Calculus anyone?
 11222002, 05:29 PM
Calculus anyone?
Someone willing to help me out here with my calculus problem:
2x(x4)^3
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!  11222002, 06:24 PM
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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
ManBeast  11222002, 08:49 PM
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!

 11222002, 10:38 PM
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Agreed, I still suggest (mostly just for knowledge's sake) that you doublecheck me on the derivatives
ManBeast  11222002, 10:40 PM
Course, I just wanted a second opinion, check myself! Never can be too careful ya know!
 11222002, 10:43 PM
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Agreed. I just wanted to make sure you weren't taking my word as 100% right without checking... I know me
ManBeast  12032002, 03:38 AM
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=(3x4)^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= (x2)(x4)
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)
concavity:
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  12032002, 07:42 PM
I did well in Math 31 (high school), but badly in Calculus 201 (University). I don't remember any of it. Now, how can we apply what we learned in high school/universtiy math to our daily lives?
 12032002, 09:16 PM
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Thanks for catching me on that Pete, I told ya'll I'm not the brightest crayon in the box!
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