Question

Seja f(x)=2x^2-3x+1,calcule f(√2/3).

Answer 1

$\displaystyle \text{f(x)}=2\text x^2-3\text x+1 \ ; \ \text {f}(\frac{\sqrt{2}}{3}) = \ ? \\\\\\ \text{f}(\frac{\sqrt 2}{3}) = 2(\frac{\sqrt{2}}{3})^2-\frac{3\sqrt{2}}{3}+1 \\\\\\ \text{f}(\frac{\sqrt{2}}{3})=\frac{4}{9}-\sqrt{2}+1 \\\\\\ \text{f}(\frac{\sqrt 2}{3}) = \frac{4+9}{9}-\sqrt{2} \\\\\\ \huge\boxed{\text f(\frac{\sqrt 2}{3}) = \frac{13}{9} -\sqrt 2\ }\checkmark \\\\\\ \text{ou} \\\\\ \huge\boxed{\text f(\frac{\sqrt2}{3})\approx 0,3\ }\checkmark$