1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Absolute Maximum and Minimum Values over a given interval

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the absolute maximum and absolute minimum values of the function

    on each of the following interval

    3. The attempt at a solution
    I got this and its saying its wrong.
    y'= (3x^2)+(24x)-27
    factor: 3(x^2+8x-9) so x=-9 and x=-1
    I plug in -9 into there and come up with -30 as the minimum which is not correct?
    Anyone have some input?
    Thanks! Dan
  2. jcsd
  3. Jul 17, 2011 #2
    You need to plug those numbers into the original equation. -9 is actually going to be a max value over your interval, not minimum. What do you get when you solve [itex]y = x^3 + 12x^2 - 27x + 9[/itex] for when x = -9?

    Also, -1 is not a solution. Check your factoring again:

    [itex]0 = x^2 + 8x - 9[/itex]
    [itex]0 = ( x - 1 )( x + 9 )[/itex]
    Last edited: Jul 17, 2011
  4. Jul 17, 2011 #3
    Thanks! I knew I had to plug it back in but for some reason I was getting the wrong answer when I plugged it in the equation the first time hahah. I managed to get all the other questions done except for this, I know I need to do the quotient rule but the equation I get for y' doesnt make any sense.
    Find y' and the critical numbers for
    (4x+1)/(x^2+x+1) I end up getting (4x^2+12x+2)/(x^2+x+1)^2 which I can factor the top and get (2(2x^2+12x+2))/(x^2+x+1)^2.

    To me this is y' but I dont quite see how to get the critical numbers?
  5. Jul 17, 2011 #4
    Critical numbers come from multiple places. A critical number is a location where min/max is possible. Ergo, max/min is always a critical number. By finding all critical x points, you can find the max/mins of the function by finding each corresponding y value. Above, our critical numbers were -9, 1, -10, and 0.

    -A critical number is at the start and end of your interval, but you didn't list an interval. If they didn't give you one, then ignore that.

    -A critical number is also any point where y' = 0, which we found above.

    You can find the critical numbers by finding y' and setting y' = 0. However, your derivative is incorrect. Make sure you are applying the quotient rule correctly, or you can show your work if you want and I'll see if I can find the error.

    So, just try again at finding the derivative of y = (4x+1)/(x^2+x+1), then find the x values where y' = 0. You'll need to use the quadratic formula for this one.
    Last edited: Jul 17, 2011
  6. Jul 17, 2011 #5
    for the derivative per the quotient rule I got [(4x+6)(2x+1)-(4)(x^2+x+1)]/(x^2+x+1)^2 from there I multiplied out the top and got (4x^2+12x+2)/(x^2+x+1)^2 then I factored out a 2 from the top to give the equation i gave in my last post. I'm a little weary about factoring out that 2 though as in is it correct? Or should I just use the quadratic formula on (4x^2+12x+2)/(x^2+x+1)^2? If so wouldnt I use the quadratic on top, get the possible values of x then set the bottom equal to 0 and solve for the bottom? Uggh this problem is driving me nuts lol
  7. Jul 17, 2011 #6
    never mind lol I got it I used the quaratic on the top part of (4x^2+12x+2) and got the values and those ended up being my critical numbers. One thing though how come I didnt need to factor and solve the bottom? I thought that when a derivative was in the form of a fraction the top and bottom had to be solved for to get all the critical zeros?
  8. Jul 17, 2011 #7
    Your mistake on the quotient rule is only that you must put the nonderived denomenator times the derived numerator first, the order matters since you're subtracting.

    You have:

    Try instead:
    [(4)(x^2+x+1) - (4x+6)(2x+1)]/(x^2+x+1)^2

    You don't need to factor and solve the bottom because you're setting everything equal to 0. In order for a ratio to be equal to zero, the numerator must be equal to zero, so we just set the numerator to be zero and discard the bottom (alternatively, we can multiply both sides by the denominator, which is the same thing.)

    [itex]y = \frac{4x+1}{x^2+x+1}[/itex]

    [itex]y' = \frac{(4)(x^2+x+1) - (4x+1)(2x+1)}{(x^2+x+1)^2}[/itex]
    [itex]y' = \frac{4x^2+4x+4-8x^2-4x-2x-1}{(x^2+x+1)^2}[/itex]
    [itex]y' = \frac{-4x^2-2x+3}{(x^2+x+1)^2}[/itex]

    Always remember that! For the quotient rule, the denominator times the derivative of the numerator comes first, always! It's a very common error.

    [itex]\frac{-4x^2-2x+3}{(x^2+x+1)^2} = 0[/itex]
    [itex]-4x^2-2x+3 = 0[/itex]
    Last edited: Jul 17, 2011
  9. Jul 17, 2011 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The critical values of y' are at x = -9 and x = 1. However, x = 1 is outside your stated interval [-10,0]. To find max/min in a case like this (i.e., with *bound* constraints on x) you need to examine both interior stationary points, and the end points as well. Here, there are two local minima (at x = -10 and at x = 0) and one local maximum (at x = -9). The left-hand end-point x = -10 is a local mininum because y' > 0 there; the right-hand end-point x = 0 is a local minimum because y' < 0 there. Point x = 0 is the absolute minimum on the interval [-10,0] because y(0) < y(-10); that is, y(0) is smaller than any other value y(x) in the interval. The point x = -9 is the absolute maximum on the interval.

    The moral of the story is: in a *constrained* problem, the derivative does not need to vanish at a max or min that lies on the boundary of the constraint region. (For higher dimensions, we get involved with the so-called Karush-Kuhn-Tucker conditions, which are basically multivariate versions of the type of end-point conditions above.)

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Absolute Maximum and Minimum Values over a given interval