If #f(x)# is of first degree its second derivative is identically null, so also #f(x)# would have to be identically null. to satisfy the equation #f(3x) = f'(x)f''(x)#
Let then #f(x)# be a generic polynomial of degree #n >= 2#. Then #f'(x)# will have degree #(n-1)# and #f''(x)# degree #(n-2)#
Now, the product #f'(x)*f''(x)# is a polynomial of degree #(n-1)+(n-2)# and as two polynomials can be equal for every #x# only if they have the same degree:
#(2) " " f(3x) = f'(x)*f''(x)#
implies:
#n = (n-1)+(n-2)#
#n = 2n -3#
#n=3#
that is #f(x)# must be of third degree:
#f(x) =ax^3+bx^2+cx+d#
#f'(x) = 3ax^2+2bx+c#
#f''(x) = 6ax+2b#
then #(2)# becomes:
#27ax^3+9bx^2+3cx+d = (3ax^2+2bx+c)(6ax+2b)#
#27ax^3+9bx^2+3cx+d = 18a^2x^3+12abx^2+6acx+ 6abx^2+4b^2x+2bc #
#27ax^3+9bx^2+3cx+d = 18a^2x^3+18abx^2+(6ac+4b^2)x+ 2bc #
Equating the coefficients of the same degree we get:
#27a = 18a^2#
and so as #a!=0#
#a=3/2#
Then:
#9b = 18ab = 27b#
#b=0#
at the first degree:
#3c = 6ac+4b^2#
#3c = 9c #
#c = 0#
and finally:
#d= 2bc = 0#
The polynomial which satisfies the equation is then:
#f(x) = 3/2x^3#
so that:
#f'(x) = 9/2x^2#
#f''(x) = 9x#
#f(2) = 12#
#f'(2) = 18#
# f''(2) = 18#
and only the second statement is correct.
