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Thread: Lots Of Help Please!! Asap!

  1. May 11th 2008, 08:26 PM #1
    chousta
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    Lots Of Help Please!! Asap!

    hi i need help with these asap


    please...


    1.
    If , then is continuous at


    Choice

    0

    everywhere


    2.
    The equation
    implicitly defines as a function of . Find .



    3.

    is discontinuous at ____ because _________________
    (More than one selection may be correct. Keep in mind that if a function tends to infinity as , the limit does not exist at .)
    Choice
    because does not exist.
    because does not exist.
    because does not exist.
    because does not exist.
    because
    Number of available correct choices: 2

    4.

    If ,
    then we have

    (Select the correct statement(s).)

    Choice
    f is both continuous and differentiable at x = 0
    f '(x) exists for all x except x = 0 f is not continuous at x = 0
    f '(x) is continuous for all x
    f '(x) does not exist anywhere
    f '(x) exists for all x but is not continuous for all x
    f is continuous at x = 0 but not differentiable at x = 0
    Number of available correct choices: 2

    5.

    Above is the graph of a function . Sketch the graph of ?
    6.

    If ,
    then at 0, is??



    7.


    Find if .

    8.



    . Select the intervals on which the function is continuous.
    Choice






    Number of available correct choices: 3
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  2. May 11th 2008, 08:32 PM #2
    Mathstud28
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    Quote Originally Posted by chousta View Post
    hi i need help with these asap


    please...


    1.
    If , then is continuous at


    Choice

    0

    everywhere


    2.
    The equation


    implicitly defines as a function of . Find .



    3.

    is discontinuous at ____ because _________________
    (More than one selection may be correct. Keep in mind that if a function tends to infinity as , the limit does not exist at .)
    Choice
    because does not exist.
    because does not exist.
    because does not exist.
    because does not exist.
    because
    Number of available correct choices: 2

    4.

    If ,
    then we have

    (Select the correct statement(s).)

    Choice
    f is both continuous and differentiable at x = 0
    f '(x) exists for all x except x = 0 f is not continuous at x = 0
    f '(x) is continuous for all x
    f '(x) does not exist anywhere
    f '(x) exists for all x but is not continuous for all x
    f is continuous at x = 0 but not differentiable at x = 0
    Number of available correct choices: 2

    5.

    Above is the graph of a function . Sketch the graph of ?
    6.

    If ,
    then at 0, is??



    7.


    Find if .

    8.



    . Select the intervals on which the function is continuous.
    Choice






    Number of available correct choices: 3
    Is this a test?

    I will do ONE until you say it is not

    8. y=\tan\bigg(\frac{1}{(x+1)^4}\bigg)

    So arctan(y)=\frac{1}{(x+1)^4}

    Differentiating we get \frac{y'}{1+y^2}=\frac{-4}{(x+1)^5}

    Seeing that y=\tan\bigg(\frac{1}{(x+1)^4}\bigg)

    wee see that y'=\frac{-4\bigg[1+\tan^2\bigg(\frac{1}{(x+1)^4}\bigg)\bigg]}{(x+1)^5}
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  3. May 11th 2008, 09:03 PM #3
    chousta
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    No

    Hi.

    no its not a test. out of a multiple choice worksheet with no answers. and i have an exam very very soon and im freaking out.
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  4. May 11th 2008, 09:07 PM #4
    Soroban
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    Hello, chousta!

    2) The equation: . -4xy - \frac{2}{xy} - \frac{2}{x^2y^2} \:=\:-6 .implicitly defines y as a function of x.
    Find \frac{dy}{dx}
    Multiply by -1: . 4xy + 2x^{-1}y^{-1} + 2x^{-2}y^{-2} \;=\;6

    Differentiate implicitly: . 4x\frac{dy}{dx} + 4y - 2x^{-1}y^{-2}\frac{dy}{dx} - 2x^{-2}y^{-1} - 4x^{-2}y^{-3}\frac{dy}{dx} - 4x^{-3}y^{-2} \;=\;0

    Rearrange terms: . 4x\frac{dy}{dx} - \frac{2}{xy^2}\frac{dy}{dx} - \frac{4}{x^2y^3}\frac{dy}{dx}\;=\;-4y + \frac{2}{x^2y} + \frac{4}{x^3y^2}

    Multiply by \frac{x^3y^3}{2}\!:\quad 2x^4y^3\frac{dy}{dx} - x^2y\frac{dy}{dx} - 2x\frac{dy}{dx} \:=\:-2x^3y^4 + xy^2 + 2y

    Factor: . x(2x^3y^3 - xy - 2)\frac{dy}{dx} \;=\;\text{-}y(2x^3y^3 - xy - 2)


    Therefore: . \frac{dy}{dx}\;=\; \frac{\text{-}y(2x^3y^3 - xy - 2)}{x(2x^3y^3 - xy - 2)} \;=\;\boxed{-\frac{y}{x}}

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  5. May 11th 2008, 09:08 PM #5
    Mathstud28
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    Quote Originally Posted by chousta View Post
    Hi.

    no its not a test. out of a multiple choice worksheet with no answers. and i have an exam very very soon and im freaking out.
    For number five use the fact that mins and maxs for f(x) are zeros for f'(x)

    1.its everywhere

    2. you will just have to grunt implicitly differentiate

    3. \ln(u(x)) is undefined \forall{x}\leq{0}

    4. if f(x) is continous at c then

    \lim_{x\to{c}}f(x)=f(c)

    and if it is differentiable at c then

    \lim_{x\to{c}}f'(x)=f'(c)

    6. if it is talking about continuity check out the first part of the last thing I said

    7. done

    8.check vert. asymptotes (places that make denominator equal to 0)
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  6. May 12th 2008, 02:37 AM #6
    chousta
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    Thank U

    Thank you all so much.
    but i still dont get wat the answer for 3 and 5 are. ESP 5?? how would i graph that? what is the process.


    thank you all so much.im hopin i pass this test!
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  7. May 12th 2008, 04:44 AM #7
    mr fantastic
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    Quote Originally Posted by chousta View Post
    [snip]
    5.

    Above is the graph of a function . Sketch the graph of ?
    [snip]
    Pretend it's the graph of y = sin x. Then df/dx = cos x.

    Alternatively, recall that df/dx gives the gradient of the tangent at each point of the curve .....
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