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Math Help - analysis questions

  1. #1
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    analysis questions

    please help not sure abt the anwers..
    Attached Thumbnails Attached Thumbnails analysis questions-tf.jpg  
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by bobby77 View Post
    please help not sure abt the anwers..
    a) A tangent line can be used as a linear approximation to a function near
    a point, but it is not the only such approximation, so the answer is no -
    a linear approximation to a function is not necessarily a tangent line.

    b) At least one of the definitions of a point of inflection for a function f is
    that it is a point at which the f'' changes sign so of necessity f''(x)=0 at
    a point of inflection, but it is not a sufficient condition. As Glaysher points
    out every point on f(x)=mx+c is a point where f''(x)=0, but they are not
    points of inflection. So f''(a)=0, is not sufficient to guarantee that x=a is
    a point of inflection of f.

    RonL
    Last edited by CaptainBlack; October 15th 2006 at 02:03 AM. Reason: correct an error in the definition of inflection point
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by bobby77 View Post
    please help not sure abt the anwers..
    c) f''<0, means that f' is decreasing, which is concave downwards, to be
    concave upwards we would need f''>0, so no.

    RonL
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  4. #4
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    Indeed, the limits for both f and g do equal e, hence they are both true.

    The number e is defined by those limits.
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  5. #5
    Member Glaysher's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    a) A tangent line can be used as a linear approximation to a function near
    a point, but it is not the only such approximation, so the answer is no -
    a linear approximation to a function is not necessarily a tangent line.

    b) At least one of the definitions of a point of inflection for a function f is
    that it is a point at which f''(x)=0, then yes if f''(a)=0, then x=a is a point
    of inflection of f.

    RonL
    b) Is no as it has to be a stationary point as well or else every point on a straight line would be a point of inflexion as:

    f(x) = mx + c
    f'(x) = m
    f''(x) = 0
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  6. #6
    Member Glaysher's Avatar
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    Quote Originally Posted by bobby77 View Post
    please help not sure abt the anwers..
    d) Yes

    e) No, depends on the shape of the graph

    eg a graph with a high positive gradient that suddenly levels off at the top can produce bad approximations

    f) No, limit is 1

    g) Yes

    h) True only if x is in radians so statement not specific enough
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Glaysher View Post
    b) Is no as it has to be a stationary point as well or else every point on a straight line would be a point of inflexion as:

    f(x) = mx + c
    f'(x) = m
    f''(x) = 0
    I was being a bit relaxed with the definition, a point of inflection is
    a point where the second derivative changes sign, so of necessity
    f''=0, but f''=0 is not a sufficient condition.

    So you are right my answer was wrong f''(a)=0 is not a sufficient
    condition to guarantee that x=a is a point of inflection.

    But a point of inflection does not have to be a stationary point.

    RonL
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  8. #8
    Member Glaysher's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    I was being a bit relaxed with the definition, a point of inflection is
    a point where the second derivative changes sign, so of necessity
    f''=0, but f''=0 is not a sufficient condition.

    It does not have to be a stationary point.

    RonL
    Never seen a proper definition for it before
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  9. #9
    Member Glaysher's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    In fact having just typed that definition it struck me that f'' can change
    sign if it has a jump discontinuity at a point, is this still a point of inflection?

    If so f'' !=0 at this point?

    RonL
    I suspect that the full formal definition would need the function to be continuous over some interval containing the point.

    But just to confirm would there be a point of inflexion for a function that starts with a decreasing positive gradient which later turns into an increasing positive gradient before the gradient is zero?
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by Glaysher View Post
    I suspect that the full formal definition would need the function to be continuous over some interval containing the point.

    But just to confirm would there be a point of inflexion for a function that starts with a decreasing positive gradient which later turns into an increasing positive gradient before the gradient is zero?
    Yes.

    (I wouldn't pay too much attention to the post of mine that you quoted,
    as I think a jump discontinuity in the second derivative implies a corner
    in the derivative at the point, which is a point at which the second derivative
    does not exist. My comment was an unjustified outbreak of mathematical
    paranoia )

    RonL
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  11. #11
    Member Glaysher's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Yes.

    (I wouldn't pay too much attention to the post of mine that you quoted,
    as I think a jump discontinuity in the second derivative implies a corner
    in the derivative at the point, which is a point at which the second derivative
    does not exist. My comment was an unjustified outbreak of mathematical
    paranoia )

    RonL
    I was thinking that it would be unusual to have a point of inflexion on a point that was not part of the original graph
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  12. #12
    Grand Panjandrum
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    Quote Originally Posted by Glaysher View Post
    I was thinking that it would be unusual to have a point of inflexion on a point that was not part of the original graph
    You initial beleif that a point of inflection has to be a stationary point of
    the function is quite interesting, as it is what I thought up to only a few
    years ago. I suspect it is something left over as a half memory from
    A-level maths 35+ years ago.

    It just goes to show that one always need to be verifying what one
    thinks one knows

    RonL
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  13. #13
    Member Glaysher's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    You initial beleif that a point of inflection has to be a stationary point of
    the function is quite interesting, as it is what I thought up to only a few
    years ago. I suspect it is something left over as a half memory from
    A-level maths 35+ years ago.

    It just goes to show that one always need to be verifying what one
    thinks one knows

    RonL
    True as it was my A level teaching that made me think that too and the textbooks I currently teach from too. Though to be fair it is never explicitly said, just implied.
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  14. #14
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    Quote Originally Posted by Glaysher View Post
    d) Yes

    e) No, depends on the shape of the graph

    eg a graph with a high positive gradient that suddenly levels off at the top can produce bad approximations

    f) No, limit is 1

    g) Yes

    h) True only if x is in radians so statement not specific enough
    How is the limit in part f 1? Care to explain? I got e, although I might have done some careless mistake.
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  15. #15
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    Quote Originally Posted by AfterShock View Post
    How is the limit in part f 1? Care to explain? I got e, although I might have done some careless mistake.
    If the problem is:
    limit(x->0) (1+x)^(1/x) (it was hard to read the exponent, is this right?)

    then the answer is indeed "e" not 1.

    We can change the limit variable to n = 1/x, then the limit becomes:

    limit(n->infinity) (1 + [1/n])^n which is one of the definitions of e.

    -Dan
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