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The function \\(f(x)=2x^3-\\simplify{3*({a}+{b})x^2+6*{a}*{b}x}+\\var{c}\\)  has two turning points.

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\\(f(x)=2x^3-\\simplify{3*({a}+{b})x^2+6*{a}*{b}x}+\\var{c}\\)

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To locate a turning point, we need to find points where the derivative $f'(x)=0$

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First we find $f'(x)$:

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\\(f'(x)=6x^2-\\simplify{6*({a}+{b})x+6*{a}*{b}}\\)

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To find the turning points, we must set this to be equal to zero and then solve for $x$. i.e. 

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Solve $6x^2-\\simplify{6*({a}+{b})x+6*{a}*{b}}=0$

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Divide across by 6 to get the quadratic equation

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\\(x^2-\\simplify{({a}+{b})x+{a}*{b}}=0\\)

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This has factors

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\\((x-\\var{a})(x-\\var{b})=0\\)

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\\(x-\\var{a}=0\\)     or     \\(x-\\var{b}=0\\)

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\\(x=\\var{a}\\)     or     \\(x=\\var{b}\\)

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Input the smaller of the two \\(x\\) values. 

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\\(x=\\) [[0]]

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Input the larger of the two \\(x\\) values.

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\\(x=\\) [[1]]

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Turning points of a cubic function

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