Curve $C$ - how can calculus help?

Curve \(C\), the red curve, does not look like one of the typical cubic shapes we have seen before. However, it has rotational symmetry about \((0,1)\).

Plot of the red curve. It intersects the y axis at 1.

It has some similarities to the curve \(y=x^3+1\) but it does not have the same key feature at \(x=0\).

What are the features of this curve compared to \(y=x^3\) or a translation of it?

Curve \(C\) intersects the \(x\)-axis once and the \(y\)-axis at \((0,1)\).

As the value for \(x\) increases, the value for \(y\) increases.

However, it does not appear to have the same shape as the \(y=x^3+1\) curve at the point \((0,1)\).

We could try to form a set of simultaneous equations to find a cubic equation, but would this be an efficient way to find an equation?

Calculus of curve \(C\)

The curve has no stationary points.

If a cubic curve given by a function \(y=g(x)\) has no stationary points then what do we know about \(g'(x)\)?

We know that the equation \(g'(x)=0\) has no real solutions. Since \(g(x)\) is a cubic function, \(g'(x)\) will be a quadratic expression that forms an equation \(g'(x)=0\) with no real solutions.

So we can use our knowledge of calculus and quadratic equations to help construct a cubic curve with the same properties as curve \(C\).

We know that the quadratic equation \(x^2+1=0\) has no real solutions, so we could take \(g'(x)=x^2+1\).

Using integration we find that \(g(x)=\dfrac{1}{3}x^3+x+k\), where \(k\) is an arbitrary constant.

We could therefore choose \(y=\dfrac{1}{3}x^3+x+1\) as our cubic curve equation, since the curve \(C\) cuts the \(y\)-axis at \(y=1\).

Is this an equation of curve \(C\)? We may need to manipulate \(y=\dfrac{1}{3}x^3+x+1\) so that it goes through the points \((-1,-2)\) and \((1,4)\) as the curve \(C\) does.

Have a look back at where the coefficients for your cubic have come from. Can we alter the coefficients without the argument above breaking down?

After using this resource you could take a look at Can you find… curvy cubics edition.