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I am struggling with this question please help..

Suppose $f(x)$ is a polynomial function as well as continuous in $\mathbb{R} \to \mathbb{R}$. Given that $f(2x)=f'(x) f''(x)$, then find $f(3)$.

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    Notice that if $f$ is a polynomial of degree $n$, $f'\cdot f''$ is of degree $2n-3$, hence $n=2n-3$ and $n=3$. Therefore, $f(x)=ax^3+bx^2+cx+d$. Now find $a,b,c$ and $d$. – 2017-02-05
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    @wckronholm please edit the title as well [continuos] – 2017-02-05
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    If you could show uniqueness then $f(x) = 0$ is a polynomial function satisfying all hypotheses and $f(3)=0$ :P but apparently $f(x) = \frac{4}{9}x^3$ satisfies the hypotheses as well. So $f(3) = 12$. Hence, there are two solutions. – 2017-02-05
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    Erratum to my comment above: true only if neither $f'$ nor $f''$ is zero (their degree would not be $n-1$, $n-2$). Hence one solution is zero and the other has degree three, as Martingalo points out. – 2017-02-05

2 Answers 2

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Follow-up on my comment. We know that $f(x)=ax^3+bx^2+cx+d$, if neither $f'$ nor $f''$ is zero. Otherwise, you would have $f(x)=0$ for all $x$, which is also a solution.

Expand $f'(x)\cdot f''(x)$:

$$f'(x)\cdot f''(x)=(3ax^2+2bx+c)(6ax+2b)=18a^2x^3+18abx^2+2(3ac+2b^2)x+2bc$$

Equate this to $8ax^3+4bx^2+2cx+d$.

  • Coefficient of $x^3$: $18a^2=8a$ hence $a=\frac{4}{9}$
  • Coefficient of $x^2$: $18ab=4b$, hence a contradiction with the preceding unless $b=0$
  • Coefficient of $x$: $c=3ac+2b^2=3ac$, hence $c=0$
  • $d=2bc=0$

You have thus two solution to the equation $f(2x)=f'(x)\cdot f''(x)$, one is $f(x)=\frac49x^3$ and the other is $f(x)=0$.

Therefore, you can't know $f(3)$ for sure.

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    How does one arrive at the solution $f(x) =0$ without observation? If we look at a synonymous example – say, solving $x^2-x=0$ – then we can factor to get $x(x-1)=0$ and divide by $x$ to get $x=1$, *assuming $x \neq 0$*. It's that assumption that makes us look at the condition $x=0$. Is that what's happening here? I'm trying to see how the condition $n=2n-3$ *naturally* arises with the condition that $n\neq0$ (*naturally* meaning without observation, coming about in the same sense $x=0$ appears in the quadratic above) – 2017-02-05
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    @BrevanEllefsen See my two comments under the question. One gets that $f$ has degree $3$ only if neither $f'$ nor $f''$ is zero (because the equation on degree is not valid then). And if one of them is zero, $f=0$. – 2017-02-05
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if f is a polynomial of the degree $n$ ($n>1$), then $f'$ and $f''$ will have the degrees $n-1$ and $n-2$. So, we'll have: $$n=n-1+n-2\Rightarrow n=3$$. So our polynomial will be something like this: $$f(x)=ax^3+bx^2+cx+d$$. By $f(2x)=f'(x)f''(x)$, we will have: $$8ax^3+bx^2+cx+d=(3ax^2+2bx+c)*(6ax+2b) \Rightarrow 8ax^3+bx^2+cx+d= 18a^2x^3+(6ab+12ab)x^2+(6ac+4b^2)x+2bc \Rightarrow (8a-18a^2)x^3+(b-18ab)x^2+(c-6ac-4b^2)x+(d-2bc)=0 $$. Thus, we will have the following equations: $$8a-18a^2=0 \Rightarrow a=\frac{8}{18}=\frac{4}{9}$$ Note that $a$ cannot be zero due to the degree of f. $$b-18ab=0\Rightarrow b=0$$ $$c-6ac-4b^2=0\Rightarrow c=0$$ $$d-2bc=0\Rightarrow d=0$$. Thus, $$f(x)=\frac{4}{9}x^3\Rightarrow f(3)=\frac{4}{9}*3^3=12$$ if $n<2$, then $f''=0$;hence, $f(2x)=0\Rightarrow f(3)=0$.

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    this was the only thing given in the question i never thought it wpuld go so deep – 2017-02-17
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    Simple questions don't have a simple answer always. – 2017-02-18