# What is the function f(x) if f(x)=f'(x)*f"(x), x is in the real number set?

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### 1 Answer

We'll consider the degree of the function f(x) as being n.

If we'll multiply 2 polynomials, the exponents of matching variables are adding.

If the grade of f(x) is n, then the grade of f'(x) is (n-1) and the grade of f"(x) is (n-2) => n= n-1+n-2 => n=3

Therefore the order of the function f(x) is n=3.

f(x)= ax^3 + bx^2 +cx+d

f'(x)=3ax^2+2bx+c

f"(x)=6ax+2b

f(x)=f'(x)*f"(x)

ax^3 + bx^2 +cx+d=(3ax^2+2bx+c)(6ax+2b)

Comparing, we'll get:

a=18a^2, a=1/18

b=18ab

b=18*(1/18)*b, therefore b may be any real number

c=4b^2+6ac => c=6b^2

d=2bc => d=12b^3

**The requested function f(x) is f(x)= (1/18)x^3 + bx^2 +6bx+12b^3.**