Answer
Increasing: $(0,\tfrac{8}{27}),$
Decreasing: $(-\infty,0)\cup(\tfrac{8}{27},\infty),$
Concave up: $\varnothing,$
Concave down: $(-\infty,0)\cup(0,\infty),$
Inflection points: none.
Work Step by Step
$\textbf{Given:}\qquad f(x)=x^{2/3}-x.$
$\text{Compute derivatives:}$
\begin{align*}
f'(x)&=\frac{2}{3}x^{-1/3}-1
=\frac{2-3x^{1/3}}{3x^{1/3}},\\[6pt]
f''(x)&=-\frac{2}{9}x^{-4/3}.
\end{align*}
Note: $f'$ and $f''$ are not defined at $x=0$, while $f$ is defined there with $f(0)=0$.
$\textbf{(a) Intervals where $f$ is increasing.}$
Let $t=x^{1/3}$ (the real cube root), so $f'(x)=\dfrac{2-3t}{3t}$. The critical value from the numerator is $t=\tfrac{2}{3}$, i.e. $x=t^3=\left(\tfrac{2}{3}\right)^3=\dfrac{8}{27}$. Analyze the sign of $f'$:
\begin{array}{ll}
t<0 &\Rightarrow 2-3t>0,\;3t<0 \Rightarrow f'<0,\\[4pt]
00,\;3t>0 \Rightarrow f'>0,\\[4pt]
t>\tfrac{2}{3} &\Rightarrow 2-3t<0,\;3t>0 \Rightarrow f'<0.
\end{array}
Hence $f'(x)>0$ only for $0\dfrac{8}{27}$.
$\boxed{\text{Decreasing on }(-\infty,0)\cup(\tfrac{8}{27},\infty).}$
$\textbf{(c) Open intervals where $f$ is concave up.}$
Since
$f''(x)=-\frac{2}{9}x^{-4/3},$
and $x^{-4/3}>0$ for all $x\neq0$, we have $f''(x)<0$ for every $x\neq0$. Thus there are no intervals where $f$ is concave up.
$\boxed{\text{Concave up: }\varnothing.}$
$\textbf{(d) Open intervals where }$ $f\textbf{ is concave down.}$
Because $f''(x)<0$ for all $x\neq0$, $f$ is concave down on both sides of $0$ (excluding $0$ itself where $f''$ is undefined):
$\boxed{\text{Concave down on }(-\infty,0)\cup(0,\infty).}$
$\textbf{(e) $x$-coordinates of all inflection points.}$
There is no change of concavity (it is negative on both sides of $0$), so $f$ has no inflection points.
$\boxed{\text{No inflection points (no }x\text{-coordinates).}}$
$\textbf{Summary:}$
Increasing: $(0,\tfrac{8}{27}),$
Decreasing: $(-\infty,0)\cup(\tfrac{8}{27},\infty),$
Concave up: $\varnothing,$
Concave down: $(-\infty,0)\cup(0,\infty),$
Inflection points: none.
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