Questão 42 - Unicamp - T_X - Unicamp 2027

Determinação do ponto $\mathrm{Q}$:

${\mathrm{x}}^3<\mathrm{f}\left(\mathrm{x}\right)<{\mathrm{x}}^2$ $\to$${\left(\frac{1}{2}\right)}^3<\mathrm{f}\left(\frac{1}{2}\right)<{\left(\frac{1}{2}\right)}^2 \to \frac{1}{8}<\mathrm{f}\left(\frac{1}{2}\right)<\frac{1}{4}$

Assim, tem-se o ponto $\mathrm{Q}$ variando entre ${\mathrm{Q}}_{\mathrm{a}}\left(\frac{1}{2};\frac{1}{8}\right) \mathrm{e} {\mathrm{Q}}_{\mathrm{b}}\left(\frac{1}{2};\frac{1}{4}\right)$, gerando os triângulos a seguir:

Assim, as áreas máxima e mínima são:

$$\begin{gathered} {\mathrm{A}}_{mín}=\frac{\mathrm{b}·\mathrm{h}}{2}\to {\mathrm{A}}_{mín}=\frac{\left({\displaystyle \frac{1}{2}-\frac{1}{4}}\right)·{\displaystyle \frac{1}{2}}}{2}\to {\mathrm{A}}_{mín}=\frac{1}{16} \\ {\mathrm{A}}_{máx}=\frac{\mathrm{b}·\mathrm{h}}{2}\to {\mathrm{A}}_{máx}=\frac{\left({\displaystyle \frac{1}{2}-\frac{1}{8}}\right)·{\displaystyle \frac{1}{2}}}{2}\to {\mathrm{A}}_{máx}=\frac{3}{32} \end{gathered}$$

Portanto, $\frac{1}{16}<\mathrm{A}<\frac{3}{32}$.