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I have to prove that there exists a unique function $f:\mathbb{R}\rightarrow \mathbb{R}$ that satifies the following $\forall x \in \mathbb{R}$

$x^2f(x)^3+2f(x)=\cos(f(x))$

My attempt:

Seeing as I have literally been shown two or three simple problems concerning this theorem I'm not yet trained in the art of using this theorem.

So, I define a new function as $F(x,y) = x^2y^3+2y-\cos(y)$

To apply the theorem I should find $(x_0,y_0)$ such that $F(x_0,y_0)=0$ and the matrix that consists of partial derivatives by $y$ is regular. $1x1$ matrix in this case: $3x^2y^2 + 2 + \sin(y) \neq 0$

Then I can conclude that such a function exists, of course $:A \rightarrow B$ where $A$ is an open set around $x_0$ and $B$ is an open set around $y_0$.

I know that I don't need to explicitly determine the function, just prove that it exists. But I'm not sure how to go about that, any hint would be appreciated! Thanks in advance!

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1 Answers1

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We have to prove that, for a fixed $x$, the equation in $y$: $$x^2y^3 + 2y -\cos (y) =0 $$ has a unique solution.

Let's take $f(y)=x^2y^3 + 2y -\cos (y)$. Then the derivative $f'(y)=3x^2y^2 + 2 + \sin (y) \gt 0$ therefore $f$ is strictly increasing, so the equation $f(y)=0$ can have one solution at most. Now just notice that the limits at $-\infty$ and $+\infty$ of $f$ are $-\infty$ and $+\infty$ respectively to conclude there is a solution for $f(y)=0$.