Answer
concave up $0$ < $x$ < $1$
concave down $x$ < $0$, $x$ > $1$
inflection point is at $x$ = $0$, $x$ = $1$
Work Step by Step
Let
$f(x)$ = $(x-2)(1-x^{3})$
$f'(x)$ = $(x-2)(-3x^{2})+(1-x^{3})(1)$ = $-3x^{3}+6x^{2}+1-x^{3}$ = $-4x^{3}+6x^{2}+1$
$f''(x)$ = $-12x^{2}+12x$
concave up
$-12x^{2}+12x$ > $0$
$-12x(x-1)$ > $0$
$12x(x-1)$ < $0$
$0$ < $x$ < $1$
concave down
$-12x(x-1)$ < $0$
$12x(x-1)$ > $0$
$x$ < $0$, $x$ > $1$
inflection point is at
$x$ = $0$, $x$ = $1$
