# Establish all polynomial functions, which have the property f(x) = f'(x)*f"(x), x is in the real number set.

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### 2 Answers

We will take tentatively : f(x) = ax^n.

f'(x) = a*nx^(n-1)

f"(x) = an(n-1)x^(n-2)

So f(x) = f'(x)*f"(x) gives:

a x^n = a * n x^(n-1)}{an(n-1) x^(n-2)}

ax^n = a^2* n^2(n-1) x^(n+n-2) = a^2*n^2(n-1)x^(2n-3)

So equating the powers and coefficients,

n = 2n-3.

Or 2n-n = 3

n =3.

a = a^2n^2(n-1)

1 = a*3^2(3-1) .

a = 1/18.

So of f(x) = (1/18)x^3

f'(x) = 3/18x^2 = (1/6)x^2

f"(x) = (2/6)x = (1/3)x

f(x) = f'(x)*f"(x) = (1/6)x^2 * (1/3)x^2 = (1/18)x^3

Now again we assume that

f(x) = 1/18x^3+bx^2+cx+d.

f'(x) = (3/18)x^2+2bx+c .

f"(x) = (6/18)x+2b.

So f'(x)*f"(x) = {(3/18)x^2 + 2bx+c}{(6/18)x+2b}

= x^3+{(6b/18)+(12b/18)}x^2+{4b^2+6c/18)x+2bc . This should be identitically equal to f(x). Now equating the coefficients of the like powers, we get:

x^3 : both sides agree and is equal to 1

x^2: both sides agree and equal to b.

x : (4b^2+6c/18) = c,

4b^2+c/3 =c

12b^2 +c =3c, Or 2c= 12b^2, c = 6b^2.

Constant term: 2bc = d. So d = 2b(6b) = 12b^2.

Therefore the required polynomial is

f(x) = (/18)x^2+bx^2+6b^2*x+12b^2 which satisfies

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

Assuming that f(x)'s grade is n and based on the fact that if we multiply 2 polynomials, their grades are adding, in order to find the resulting grade, we'll calculate:

n= n-1+n-2, n=3

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).

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

Let's calculate the first derivative:

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

We'll calculate the second derivative:

f"(x)=6ax+2b

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

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

The expressions above are equal if and only if the corresponding quotients are equals.

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

b=18ab=18*(1/18)b, so **b could be any real number**

c=4b^2+6ac, **c=6b^2**

d=2bc, **d=12b^3**

**f(x)= (1/18)x^3 + bx^2 +6bx+12b^3**