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Math Help - Critical points/ Abs. max/min values

  1. #1
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    Critical points/ Abs. max/min values

    Hey guys so I need help on two questions that I'm sort of stuck on.

    1. Find the critical numbers of the given function: f(x) = 2x3+x2+2x


    So I got f'(x) = 6x2+2x+2

    Factored out the 2(3x2+x+1)

    I can't factor this out anymore, so is the answer there are no critical numbers?



    2. Find the absolute max and min values of f on the given interval.


    √ = square root

    f(t) = 3√t * (8-t), [0,8]

    So I used product rule on this one and got

    f'(t) = 3/2√t * (8-t) - 3√t = 0

    3(8-t)-3√t = 0

    So, I'm uncertain about the critical numbers here as well are they 8 and 0?

    Last edited by Oldspice1212; May 25th 2013 at 03:32 PM.
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  2. #2
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    Re: Critical points/ Abs. max/min values

    1) If you sketch a cubic equation you will see that it has no maximum or minimum.

    2) There's a mistake in your differentiation of t1/3
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    Re: Critical points/ Abs. max/min values

    Quote Originally Posted by Shakarri View Post
    1) If you sketch a cubic equation you will see that it has no maximum or minimum.

    2) There's a mistake in your differentiation of t1/3

    Yeah I just noticed the derivative was wrong for number 2.

    For 1 do I just leave my answer as it is? and write no critical points then?
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    Re: Critical points/ Abs. max/min values

    For the first one the reason is that the quadratic 3x^2+x+1 does not have real roots. Its discriminant is negative so it has imaginary roots and hence no critical points.
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    Re: Critical points/ Abs. max/min values

    Critical points/ Abs. max/min values-26-may-13.png
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    Re: Critical points/ Abs. max/min values

    Quote Originally Posted by ibdutt View Post
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    Quote Originally Posted by ibdutt View Post
    For the first one the reason is that the quadratic 3x^2+x+1 does not have real roots. Its discriminant is negative so it has imaginary roots and hence no critical points.



    Have I told you how much I love you?

    Haha, thanks a lot for both of the questions. You are the best!
    Last edited by Oldspice1212; May 25th 2013 at 09:53 PM.
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    Re: Critical points/ Abs. max/min values

    So the only critical point for question 2 will be 0, and when looking for absolute max/ min, with the intervals, the f(x) would give 0 for both the numbers in the interval. So, is it possible to have a abs. max/min on the same point?
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    Re: Critical points/ Abs. max/min values

    Quote Originally Posted by Oldspice1212 View Post
    So the only critical point for question 2 will be 0, and when looking for absolute max/ min, with the intervals, the f(x) would give 0 for both the numbers in the interval. So, is it possible to have a abs. max/min on the same point?
    No same point can't be both absolute min and max at the same time.

    First you find the critical points c = \{c_1, c_2, c_3, ...\} by letting \frac{d(f(x))}{dx} = 0 and solving for x

    then take one critical point and say it's c and do second derivative test. Second derivative test is:

    if \frac{d(f(c))}{dx^2} > 0 then c is absolute/local minimum.
    or on the other hand if \frac{d(f(c))}{dx^2} < 0 then c is absolute/local maximum.

    Now if this second derivative test shows that you've multiple mins and/or maxes or you're asked to find local min/max in an interval ( x = a to x = b ) then plugin the values of critical points and/or both ends of interval into f(x) and see which one is max and which one is min.

    If second derivative of a function doesn't exist then use First derivative test which you can find by using Google.

    Hope this helps.
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