Answer: (a) \(f'(x)=2(3x-2)(9x+1)\); (b) \(\left(\frac23,0\right)\), \(\left(-\frac19,\frac{343}{81}\right)\); (d) \(0\lt k\lt \frac{343}{81}\).
Differentiate first. Stationary points occur where the derivative is zero, while tangent and normal problems use the gradient at the given point.
(a) Differentiate using the product rule.
\(f(x)=(2x+1)(3x-2)^2\).
So
\(f'(x)=2(3x-2)^2+(2x+1)\cdot2(3x-2)\cdot3\).
Factor out \(2(3x-2)\):
\(f'(x)=2(3x-2)\big((3x-2)+3(2x+1)\big)\).
Inside the bracket,
\((3x-2)+3(2x+1)=3x-2+6x+3=9x+1\).
Therefore
\(f'(x)=2(3x-2)(9x+1)\).
Hence \(p=9\) and \(q=1\).
(b) Stationary points occur when \(f'(x)=0\).
So
\(2(3x-2)(9x+1)=0\).
Thus
\(x=\frac23\) or \(x=-\frac19\).
When \(x=\frac23\),
\(f\left(\frac23\right)=\left(\frac43+1\right)(0)^2=0\).
When \(x=-\frac19\),
\(2x+1=-\frac29+1=\frac79\),
and
\(3x-2=-\frac13-2=-\frac73\).
Therefore
\(f\left(-\frac19\right)=\frac79\left(-\frac73\right)^2=\frac79\cdot\frac{49}{9}=\frac{343}{81}\).
The stationary points are
\(\left(\frac23,0\right)\) and \(\left(-\frac19,\frac{343}{81}\right)\).
(c) The roots of \(f(x)\) are found from
\((2x+1)(3x-2)^2=0\).
So the graph crosses the \(x\)-axis at \(x=-\frac12\), and touches the \(x\)-axis at \(x=\frac23\), since this is a repeated root.
The \(y\)-intercept is
\(f(0)=(1)(-2)^2=4\).
The cubic has positive leading coefficient, so it goes from lower left to upper right, with a local maximum at \(\left(-\frac19,\frac{343}{81}\right)\) and a local minimum at \(\left(\frac23,0\right)\).
(d) The equation \(f(x)=k\) has 3 distinct real roots when the horizontal line \(y=k\) cuts the cubic in 3 distinct places.
This happens when \(k\) is strictly between the local minimum value and the local maximum value.
The local minimum value is \(0\), and the local maximum value is \(\frac{343}{81}\).
Therefore
\(0\lt k\lt \frac{343}{81}\).
