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Suppose $f'(x)$ = 2$x^4$ $f(x^2)$ and $f(1)$ = 2.

Find $f''(x)$ = ${d}/{dx}$ $f'(x)$

I'm having trouble starting this. I was wondering if I work backwards from 𝑓(1) = 2 OR use $f(x^2)$ somehow.

How can I solve this?

EDIT: So it looks like my friend miswrote the question, it's now edited. It should be $f'(x)$ = 2$x^4$ $f(x^2)$ NOT $f(x)$ = 2$x^4$ $f(x^2)$

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    $\begingroup$ It's impossible to have $f(1)=2$ since $f(1) = 2\cdot 1^4 \cdot f(1) \iff f(1) = 0$. $\endgroup$
    – Sávio
    Jun 16 '20 at 19:59
  • $\begingroup$ Yeah, this doesn't make sense. Where did you get this question from? $\endgroup$ Jun 16 '20 at 20:02
  • $\begingroup$ @Savio How did you get 𝑓(1) = 0? $\endgroup$ Jun 16 '20 at 20:07
  • $\begingroup$ @Riemann'sPointyNose The question was one my friend sent me from a tutorial he was doing. The question is word for word. $\endgroup$ Jun 16 '20 at 20:07
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    $\begingroup$ Subtracting ${f(1)}$ from both sides gives ${f(1)=0}$, as Savio said. A lot of "paradoxes" rely on the fact that the only solution to this equation is ${0}$ to try and prove ${1=2}$ (notice dividing through by ${f(1)}$ yields ${1=2}$ - but this is not allowed, since we have not checked ${f(1)}$ is $0$, which it is - so maths is not broken) $\endgroup$ Jun 16 '20 at 20:20
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Cool, so as we can see ${f''(x) = \frac{d}{dx}\left(2x^4f(x^2)\right)=8x^3f(x^2)+4x^5f'(x^2)}$, but ${f'(x)=2x^4f(x^2)}$ and hence ${f'(x^2)=2x^8f(x^4)}$, and so overall

$${f''(x)=8x^3f(x^2)+4x^5\left(2x^8f(x^4)\right)=8x^3f(x^2)+8x^{13}f(x^4)}$$

But yeah, I'm not sure what relevance ${f(1)=2}$ has in this particular case

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  • $\begingroup$ It seems to be right, though I wonder what relevance $f(1)$ = 2 had? $\endgroup$ Jun 16 '20 at 20:41
  • $\begingroup$ Hmm yeah exactly. It seems redundant if this is the answer it wanted $\endgroup$ Jun 16 '20 at 20:41

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