx

    ⇒   f ′(x) = (ln x + 1) xx.

    Problem 28: (#6.4.52 from [Stewart].)

    Use logarithmic diff erentiation to diff erentiate the function

    f (x) = (sin x)lnx.

    Solution

    Compute:

    f (x) = (sin x)lnx

    ⇒   ln f (x) = ln sin xlnx⇒   ln f (x) = ln x ∗ ln (sin x)⇒   f ′(x)

    f (x)   =  1x ln (sin x) + ln x ∗   1sinx ∗ cos x

    ⇒  f 

    ′(x) = ln (sin x)

    x  +   lnx cosx

    sinx (sin x)ln x .

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