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Math Help - More calculus questions

  1. #1
    Newbie calculus-geeks09's Avatar
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    Exclamation More calculus questions

    here's some more problem that I am on right this moment:

    1. determine the derivative of F(x)=(cos(2x-4))^3 at x=pie/b
    2. determine d/dx 3x^4-3x/3x^4+3x
    3. compute -3x^2x^3+3dx
    4. an open rectangular box with volume 6m^3 has a square base. Express the surface area of the box as a function s(x) of the length x of a side of the base .
    5. a box with an open top is to be constructed from a rectangular peice of cardboard with dmensions. b=5in. a=28in by cutting out equal squares of side x at each corner and then folding up the sides. find the maximum volume.
    6. a spherical balloon with raduis r inches has volume 4/3pie^3. Find a function that represents the amount of air required to inflate the balloon from a radius of r innches to a radius of r+1 inches.
    7. Lim as t approaches 2= t^2-4/t^3-8
    8. absolute max. y=36-x (-6,6)
    9. How many points of inflection in this graph
    f(x)=12x^3=14x^2-7x-p

    10. estimate the area from 0 to 5 under the graph
    f(x)=81-x^2 using 5 rectangles from right end points.


    thank you so much for the help!!!!! my life is on the line!!! i'm struggling in my AP calulus class and i have to pass this final. it's very urgent!!! thanks to anyone who helps.
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  2. #2
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    10. estimate the area from 0 to 5 under the graph
    f(x)=81-x^2 using 5 rectangles from right end points.
    First you determine the width of each rectangle by using this formula
    \Delta x = \frac{b-a}{n}
    Where n is the number of rectanges.
    \Delta x = \frac{5-0}{5} = 1


    A = \Delta x(f(x1)+f(x2)+f(x3)+f(x4)+...)

    Since we are using right end-points triangles for f(x) = -x^{2}+81


    f(1)=80 , f(2)=77, f(3)=72, f(4)=65, f(5)=56<br />

    A = 80 + 77 + 72 + 65 + 56  = 350
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  3. #3
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    Question#7

    \lim_{x \rightarrow 2}(x^{2} - \frac{4}{x^{3}} - 8) = 2^{2} - \frac{4}{2^{3}}-8 = -\frac{9}{2}

    Question#3

    \int -3x^{2} \sqrt{x^{3}+3}dx

    Let u =  x^{3}+3

    du = 3x^{2} dx

    -du = -3x^{2} dx

    \int -\sqrt{u}du = -\frac{2(u)}{3}^{\frac{3}{2}}

    Now you replace the u by x^{3} +3 and you get

    -\frac{2(x^{3}+3)}{3}^{\frac{3}{2}}
    Last edited by fonso_gfx; December 18th 2008 at 03:49 PM.
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  4. #4
    Newbie calculus-geeks09's Avatar
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    thanks
    Last edited by mr fantastic; December 18th 2008 at 09:50 PM. Reason: Changed the 5 point font size and excessive smilies
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  5. #5
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    For ur 4th question local maximum of the below equation.. I think we need to differentiate the equation and solve that for x by assigning the equation to zero

    That means f ' (x) = 0

    f(x)=x^3-9x^2+24x+4
    3*x^2 - 18*x + 24 = 0
    x^2 - 6*x + 8 = 0
    (x - 2) * (x-4) = 0

    either x = 2 or 4

    so find out F(2) and f(4) and see which one is greater..
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  6. #6
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    Quote Originally Posted by calculus-geeks09 View Post
    help
    #8 - A function has an absolute maximum at c if f(c) is greater than or equal to f(x) for all x values in the domain which in your case is (-6,6)

    #9 - get the second derivative of f(x)

    d/dx f(x) = f'(x)
    d/dx f'(x) = f"(x)

    set f"(x) = 0 and solve for x, and that is your inflection point.
    Last edited by mr fantastic; December 18th 2008 at 09:51 PM. Reason: Edited the quote.
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  7. #7
    Newbie calculus-geeks09's Avatar
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    Quote Originally Posted by fonso_gfx View Post
    #8 - A function has an absolute maximum at c if f(c) is greater than or equal to f(x) for all x values in the domain which in your case is (-6,6)
    i don't get what you mean by this. can you explain it further pleasee.
    i'm a dummie at this. LOL

    thank u so much.
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