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Calculus, derivatives (curve sketching)

  1. Jan 5, 2014 #1
    1. Find a cubic function f(x)=ax3+bx2+cx+d that has a local maximum value of 3 at x=-2 and a local minimum value of 0 at x=1.

    2. Relevant equations $$f'(x)=0$$
    3. The attempt at a solution

    The first thing I did was taking the derivative of f(x). $$f'(x)=3ax^2+2bx+c$$ I know that you can get the critical numbers by setting it to zero, and those values should equal the local max and min values. So, if $$f'(x)=3ax^2+2bx+c=0$$, then that would mean $$x=-2, 1$$ $$(x+2)(x-1)=0$$ $$x^2+x-2=0$$ Thus, a=[itex]\frac{1}{3}[/itex], b=[itex]\frac{1}{2}[/itex], and c=-2. Putting that back into f(x) I get $$f(x)=\frac{1}{3}x^3+\frac{1}{2}x^2-2x+d$$ I know that a minimum value is at (1,0), so plugging that in I get $$0=\frac{1}{3}+\frac{1}{2}-2+d$$ $$d=\frac{7}{6}$$ My final answer being $$f(x)=\frac{1}{3}x^3+\frac{1}{2}x^2-2x+\frac{7}{6}$$ However, my answer key says that it should be $$f(x)=\frac{1}{9}(2x^3+3x^2-12x+7)$$ What did I do wrong? Any help is much appreciated.
     
    Last edited: Jan 5, 2014
  2. jcsd
  3. Jan 6, 2014 #2

    Simon Bridge

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    Note: your equation is $$f(x)=\frac{1}{6}(2x^3+3x^2-12x+7)$$... spot the similarities?

    Check by putting x=-2 to see who is right.

    I think you may have made your mistake when you compared ##y'=3ax^2+2bx^2+c=0## with ##(x+2)(x-1)=0##.

    This does not uniquely determine the constants.
    Consider:

    $$x^2+\frac{2b}{3a}x + \frac{c}{3a}=0$$ ... is y'=0 too.

    Compare with ##x^2+x-2=0## tells you that ##2b=3a## and ##c=-6a## ...
     
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