For the following two (02) items: Consider the function f(x) = 1 -...

Given $f(x) = 1 - (x-1)^{2/3}$

Differentiating with respect to $x$:

$f'(x) = -\dfrac{2}{3}(x-1)^{-1/3} = \dfrac{-2}{3(x-1)^{1/3}}$

For $x < 1$, $x-1 < 0 \Rightarrow f'(x) > 0$. Thus, $f(x)$ is strictly increasing.

For $x > 1$, $x-1 > 0 \Rightarrow f'(x) < 0$. Thus, $f(x)$ is strictly decreasing.

Since $f'(x)$ changes sign from positive to negative as $x$ passes through $1$, $f(x)$ has a local maximum at $x = 1$.

Alternatively, $(x-1)^{2/3} \geq 0$ for all real $x$.

$1 - (x-1)^{2/3} \leq 1$

$f(x) \leq f(1)$ for all $x$.

Hence, $f(x)$ has a maximum at $x = 1$.

Answer: a maximum at $x = 1$