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Let's f be define as $F(x,y)=y^3 + 2x^3y^2 + 3x^2y^2 + 6xy^2 + 4y.$ such that F is going from $R^2$ to $R$ (R is a set of real numbers).

Prove that eqsist differentiable function $f$ from $R$ to $R$, such that for every $x$. $$F(x,f(x))=1$$ My only clue is to use Implicit multivarible theorem but I really don't know how.

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    Do you mean $F(x(f(x,y))=1$? – 2017-01-23
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    No because f is real function of real variables(one variable) – 2017-01-23
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    So what is $y$ when you say $f(x,y)$? – 2017-01-23
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    Sorry, I wrong rewrite the task., but I fix now. – 2017-01-23

3 Answers 3

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The conjecture doesn't hold.

The reason is that there are (many) values of $x$ for which $f(x)$ will be multivalued, which contradicts the requirement that $f(x)$ is to be a function.

For example, taking $x=1$ and requiring $F(x,y=f(x))=1$ has three results. Numerically, they can be obtained approximately:

$f_1(1) = 0.1696$

$f_2(1) = -0.5554$

$f_3(1) = -10.614$

REMARK:

I conjecture that there exists a differentiable function $g(y)$ from R \ {0} to R, such that for every y, $F(g(y),y)=1$.

$g(y)$ is the inverse of $f(x)$, so this might help with your problem.

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$F(x,y)=y^3 + 2x^3y^2 + 3x^2y^2 + 6xy^2 + 4y$

$G(x,y)=F(x,y)-1$

It can not be $G_{y}(x,y)=3y^2+4yx^3+6x^2y+12xy+4\neq0$for all $x,y\in\mathbb{R}$

there is not a function$ f(x) $make $F(x,f(x))=1$

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The fact that $F(x,y)=1$ has three roots does not mean one cannot select continuous and differentiable $f(x)$ out of the three roots such that $F(x,f(x))=1$.

However, it looks like your $f(x)$ does not exist. The two figures below plot roots of $F(x,y)=1$.

enter image description here enter image description here

The second figure shows what is easy to show analytically, that for $x=0$, there is only one root $F(0,y)=1$ and the root is strictly positive. Also, there is no $x$ such that $F(x,0)=1$, which again can be show formally. Hence if your $f(x)$ exists, then $f(x)>0$ for any $x$ by continuity of $f$. But the second figure shows that such $f(x)$ does not exist.

Update (more formal but still incomplete argument): Solving $F(x,y)=1$ for $x$ gives $x=-\frac{1}{2}-\frac{9y^2}{2^{\frac{2}{3}}D^{\frac{1}{3}}}+\frac{D^{\frac{1}{3}}}{6\cdot2^{\frac{1}{3}}y^2}$, where $D=108y^4-432y^5+270y^6-108y^7+\sqrt{78732y^{12}+(108y^4-432y^5+270y^6-108y^7)^2}$ and two other roots.

Now suppose you prove that the root given is the only real root. I argued above that $f(x)>0$ for any $x$ for your $f$ if it exists. In terms of the root for $x$, this implies that the root above has to be monotone in $y$ over $\mathbb{R}_{++}$. And it is not (might be provable analytically).

So the last missing bit is to argue that the root above is the only real root. There is no root if $y=0$. Using Descarte's rule of signs, I was able to prove that there is i) one positive root for $y<0$, ii) one positive root for $y>0$ when $y^3+4y-1<0$, iii) no positive root for $y>0$ when $y^3-4y-1>0$. For negative roots the Descarte's rule of signs does not help and I suggest checking this book, especially section 1.4, section 1.4.2 in particular.