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Have a few questions, can you look them over?

  1. Nov 4, 2007 #1
    1) integral from -4 to 2 of 2/sqrt(x) dx

    I ended up with 4(sqrt2) - 4(sqrt-4); so I concluded it doesn't exist, correct?

    2) integral (1+3x)x^2
    This just seemed too simple to me, I ended up with f(x)=1/3x^3+3/4x^4+c, how's that look?

    3) fifth derivative=0, find f(x); I ended up with C1x^4+C2X^3+C3X^2+C4X+C5
    Does that seem okay?

    4)I didn't finish this because I'm stuck, and not sure what the graph will look like, but the question is: Sketch the graph of a function that satisfies all of the given conditions.
    f'(-1)=f'(1)=0
    this means that there are tangent lines at x=1 and x= -1?
    f'(x) <0 if |x|<1
    this means graph decreases on all intervals when x<1?
    f'(x)>0 if |x|>1
    this means the graph increases on all intervals when x>1?
    f(-1)=4, f(1)=0
    these are critical points, local max's or min's?
    f''(x) < 0 if x < 0
    meaning concavity is down on all intervals where x is less than 1?
    f''(x) > 0 if x>0
    meaning concavity is up on all intervals where x is greater than 1?

    If these are right, i'm not sure what the graph will look like, can someone give me a hand?

    5) This last one seems simple, but I can't seem to figure out where to begin.
    Question: Find the point of the graph of the function that is closest to the given point (2, 1/2) and the function: f(x)=x^2

    EDIT: Think I got it, used the distance formula and solved through by optimization. Ended up with the points being (1,1)
    Does that seem correct? Is there any way to check it?


    Sorry for so many questions, but this is around 2 chapters worth of work that i'm trying to get caught up on and these are the ones that I got stumped or am unsure of.

    Thanks a lot.
     
    Last edited: Nov 4, 2007
  2. jcsd
  3. Nov 4, 2007 #2
    anybody have some time to give me a hand? lol
     
  4. Nov 4, 2007 #3
    I think I solved question 5.
     
  5. Nov 4, 2007 #4
    can someone at least take a look at number 4 for me?
    I think the others are okay, but I need to get 4 done tonight.

    thanks..
     
  6. Nov 4, 2007 #5

    rock.freak667

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    Homework Helper

    At a stationary point f'(x)=0 so then f'(-1)=f'(1)= 0 means that at x=-1 and x=1 there is a stationary point...now they gave you f(1) and f(-1). Thus (-1,4) and (1,0) are stationary points...they told you what f''(x) would be if x>0 and x<0...check the signs of the x-coordinates of the stationary points and see if they are max or min points

    and |x|<1 => -1<x<1
     
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