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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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\nTo locate a turning point, we need to find points where the derivative $f'(x)=0$
\nFirst we find $f'(x)$:
\n\\(f'(x)=6x^2-\\simplify{6*({a}+{b})x+6*{a}*{b}}\\)
\n\nTo find the turning points, we must set this to be equal to zero and then solve for $x$. i.e.
\nSolve $6x^2-\\simplify{6*({a}+{b})x+6*{a}*{b}}=0$
\nDivide across by 6 to get the quadratic equation
\n\\(x^2-\\simplify{({a}+{b})x+{a}*{b}}=0\\)
\nThis has factors
\n\\((x-\\var{a})(x-\\var{b})=0\\)
\n\\(x-\\var{a}=0\\) or \\(x-\\var{b}=0\\)
\n\\(x=\\var{a}\\) or \\(x=\\var{b}\\)
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\nInput the larger of the two \\(x\\) values.
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