Can you find... cubic curves
Can you find equations for cubic curves that have specific features?
Age
16 to 18
Challenge level
Problem
Can you find a cubic curve that...
(a) ... has no stationary points?
(b) ... has two stationary points: one when $x=2$ and one when $x=5$?
(c) ... has a local minimum when $x=-1$?
(d) ... has a local minimum when $x=-2$ and a local maximum when $x=4$?
It would be a good idea to try and sketch some of the cubics first before trying to form an equation.
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Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Student Solutions
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.