I'm working on the following problem:
Let $f(x) = x^4 - x^3 + x$, for $x \geq 0$.
Show that (i) $f$ is strictly increasing on $[0,\infty)$, (ii) $f$ is a one-to-one and onto function on $[0,\infty)$, and (iii) if $g$ denotes the inverse function of $f$, $g''$ vanishes exactly once on $(0,\infty)$.
My attempt:
To show that $f$ is strictly increasing on $[0,\infty)$, it suffices to show that $$f'(x) > 0 \qquad \forall x \in [0,\infty).$$ Accordingly, $f'(x) = 4x^3 - 3x^2 + 1$, and $f''(x) = 12x^2 - 6x$. Analysis of $f''$ yields that $f'$ has a global minimum on $[0,\infty)$ of $f'(\frac{1}{2}) = \frac{3}{4} > 0$, so $f$ is strictly increasing on $[0,\infty)$.
To prove $f$ is one-to-one, we proceed via contrapositive. Take $x_1, x_2 \in [0, \infty)$ with $x_1 < x_2$ (i.e. $x_1 \neq x_2$). Then, since $f$ is strictly increasing, it follows that $f(x_1) < f(x_2)$ (i.e. $f(x_1) \neq f(x_2)$). Thus, if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$, or equivalently, if $f(x_1) = f(x_2)$, then $x_1 = x_2$. Thus, $f$ is one-to-one. To prove that $f$ is onto, obverse that $f(0) = 0$ and $\lim_{x \rightarrow \infty} f(x) = \infty$. Since $f$ is continuous as a polynomial and strictly increasing, $\forall y \in (0,\infty)$, $\exists x \in (0,\infty)$ such that $f(x) = y$. Thus $f$ is onto.
Let $x \in (0,\infty)$. Then, $[0,\infty)$ is a neighborhood of $x$, and from part's $(a)$ and $(b)$, $f:[0,\infty) \rightarrow [0,\infty)$ is monotone and continuous, and $f'(x) \neq 0$. Thus, by the Inverse Function Theorem,
$$g'(f(x)) = \frac{1}{f'(x)}.$$ Further, \begin{align*} g''(f(x)) &= -\frac{f''(x)}{(f'(x))^2} \\ &= -\frac{12x^2 - 6x}{(4x^3 - 3x^2 + 1)^2} \\ \end{align*} so $g''$ vanishes only when $x = \frac{1}{2}$ ($x \neq 0$ by assumption).
Is this a valid solution? I'm getting tangled in the details on the inverse function theorem. I appreciate the feedback!