Results 1 to 7 of 7

Math Help - Derivative question

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    42

    Derivative question

    f(x)=\frac{2x^2+4x+2}{x^2+1}

    \lim_{x\ \rightarrow\ \infty}f(x)=

    (I am not familiar with this notation at all... )

    f'(x)=

    f'(x)=0 when x=A,\:B

    A=\:\:,\:B=

    \int_A^Bf(x)dx=

    Please help

    I'm not even sure where to start. I mean, I can factorise the equation, but would that help? I know that the derivative of the equation is the slope at any given value of x, and the slope is 0 for each minima/maxima. But I don't know how to answer these questions...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2009
    Posts
    128
    Quote Originally Posted by davidman View Post
    f(x)=\frac{2x^2+4x+2}{x^2+1}

    \lim_{x\ \rightarrow\ \infty}f(x)=

    The limit as x goes to infinity of the function said above.

    How to do this problem? Divide every term by x^{2}

     \lim_{x\ \rightarrow\ \infty} f(x)=\frac{\frac{2x^2}{x^2}+\frac{4x}{x^2}+\frac{2  }{x^2}}{\frac{x^2}{x^2}+\frac{1}{x^2}}

    Do you see how to solve the problem? Remember that as x goes to infinity, denominators become INFINITELY large ... so what does that do to numbers/fractions? It makes them zero.

    Buuuuuuuuut ... don't forget to do any necessary "cancellations."
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,454
    Thanks
    1868
    Quote Originally Posted by davidman View Post
    f(x)=\frac{2x^2+4x+2}{x^2+1}

    \lim_{x\ \rightarrow\ \infty}f(x)=

    (I am not familiar with this notation at all... )

    f'(x)=

    f'(x)=0 when x=A,\:B

    A=\:\:,\:B=

    \int_A^Bf(x)dx=

    Please help

    I'm not even sure where to start. I mean, I can factorise the equation, but would that help? I know that the derivative of the equation is the slope at any given value of x, and the slope is 0 for each minima/maxima. But I don't know how to answer these questions...
    If the only thing you know about the derivative is that it is the slope (strictly speaking it is the slope of the tangent line to the graph- only lines have "slope".) then these questions are way over your head. Go to your teacher and make sure he/she knows your difficulty.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2010
    Posts
    42
    Quote Originally Posted by Lord Darkin View Post
    Divide every term by x^{2}
    Thanks... so

    \lim_{x \rightarrow\ \infty}f(x)=\lim_{x \rightarrow\ \infty}\frac{2+\frac{4}{x}+\frac{2}{x^2}}{1+\frac{  1}{x^2}}=2

    Although, I'm not entirely sure why to divide every term by x^2 specifically... I mean, I can see that it's the largest exponent of x. Is that how you decide what to divide by?

    Quote Originally Posted by HallsofIvy
    [...] these questions are way over your head.
    This is true, but I actually understated my ability/experience with derivatives a bit.

    I don't remember much at all of what I studied (several years ago), but I think for instance if F(x)=2x^2+4x+2\:\:,\:\:F'(x)=4x+4

    How to get the derivative of f(x)=\frac{2x^2+4x+2}{x^2+1} I forgot.

    And lastly, unfortunately, I have no teacher, just books and the internet. Really grateful for all the help though.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    Posts
    128
    Yes, I divide by the largest exponent of x. That's how I decide.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,454
    Thanks
    1868
    Then you had better keep reading those books! Hopefully, just after telling you that the derivative is the slope of the tangent line, they will have a bunch of "rules" for differentiating.

    For all of these problems, it helps a lot to recognise that
    \frac{2x^2+ 4x+2}{x^2+ 1}= \frac{2x+2}{x^2+ 1}+ \frac{4x}{x^2+ 1}= 2+ \frac{4x}{x^2+1}. Since the denominator in that fraction has a higher power than the numerator, it goes to 0 as x goes to infinity.

    To differentiate that, use the quotient rule. Do you know what that is?

    For the second question, set the derivative equal to 0 and solve for x. Since the derivative is also a fraction, the first thing you can do is multiply both sides of the equation by the denominator.

    Integrating, use the substitution u= x^2+ 1 for the fraction. I would hope you would not have any problem integrating "2" but integration, in most books, comes well after the definition of the derivative!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2010
    Posts
    42
    Thank you!

    f(x)=\frac{2x^2+4x+2}{x^2+1}

    f'(x)=\frac{(2x^2+4x+2)'(x^2+1)-(2x^2+4x+2)(x^2+1)'}{(x^2+1)^2}

    =\frac{(4x+4)(x^2+1)-2x(2x^2+4x+2)}{(x^2+1)^2}

    =\frac{4x^3+4x+4x^2+4-4x^3-8x^2-4x}{(x^2+1)^2}

    =\frac{-4x^2+4}{(x^2+1)^2}

    =\frac{-4(x^2-1^2)}{(x^2+1)^2}

    =\frac{-4(x+1)(x-1)}{(x^2+1)^2}

    f'(x)=0\:\:,\:\:x=-1\:\:,\:\:1

    I notice that the denominator isn't going to become equal to zero if the x there is squared.

    Anyways, this gives me the limits of the integration to be done. Or whatever it's called... xD Next step,

    \int_{-1}^1f(x)dx=

    As pointed out; f(x)=2+\frac{4x}{x^2+1}=2+\frac{4x}{u}

    \int_{-1}^1f(x)dx=\int_{-1}^1\left(2+\frac{4x}{u}\right)dx

    \left[2x\right]_{-1}^1+\int_{-1}^1\left(\frac{4x}{u}\right)dx=

    and that is where I get stuck

    thanks for all the help. Not quite sure how to do the following step(s), but I'll try and look it up.

    Edit: found a formula that looked relevant \int{af(x)}dx=a\int{f(x)}dx

    =\left[2x\right]_{-1}^1+4\int_{-1}^1\left(\frac{x}{u}\right)dx

    Edit 2: ok, I think I got to the answer in my book somehow

    another formula \int_a^bf(x)dx=F(b)-F(a) inspired me to try and plug in the two limits and simplify, but I'm not entirely sure why it works.

    \left[2x\right]_{-1}^1+4\int_{-1}^1\left(\frac{x}{x^2+1}\right)dx=\: \color{red}2(1)-2(-1)+4[\frac{1}{1^2+1}-\frac{-1}{-1^2+1}]

    =4(\frac{1}{2}+\frac{1}{2})=4

    \int_{-1}^1f(x)dx=4

    What I'm really wondering now is why I can stop simplifying when I have \frac{x}{x^2+1}

    And is it really necessary to move the 4 outside the \int?

    Edit 3: Noticed my calculation was off... 2(1)-2(-1)=4\neq0

    Ok, seems I'm only going to bury myself deeper if I keep editing here. Anyone care to help me out?

    Edit 4: Ok, last edit

    didn't realise I was posting this in the wrong subforum. Sorry about that. Anyways, I tried graphing the bit I couldn't find the integral for. It's pretty obvious why the integral of that on a symmetrical space across the origin would be equal to zero... so I guess I'll keep that in mind when I can't figure out the integral.

    So thanks again everyone.
    Last edited by davidman; January 28th 2010 at 02:37 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. question on derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 27th 2010, 11:35 PM
  2. Replies: 19
    Last Post: October 19th 2009, 06:10 PM
  3. Derivative question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 30th 2009, 05:11 PM
  4. help with derivative question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 20th 2008, 07:01 PM
  5. Derivative question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 24th 2007, 03:33 PM

Search Tags


/mathhelpforum @mathhelpforum