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Can You Find... Cubic Curves
Age 16 to 18
Challenge Level 
Kieran Fitzgerald fro St Stephens Carramar in Australia got us started on the first parts of the problem using an algebraic approach:
a) For the cubic to have no stationary points the derivative cannot equal $0$
($dy/dx \neq 0$).This means that the discriminant must equal less than $0$. For
this to occur, assuming $y=ax^3+bx^2+cx+d$, both $a$ and $c$ must be either + or -
but they cannot be different. $b$ must also be considerably lower than $a$ and
$c$ as $b^2$ must be less than $4ac$. An example would be $y=4x^3-2x^2+3x$.
b) For a cubic to have stationary points at $x=2$ and $5$ the derivative must
equal $(x-2)(x-5)$. This can then be expanded and anti-differentiated to
become $y=x^3/3-3.5x^2+10x+d$ however $d$ can be any number.
c) In this case the derivative must equal $(x+1)(x+d)$. This is because there
will be a stationary point at $x= -1$ and another at $d$. To make $x=-1$ the
local minimum $d$ must be more than $1$ so that the other turning point is a
maximum. An example would be $y=x^3/3+2x^2+3x$.
Thomas from BHASVIC Sixth Form College Brighton used similar reasoning on the last part of the question:
d) We require $\frac{dy}{dx}=3(x+2)(x-4)$ for one max and one min at $x=-2,4$ in some order. Then integrating, $y=x^3-3x^2-24x$. This has the min and max the wrong way round so use transformation $y \rightarrow -y$ to reflect in $x$-axis, giving $$y=-x^3+3x^2+24x$$ as a possible cubic.
a) For the cubic to have no stationary points the derivative cannot equal $0$
($dy/dx \neq 0$).This means that the discriminant must equal less than $0$. For
this to occur, assuming $y=ax^3+bx^2+cx+d$, both $a$ and $c$ must be either + or -
but they cannot be different. $b$ must also be considerably lower than $a$ and
$c$ as $b^2$ must be less than $4ac$. An example would be $y=4x^3-2x^2+3x$.
b) For a cubic to have stationary points at $x=2$ and $5$ the derivative must
equal $(x-2)(x-5)$. This can then be expanded and anti-differentiated to
become $y=x^3/3-3.5x^2+10x+d$ however $d$ can be any number.
c) In this case the derivative must equal $(x+1)(x+d)$. This is because there
will be a stationary point at $x= -1$ and another at $d$. To make $x=-1$ the
local minimum $d$ must be more than $1$ so that the other turning point is a
maximum. An example would be $y=x^3/3+2x^2+3x$.
Thomas from BHASVIC Sixth Form College Brighton used similar reasoning on the last part of the question:
d) We require $\frac{dy}{dx}=3(x+2)(x-4)$ for one max and one min at $x=-2,4$ in some order. Then integrating, $y=x^3-3x^2-24x$. This has the min and max the wrong way round so use transformation $y \rightarrow -y$ to reflect in $x$-axis, giving $$y=-x^3+3x^2+24x$$ as a possible cubic.
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