Sunday, September 20, 2009

week 5 post

This week in calculus, we mostly just practiced stuff we already knew so that we could do those study guides and take home tests and stuff... but I still struggled a lot with it. But anyway, one thing that I specifically learned how to do this week was finding an absolute max or min. steps and procedure to do this:

1.First derivative test.
2.Plug critical values into original function to get y values.
3.Plug endpoints into original function to get y values.
4.Highest y value will be absolute max. – Lowest y value will be absolute min.

Note: Absolute max's and min's are written as a point or a simple y value.

Example: Find the absolute max or min of f(x)= 3x^(4)-4x^(3) on [-1,2] (-1) less than or equal to (x) less than or equal to (2)

First, you take the derivative.

F' (x)= 12x^(3)-12x^(2) = 0 which simplifies to (12x^(2))(x-1)=0. So x=1,0

so you have to plug in numbers between these points (-1,0) u (0,1) u (1,2) into the first derivative:

so, f'(-.5)= negative # f'(.5)= negative # f'(1.5)= positive #

so then you plug in your endpoints into the original function:

3(1)^(4)-4(1)^(3)= -1 so, (1,-1)
3(-1)(4)-4(-1)^(3)=7 so, (-1,7)
3(2)^(4)-4(2)^(3)= 3(16)-4(8)=16

so your absolute min is (1,-1), or -1, or -1 at x=1. They all mean the same thing.

And your absolute max is (2,16) or 16, or 16 at x=2. Once again, they all mean the same thing.

NOTE: there is a shortcut to this, but I don't completely understand it.

So my question is, can someone show me the shortcut to finding the absolute max's and min's?

Ok thanks and good night everyone. God help us on tomorrow's test...

3 comments:

  1. ok, to find absolute max and min...you do the same thing as for first derivative test.

    so 3x^3 - 36x on (-1,1)

    you would take the derivative first...
    which is
    9x^2-36

    when you solve after setting equal to zero you'll get:
    x^2 = 4
    then x=2

    now that you have x=2 and in the problem it says (-1,1), you'll plug in those three numbers into the original

    so when you plug in -1 you get 33
    when you plug in 2 you get -48
    and when you plug in 1 you get -33

    so looking at the answers that you got when you plugged in your ABSOLUTE MIN would be (2,-48)
    and your ABSOLUTE MAX would be (-1,33)

    but remember if you have a problem that you end up with like...1,1...2,4...and -1,1 then there is no ABSOLUTE MIN because there is two smallest numbers..and the ABSOLUTE means only one!!!

    HOPE THIS HELPED!

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  2. ok lets say we have the function 4x^2-6x. on (-2,3). First we take the derivative of the function, which would be

    8x-6

    next we solve for x= 3/4

    Now we set that up into intervals:
    (-2, 3/4) u (3/4, 2)

    now we plug in our critical number and our endpoints into the original function.

    so for the first interval lets plug in 3/4.
    f(3/4)= 4(3/4)^2-6(3/4), that gives us -9/4 so so far we have (3/4, -9/4)

    now lets plug in -2
    f(-2)=4(-2)^2-6(-2) which gives us 28. so we have (2,28)

    and finally we plug in 3.
    f(3)=4(3)^2-6(3)which gives us 18, so we have (3,18).

    so the intervals we found by plugging in are
    (3/4, -9/4)(2,28) (3,18)
    so by looking at this we have an absolute max at (2,28) and an absolute min at (3/4, -9/4) hope this helped!

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  3. The shortcut for absolute maximums and minimums is really easy if you just remember the steps.

    First you take the first derivative, then you set the first derivative equal to zero. Once you have your critical points from the first derivative, instead of creating intercepts with them, you simply take the second derivative and plug the critical points into it. Be sure if there are endpoints you plug these into your second derivative as well as they may become possible maximums and minimums.

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