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About curve sketching LEAVING CERT!!!

  • 15-06-2006 9:00am
    #1
    Peter Burnes
    Closed Accounts Posts: 7 ✭


    Yo g's!
    What type of answers did you get for that curve sketching question in Paper 1? for part 1 and 2?

    The equation of a curve is 3x^4-2x^3-9x^2+8
    Part 1 Show that the curve has a local maximum point at (0,8)
    -I just plugged in the values....was that right? Because when i differentiated i was still left with a polynomial which didn't seem to factorise right.
    Any thoughts?

    Part 2 Find the co-ordinates of the two local minimum points on the curve....
    - I wasn't too sure about this... the 3x^4 put me off...How did you go about it?

    What were the thoughts on the paper.... I thought the induction part in question 6 was real mean!!

    stAy :cool:


Comments

  • #2
    seandoiler
    Registered Users, Registered Users 2 Posts: 107 ✭✭seandoiler


    f(x)=3x^4-2x^3-9x^2+8
    f'(x)=12x^3-6x^2-18x
    factor out 6x to get f'(x)=6x(2x^2-x-3), solving for zero gives x=-1,0,3/2, these are the x-values of the extremas....everything now follows from here, calculate second derivative and plug in above x-values to check max or min


  • #3
    Peter Burnes
    Closed Accounts Posts: 7 ✭Peter Burnes


    Ah that's how it's done! Dang! Thanks anyhow!!


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