Find the coordinates of each turning point on the graph of \(y=3x^4-16x^3+18x^2\) and determine in each case whether it is a maximum point or a minimum point.

We find that \(\dfrac{dy}{dx}=12x^3-48x^2+36x=12x(x^2-4x+3)=12x(x-1)(x-3)\).

Since turning points occur where \(\dfrac{dy}{dx}=0\), the turning points are at \(x=0,1\) or \(3\).

When \(x=0\) we have that \(y=0\), when \(x=1\) we find that \(y=5\), and when \(x=3\), \(y=-27\).

Therefore the turning points are \((0,0),\) \((1,5)\) and \((3,-27)\).

We can consider \(y'=\dfrac{dy}{dx}\) at different values of \(x\) to identify if a point is a minimum or a maximum.

At \(x=-\dfrac{1}{2}\) we observe that \(\dfrac{dy}{dx}<0\) and, at \(x=\dfrac{1}{2}, \dfrac{dy}{dx}>0\). This shows us that \((0,0)\) is a minimum point.

At \(x=\dfrac{3}{2}\) we can see that \(\dfrac{dy}{dx}>0\), so considering the result above at \(x=\dfrac{1}{2}\) this shows that \((1,5)\) is a maximum.

It is evident that for all \(x>3\), \(\dfrac{dy}{dx}>0\), implying that \((3,-27)\) is a minimum point.

Or else we could say that \(y'' = 36x^2-96x+36 = 12(3x^2-8x+3)\).

So \(y''(0)\) is positive, \(y''(1)\) is negative and \(y''(3)\) is positive.

This means \((0,0),\) is a minimum, \((1,5)\) is a maximum, and \((3,-27)\) is a minimum.

Sketch the graph of \(y=3x^4-16x^3+18x^2\), and state the set of values of \(k\) for which the equation \(3x^4-16x^3+18x^2=k\) has precisely two real roots for \(x\).

From the information that we have, we can sketch the graph of \(y=3x^4-16x^3+18x^2\) as follows.

By considering the values of \(k\) for which there are exactly two intersections between the graph we have just sketched and \(y=k\), we find that the values required are \[-27<k<0, \quad \text{and} \quad 5<k.\]