Results 1 to 14 of 14

Math Help - Finding the absolute max and min?

  1. #1
    Newbie KSmathstudent's Avatar
    Joined
    Jan 2010
    From
    US
    Posts
    7

    Finding the absolute max and min?

    The problem:
    Find the absolute maximum and absolute minimum of f on the interval (-1,2]:
    f(x)= (-x^3 + x^2 + 3x +1) / (x+1)
    The Answer Choices:
    a) Max: (1, -2) Min: (-1, 2)
    b) Max: (1, -2) Min: None
    c) Max: None Min: None
    d) Max: None Min: (-1,2)
    e) None of these
    My Work:
    1) I found the derivative: ( -2x^3 - 2x^2 + 2x + 2 ) / (x+1)^2
    2) I set the derivative equal to zero and solved for x, getting my critical points: x=1,x=2,x=3.
    3) Since the interval is between (-1,2], I crossed out x=3.
    4) I plugged in my critical points: x=1 and x= 2, and my end points: x=-1 and x=2 into f(x).
    5) Results: f(-1)=undefined; f(1)=(2); f(2)=1.


    So how do I determine which point is the max and which point is the min?
    I tried my best at solving this myself;however I feel stuck after going this much through it. Could anyone review my work and tell me what I did wrong or what I need to do next in order to find which are my max and min points?

    Thank you for any answers.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by KSmathstudent View Post
    The problem:


    So how do I determine which point is the max and which point is the min?
    find f''(x) to determine the nature of the stationary points.

    f''(x)>0 \implies min

    f''(x)<0 \implies max
    Last edited by pickslides; January 3rd 2010 at 04:24 PM. Reason: bad latex
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    4
    I believe your derivitive is wrong. Shouldn't it be (-3x^2+2x+3)/(x+1)^2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie KSmathstudent's Avatar
    Joined
    Jan 2010
    From
    US
    Posts
    7
    Quote Originally Posted by Helpaguyout View Post
    I believe your derivitive is wrong. Shouldn't it be (-3x^2+2x+3)/(x+1)^2
    I tried finding the derivative again & I still got the same derivative I had before.
    I used the quotient rule to find my derivative.
    The quotient rule: u=(-x^3+x^2+3x+1) and v=(x+1); so (u'v-uv')/(v^2).
    Am I perhaps doing it wrong?
    How did you do yours?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,047
    Thanks
    888
    note that on the interval (-1,2] ...

    f(x) = \frac{-x^3 + x^2 + 3x +1}{x+1} = -x^2+2x+1

    f'(x) = -2x+2
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie KSmathstudent's Avatar
    Joined
    Jan 2010
    From
    US
    Posts
    7
    Quote Originally Posted by skeeter View Post
    note that on the interval (-1,2] ...

    f(x) = \frac{-x^3 + x^2 + 3x +1}{x+1} = -x^2+2x+1

    f'(x) = -2x+2
    I am a tad bit confused.
    Why does f(x) = \frac{-x^3 + x^2 + 3x +1}{x+1} = -x^2+2x+1 ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Dear KSmathstudent,

    I think the answers to you questions lies in the attachment below. But if you find the attachment unreadable,or if you have further questions please don't hesitate to reply me.

    Hope this helps.
    Attached Thumbnails Attached Thumbnails Finding the absolute max and min?-dsc02507.jpg  
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie KSmathstudent's Avatar
    Joined
    Jan 2010
    From
    US
    Posts
    7
    Quote Originally Posted by Sudharaka View Post
    Dear KSmathstudent,

    I think the answers to you questions lies in the attachment below. But if you find the attachment unreadable,or if you have further questions please don't hesitate to reply me.

    Hope this helps.
    Thank you. That was very helpful and I can see what I did wrong.
    But I'm still a bit unsure about how to determine which is the minimum and which is the maximum.
    How did you know (1,2) was the maximum and not the minimum?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by KSmathstudent View Post
    I am a tad bit confused.
    Why does f(x) = \frac{-x^3 + x^2 + 3x +1}{x+1} = -x^2+2x+1 ?
    \frac{-x^3 + x^2 + 3x +1}{x+1} = -x^2+2x+1

    just as 6 \div 3 = 2

    You need to learn polynomial long division
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie KSmathstudent's Avatar
    Joined
    Jan 2010
    From
    US
    Posts
    7
    Quote Originally Posted by pickslides View Post
    find f''(x) to determine the nature of the stationary points.

    f''(x)>0 \implies min

    f''(x)<0 \implies max
    Using Sudharaka's post and applying this rule, the 2nd derivative would be -2. Since -2 is less than 0, is that the reason why (1,2) is the maximum?
    I think this is starting to make more sense to me now.

    Thank you all for the help.I appreciate it stupendously!
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by KSmathstudent View Post
    But I'm still a bit unsure about how to determine which is the minimum and which is the maximum.

    find to determine the nature of the stationary points.

    min

    f''(x)<0 \implies max
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Dear KSmathstudent,

    Think of it this way,
    Now you can see that when x<1; dy/dx>0 That means that the graph of
    f(x) when x<1 has a positive tangent. When x>1:dy/dx<0 that means that the graph of f(x) when x>1 has a negative tangent. So you see that clearly f(x) has a maximum at (1,2)

    The second derivative test is also correct. But I think this is the easy way.

    Please tell me whether this solves your problem.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie KSmathstudent's Avatar
    Joined
    Jan 2010
    From
    US
    Posts
    7
    Quote Originally Posted by pickslides View Post

    just as 6 \div 3 = 2

    You need to learn polynomial long division
    Thank you, but I understand it now.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie KSmathstudent's Avatar
    Joined
    Jan 2010
    From
    US
    Posts
    7
    Quote Originally Posted by Sudharaka View Post
    Dear KSmathstudent,

    Think of it this way,
    Now you can see that when x<1; dy/dx>0 That means that the graph of
    f(x) when x<1 has a positive tangent. When x>1:dy/dx<0 that means that the graph of f(x) when x>1 has a negative tangent. So you see that clearly f(x) has a maximum at (1,2)

    The second derivative test is also correct. But I think this is the easy way.

    Please tell me whether this solves your problem.
    Yes, it makes sense.
    Thank you all for the answers.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. finding r, absolute value of (sqrt3 + i)
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: January 1st 2011, 04:15 PM
  2. finding absolute max and mins
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 29th 2010, 03:22 PM
  3. finding absolute extrema
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 3rd 2010, 03:37 AM
  4. finding absolute maximum and absolute minimum
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 15th 2009, 05:22 PM
  5. Finding absolute extremaa
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 26th 2009, 05:34 PM

Search Tags


/mathhelpforum @mathhelpforum