Function problem 1
A non-constant polynomial function f ( x ) is such that f ( 2 x ) = f ′ ( x ) f ′ ′ ( x ) , then find the value of f ( 3 ) .
The answer is 12.
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1 solution
You need the condition that f ( x ) is not a constant function. Otherwise, f ( x ) = 0 clearly satisfies the condition that f ( x ) = f ′ ( x ) f ′ ′ ( x ) , and the value of f ( 3 ) = 0 .
The place where your solution breaks down, is in considering the degree of f ( x ) . If f ( x ) is a non-zero constant, then the degree is 0, and the degree of f ′ , f ′ ′ are both 0, instead of n − 1 , n − 2 . Furthermore, if f ( x ) = 0 , then the degree is taken as − ∞ .
I think the problem has multiple solutions. For instance, consider f(x)=e^x. e^x *e^x = e^2x which satisfies the parameters. So an alternate solution is e^3.
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The question states that the function is a polynomial.
Yes, there are other non-polynomial function which satisfy this differential equation.
Suppose degree of f (x) = n. Then deg f'=n - 1 and deg f"=n - 2 , so n =n - 1 + n - 2. Hence n= 3. So put f(x)= ax^3+bx^2+cx+d. (a not equal to 0). Now using f (2x) = f' (x) . f" (x) Then we have 8ax^3 + 4bx^2 + 2cx + d = (3ax^2 + 2bx +c)(6ax + 2b) = 18a^2 x^3 + 18abx-' + (6ac + 4b^2)x + 2bc. Comparing x^3, 18a^2 = 8a - (1) implies a > 4/9 Comparing x^2, 18ab > 4b -(2) implies b > 0 Comparing x, 2c = 6ac + 4b^2 -(3) implies c:=0 Comparing constant term, a = 2bc implies d = 0 f'(x)=4x^ 3 / 9 :. f(3) = 12