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$f(x)=\\simplify[all,!collectNumbers,!noleadingminus]{{a}x^3+{b}x^2+{c}x+{d}}$

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$f'(x)=$ [[2]]

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$f''(x)=$ [[3]]

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Find when $f'(x)=0$, hence find:

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$x$-coordinate of the stationary point giving a minimum $=$ [[0]]

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$x$-coordinate of the stationary point giving a maximum $=$ [[1]]

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On differentiating we get $\\displaystyle \\frac{df}{dx}=\\simplify[std]{{3*a}x^2+{2*b}x+{c}}$.

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To find the stationary points we have to solve $\\displaystyle \\frac{df}{dx}=0$ for $x$.

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So we have to solve $\\simplify[std]{{3*a}x^2+{2*b}x+{c}=0}$.

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Note that the quadratic factorises and the equation becomes $\\simplify[std]{({3a}x-{r1})(x-{r2})=0}$.

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Hence we have two stationary points: $x=\\simplify[std]{{r1}/{3a}}$ and $x=\\var{r2}$.

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To find out the types of these stationary points we look at the sign of $\\displaystyle \\frac{d^2f}{dx^2} = \\simplify{{6a}*x+{2*b}}$ at  the stationary points.

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If  $\\displaystyle \\frac{d^2f}{dx^2} \\lt 0 $ at a stationary point then it is a MAXIMUM.

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If  $\\displaystyle \\frac{d^2f}{dx^2} \\gt 0 $ at a stationary point then it is a MINIMUM.

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If  $\\displaystyle \\frac{d^2f}{dx^2} = 0 $ at a stationary point then we have to do more work!

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At $x=\\var{r2}$ we have $\\displaystyle \\frac{d^2f}{dx^2} = \\simplify{{6*a*r2+2*b}}${lg1}$0$ hence is a {type1}.

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At $\\displaystyle x=\\simplify[std]{{r1}/{3a}}$ we have $\\displaystyle \\frac{d^2f}{dx^2} = \\simplify{{2*r1+2*b}}${lg2}$0$ hence is a {type2}.

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Finding the stationary points of a cubic with two turning points

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Find the coordinates of the stationary points of the function.

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