Part (a)
g(x) = (x-a)^2(x-b)g(x)=(x−a)2(x−b)
g'(x) = 2(x-a)(x-b)+(x-a)^2(1)
= (x-a)(3x-(a+2b))
g'(x) = 0 at x=a and at x = (a+2b)/3
The stationary points are
(a,0) and ((a+2b)/3, f((a+2b)/3)) = ((a+2b)/3, (4(a-b)^3)/27)
Part (b) cannot be read and part (d) cannot be answered without knowing what p is.
Part (d)
IF we interpret "there are two solutions" to mean "there are exactly two distinct solutions", and we assume that a != b, then we have
p = 0 which occurs at (a,0) and (b ,0)
or p = (4(a-b)^3)/27 which occurs at x= (a+2b)/3 and at another value of x
That other value of x may be found by using the fact that we know one solution (in fact a solution of multiplicity 2) of the equation g(x) = (4(a-b)^3)/27. If nothing better occurs to us, we can do the division to factor g(x) - (4(a-b)^3)/27
