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!
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 $f(y)=0$.