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Math Help - Consider the cubic y = f(x) = x^2(ax-b)

  1. #1
    Zel
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    Exclamation Consider the cubic y = f(x) = x^2(ax-b)

    Consider the cubic y = f(x) = x^2(ax-b) where a > 0 and b < 0
    Sketch the graph of y = f(x)
    How would I do this?

    Find the coordinates of the local maximum, local minimum and inflection points in terms of a and b.
    I know that the first and second derivative needs to be calculated, then they are found when f'(x) and f''(x) equal 0
    Any help given will be greatly appreciated
    Thanks in advance
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  2. #2
    Member mohammadfawaz's Avatar
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    f(x) = ax^3-bx^2
    f'(x) = 3ax^2-2bx.
    f'(x)=0 gives x = 0 or x=\frac{2b}{3a} which is known to be negative since a>0 and b<0.
    If you look at the sign of f'(x), you see that it is positive before \frac{2b}{3a}, negative between \frac{2b}{3a} and 0, and then positive again after zero. This comes from the fact that the coefficient of x^2 is positive This implies that the function is increasing, decrease and then increasing. To sketch your graph, compute f at zero and at \frac{2b}{3a} and then mark the points you got as local maximum and local minimum. Now draw the graph following the variations of f.
    For the inflection point, f''(x)=6ax-2b. Setting this to zero gives x=\frac{b}{3a}. Plugging this into f gives the coordinates of the inflection point.
    hope this helps
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