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Let $p(x) = x^{3} + ax^{2} + bx + c$. Show that $p: \mathbb{R} \longrightarrow \mathbb{R}$ is a homeomorphism if, only if, $a^{2} \leq 3b$. So that $p^{-1}$ be differentiable is necessary and sufficient that $a^{2} < 3b$.

$\textbf{Solution:}$ $(\Longrightarrow)$ Suppose that $p$ be a homeomorphism. Note that $p'(x) = 3x^{2} + 2ax + b$. If $a^{2} > 3b$, $p'$ have two distinct roots $x_{1} < x_{2}$. Since $p'(x_{1}) = 0 = p'(x_{2})$, $p'(x) \leq 0$ for all $x \in [x_{1}, x_{2}]$. Thus, $p$ is decreasing in $I$. Since $p$ is not descreasing in $\mathbb{R} \setminus I$, $p$ has a point of local maximum, then, theres exist $k_{1} < k_{2}$ such that $p(k_{1}) = p(k_{2})$, an absurd.

$(\Longleftarrow)$ Suppose that $a^{2} \leq 3b$. Thus, $p$ has at most one real root. Note that $ p'(x) \geq 0$. Therefore, $p$ is not decreasing. If exist $x_{1} < x_{2}$ such that $p(x_{1}) = p(x_{2})$, we have two possibilities:

$(i)$ $p$ is constant in $[x_{1}, x_{2}]$.

$(ii)$ $p$ has a local maximum or local minimum in $[x_{1}, x_{2}]$.

Since $p$ has at most one real root and $p$ is a monotone function, $(i)$ and $(ii)$ doesn't happen. So, $p$ is increasing, then, theres exist $p^{-1}$. By the $\textbf{Auxiliar Theorem 1}$, $p^{-1}$ is a continuous function.

The hypothesys of$\textbf{Auxiliar Theorem 2}$ are satisfies, so $$p^{-1}\mathrm{\;is\;differentiable} \Longleftrightarrow p'(x) \neq 0\; \forall x \in \mathbb{R}.$$

$(\Longrightarrow)$ Suppose that $p^{-1}$ be differentiable. Thus, $p'(x) \neq 0$ for all $x \in \mathbb{R}$. If $a^{2} - 3b = 0$, theres exist $c \in \mathbb{R}$ such that $p'(c) = 0$. Therefore, $a^{2} < 3b$.

$(\Longleftarrow)$ Suppose that $a^{2} < 3b$. Thus $p'$ has no real root, then $p'(x) \neq 0$ for all $x \in \mathbb{R}$. Therefore, $p^{-1}$ is differentiable.

$\textbf{Auxiliar Theorem 1:}$ Let $f: I \longrightarrow \mathbb{R}$ a continuous injective function. Thus $f$ is monotone, $f(I)$ is an interval and $f^{-1}: f(I) \longrightarrow \mathbb{R}$ is continuous.

$\textbf{Auxiliar Theorem 2:}$ Let $f: X \longrightarrow Y \subset \mathbb{R}$ a invertible function. If $f$ is differentiable in $a \in X \cap X'$ and $f^{-1}$ continuous in $f(a)$, then $f^{-1}$ is differentiable in $f(a)$ if, only if, $f'(a) \neq 0$.

Is the correct?

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