Find $f'(0)$ if $f$ is a function such that $$1+f(x)+x^2(f(x))^3=11 \hspace{1cm}\text{and}\hspace{1cm}f(1)=2.$$

Here's what I've tried so far:

If $f$ is differentiable around $0$, then the product $x^2f(x)^3$ is zero, such that differentiating, \begin{align*} 1+f(x)+x^2(f(x))^3 &= 11\\ 0 + f'(x)+[2xf(x)^3+3x^2f(x)^2f'(x)] &= 0\\ f'(x)[1+3x^2f(x)^2] &= -2xf(x)^3\\ f'(x) &= -\frac{2xf(x)^3}{1+3x^2f(x)^2} \end{align*} so $f'(0) = -0/1$.

I don't think this is correct though, I don't see why $f(1)=2$ is a relevant fact, or maybe I have to use some theorem instead of isolating $f'(x)$. Anyway, any help or maybe a hint would be greatly appreciated. Thanks.


If the given condition is true, then $1+f(0)+0=11 \Rightarrow f(0)=10$

Let's try taking the derivative of both sides.


$f'(0)+0+0=0 \Rightarrow f'(0)=0$

  • 1
    $\begingroup$ That's the answer I arrived to too, but I'm still confused as to why $f(1)=2$ is relevant (if it is at all relevant) so it makes me think that maybe I'm missing something. $\endgroup$
    – Eduardo M.
    May 18 '16 at 23:49

I don't think there is a mistake in what you did.

If, as I assume, your $f$ is taking real values, then the condition $f(1)=2$ is not necessary: it follows from the other condition: when $x=1$, you have $$ 1+f(1)+f(1)^3=11. $$ The only real solution to this cubic is $f(1)=2$.


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