Calculus anyone?

  1. Calculus anyone?

    Someone willing to help me out here with my calculus problem:


    What I need(any or all help would be gratefully appreciated!):

    1: First and Second Derivatives
    2: First and Second Derivative Critical Numbers(zeros)
    3: Intervals of Concavity
    4: Intervals of Increasing and Decreasing
    5: Inflection Points

    *main thing I need is the first and second derivatives!*
    This is for a little take home test and I'm just wanting to check and make sure I got it all right. Thanks ahead of time!

  2. Multiplied out: 2x^4 - 24x^3 + 96x^2 -128x

    First Derivative: 8x^3 -72x^2 + 192x -128

    Second Derivative: 24x^2 -148x +192

    I really don't want to do the other stuff, I'm a lazy bitch sometimes


  3. Thank you much sir, GREATLY appreciative! The derivatives were the main things I needed because if you get those right, the rest is child's play, just takes some time!

  4. Agreed, I still suggest (mostly just for knowledge's sake) that you double-check me on the derivatives


  5. Course, I just wanted a second opinion, check myself! Never can be too careful ya know!

  6. Agreed. I just wanted to make sure you weren't taking my word as 100% right without checking... I know me


  7. hey guys, bad news, little error in second derivative ( -144x, not -148x)

    here's what i got:

    f(x)=2x^4 - 24x^3 + 96x^2 - 128x
    f'(x)=8x^3 - 72x^2 + 192x - 128
    f"(x)=24x^2 - 144x + 192

    now to find min/max:

    set f'(x)=0 and solve for x
    so we get: 0=8x^3 - 72x^2 + 192x - 128
    divide both sides by 8: 0=x^3 - 9x^2 + 24x - 16
    now rewrite with brackets: 0=x^3 - (9x^2 - 24x + 16)
    the last step is a trick i learned back in first year
    now factor: 0=x^3 - (3x - 4)^2
    this also takes some practice to see, although the quadratic equation will give you the roots
    now rewrite: x^3=(3x-4)^2
    x=1,x=4 only solutions use maple v7 (software, its great), but i got it by just plugging in numbers
    increasing/decreasing intervals:
    for all x<1, f'(x)<0 so decreasing
    for all 1<x<4, f'(x)>0 so increasing
    for all x>4, f'(x)>0 so increasing
    thus x=4 not min/max point and x=1 is local and absolute min
    plug x=1 into original: f(1)=-54
    so min point is (1,-54) and x<1 is decreasing, x>1 is increasing

    now to find points of inflection:

    set f"(x)=0 and solve for x
    so we get 0=24x^2 - 144x + 192
    divide both side by 24: 0=x^2 - 6x + 8
    factor: 0= (x-2)(x-4)
    so x=2,x=4 agrees with software
    find points of inflection by subbing into f(x):
    f(2)=-32,f(4)=0, so points are (2,-32) and (4,0)
    for all x<2, f"(x)>0, so concave up
    for all 2<x<4, f"(x)<0, so concave down
    for all x>4, f"(x)>0, so concave up

    also the graph passes through (0,0), know this, cause you can divide out the x from the 2x term

    i know this is probably late for you but it was fun anyway. you have no idea how much money one can make by doing assignments for first years, i do, been doing so for 5 yrs.

    cheers, pete

  8. I&nbsp; did well in Math 31 (high school), but badly in Calculus 201 (University).&nbsp; I don't remember any of it.&nbsp; Now, how can we apply what we learned in high school/universtiy math to our daily lives?&nbsp;

  9. Thanks for catching me on that Pete, I told ya'll I'm not the brightest crayon in the box!

    -Saving random peoples' nuts, one pair at at time... PCT info:
    -Are you really ready for a cycle? Read this link and be honest:
    *I am not a medical expert, my opinions are not professional, and I strongly suggest doing research of your own.*


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