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Math Help - [Urgent] help!! calculus, function analysis.

  1. #1
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    Exclamation [Urgent] help!! calculus, function analysis.

    Hi, I'm stuck on my calculus problem which is due on tomorrow.

    Here's the full question,

    Q. For the following functions, do a full analysis of the function itself, the first derivative, and the second derivative.

    Provide information about the
    - domain, intercepts, and vertical and horizontal asymptotes.
    Find and classify all critical points
    Find all intervals where the function is increasing or decreasing.
    Find all intervals of concavity, and inflection points.
    Finally, sketch the graph of the function, using the information you obtained.



    - Is this mean by I have to do the analysis of three function? since it says, i have to do the analysis of the function itself, the first derivative and second.









    i) f(x) = x(x^2+4)^2

    I got the first and second derivatives,
    f'(x) = (x^2+4)(5x^2+4)
    f''(x) = 4x(5x^2+12)

    Since I tried by looking at the text book and do the analysis.. I dunno where to start and what to do.. Also, for getting critical points, let say where f(x) = 0, I can't find the roots since i get no real number..


    ii) f(x) = x^(5/3) - 5x^(2/3)
    i also have the derivatives for these too.
    f'(x) = 5/3*x(2/3) - 10/3*x^(-1/3)
    f''(x) = 10/9*x^(-1/3) + 10/9*x(-4/3)

    for this one ii), I have to care for cusps or vertical tangents..

    iii) f(x) = 2cosx+sin2x
    also got the derivatives.
    f'(x) = -2sin(x)+2cos(2x)
    f''(x) = -2cos(x) - 4sin(2x) ==> is this right?

    and for this one, I have to care about whether the function is periodic or not..




    Please math people.. help me.. I'm a newb
    Last edited by Livevvire; December 3rd 2008 at 07:13 PM.
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  2. #2
    Senior Member vincisonfire's Avatar
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    (i)The second derivative is f''(x) =4x(5x^2+12)
    (ii)Everything's nice.
    (iii) \frac{d(sin(u)}{dx} = cos(u)\frac{du}{dx}
    In this case,  u = 2x \implies \frac{du}{dx} = 2
    f(x) = 2cos(x)+sin(2x) \implies f'(x) = -2sin(x)+2cos(2x)
    Change your answer for f''(x) too.
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  3. #3
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    Quote Originally Posted by vincisonfire View Post
    The second derivative is *f''(x) =4x(5x^2+12)
    ouch, i forgot to put 4

    thanks.
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    Are you sure you did the derivatives, right, I think you might be missing chain rule and product rule on some of them. For example, the derivative of sin(2x) is 2(cos(2x). The derivative of x(x^2+4)^2 would be something like (x^2+4)^2 + ((2x)(x^2+4))^2, unless I messed up somewhere.
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  5. #5
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    Quote Originally Posted by terr13 View Post
    Are you sure you did the derivatives, right, I think you might be missing chain rule and product rule on some of them. For example, the derivative of sin(2x) is 2(cos(2x). The derivative of x(x^2+4)^2 would be something like (x^2+4)^2 + ((2x)(x^2+4))^2, unless I messed up somewhere.
    the derivative of sin(2x) is 2(cos(2x)) i guess?

    but i think my derivative of x(x^2+4)^2 is right..?

    it's (x^2+4)^2 + x*2(x^2+4)(2x)..

    since i have to use power rule for the second part to inside too..
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  6. #6
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    As I expected, I messed up on that, but there is one solution for when f(x)=0, namely when x = 0.
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    Quote Originally Posted by terr13 View Post
    As I expected, I messed up on that, but there is one solution for when f(x)=0, namely when x = 0.
    could you help me doing the trig derivatives too?

    the first and the second?

    and yeah i knew that f(0) = 0, but what do i do with the imaginary roots?.. i'm sorry to say this but i didn't go to lectures.. that's why i'm screwing up what to do.. and also i dunno what should i do with the f(x) = 0 and f'(x) = 0 and stuff..
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  8. #8
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    Quote Originally Posted by vincisonfire View Post
    (i)The second derivative is f''(x) =4x(5x^2+12)
    (ii)Everything's nice.
    (iii) \frac{d(sin(u)}{dx} = cos(u)\frac{du}{dx}
    In this case,  u = 2x \implies \frac{du}{dx} = 2
    f(x) = 2cos(x)+sin(2x) \implies f'(x) = -2sin(x)+2cos(2x)
    Change your answer for f''(x) too.
    I changed and tried for f''(x) ,is this right?
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  9. #9
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    f(x) = 0 is where it crosses the x-axis.
    f'(x) = 0 is when the shape flattens out, like for a parabola
    f''(x) shows the concavity.
    The only thing for trig derivatives is to be careful with the chain rule.
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