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Local Extreme Values, Min/Max, etc.

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Find intervals on which function is (a) increasing (b) decreasing (c) concave up (d) concave down (e) local extreme values (f) inflextion points of:

    Y=1+x-x2-x4


    2. Relevant equations
    Y=1+x-x2-x4


    3. The attempt at a solution
    (a) To find where function is increasing, I tried using f'>0
    Y' = 1-2x-4x3
    So 1-2x-4x3>0 yields:
    1>2x(1+2x2)
    So x= 1/2 and 0
    But the answer in the back of the book says (-infinity, 0.385)

    (b) So decreasing is f'<0 and I did the same thing as (a) except with the sign that was different. The answer was [0.385, infinity) but again - my work didn't correspond to the answer! I have no idea how they got it.

    (c) For concave up, it's f">0
    I solved it and got x2<-1/6
    Since it's a negative, it's no solution so the answer is None.

    (d) Same thing except it's now f"<0
    The answer in the back of the book said (-infinity, infinity) ----- but where did they get this?

    (e) Local extreme values, found when f'=0 (aka: critical points)
    I get x=1/2 but.. the answer in the back said local max was at (0.385, 1.215)

    (f) Inflextion points is at f"=0
    Got x^2 = -1/6 so it's none.

    What am I doing wrong? How come my work isnt' close to the correct answer? D:
     
  2. jcsd
  3. Nov 19, 2009 #2
    a) Solve for 1-2x-4x^3 = 0, there is only 1 solution (so it only crosses the x-axis once, implying that if something on one side is positive, everything else on that "side" will be), so take some arbitrary number left or right of your solution and see if it's positive or negative.. you can see that taking the limit as 1-2x-4x^3 goes to infinity is negative infinity, so as it goes to negative infinity, it goes to positive infinity.

    b) see a)

    e) note that a critical point does NOT imply extrema value! Extrema value -> critical point is true, but not the other way. A local extreme value will be increasing on one side and decreasing on the other. Generally, you can use a first derivative check for this.
    Take a look at the graph of x^3 and you'll see what I mean

    Those are some pointers, you can try some more - I will be back if you are hopelessly stuck
     
    Last edited: Nov 19, 2009
  4. Nov 19, 2009 #3
    (a) I got stuck at [because my algebra stinks]:
    1/2 = x(1+2x^2)

    So for Y1, I put 1-2X-4x^3
    Then Y2, I put 0
    Calc>Intersect> and got X= .385

    I see that when the line is increasing, it's going to -infinity? Is that why it's -infinity (because the X is decreasing)?

    (b) Makes sense after (a).

    (e) Okay so I can't use critical points to find it. But I don't know how to use the first derivative to find the local extreme values.. ;~;
     
  5. Nov 19, 2009 #4
    the largest power term is odd, and it's -(x^odd) so large negative numbers makes that largely positive. Also, as I said before, f'(x) only crosses the x-axis once, so if you had a number to the left of f'(x) = 0 that is positive, then the more left you go, it will still be positive (since it will never go back cross the x-axis and become negative)

    you do use critical points to find the extreme points, I just meant that f'(x) = 0 is NOT ENOUGH. To show that it is a local min/max, it must be 1) critical point and 2) increasing/decreasing on alternating sides

    good luck!
     
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