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.