(Fuvest-sp) : Se f(X) = 2x²-3x+1 calcule f(√2/3)?
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Temos f(x) = 2x² – 3x + 1.
Só precisamos substituir x por √2/3:
![\displaystyle f \left( \frac{\sqrt[]{2}}{3} \right) = 2 \cdot \left( \frac{\sqrt[]{2}}{3} \right)^2 - 3 \cdot \frac{\sqrt[]{2}}{3} + 1](https://tex.z-dn.net/?f=+%5Cdisplaystyle+f+%5Cleft%28+%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B3%7D+%5Cright%29++%3D+2+%5Ccdot+%5Cleft%28+%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B3%7D+%5Cright%29%5E2+-+3+%5Ccdot+%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B3%7D+%2B+1++ "\displaystyle f \left( \frac{\sqrt[]{2}}{3} \right) = 2 \cdot \left( \frac{\sqrt[]{2}}{3} \right)^2 - 3 \cdot \frac{\sqrt[]{2}}{3} + 1")
![\displaystyle f \left( \frac{\sqrt[]{2}}{3} \right) = 2 \cdot \left( \frac{2}{9} \right) - \not3 \cdot \frac{\sqrt[]{2}}{\not3} + 1](https://tex.z-dn.net/?f=+%5Cdisplaystyle+f+%5Cleft%28+%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B3%7D+%5Cright%29++%3D+2+%5Ccdot+%5Cleft%28+%5Cfrac%7B2%7D%7B9%7D+%5Cright%29+-+%5Cnot3+%5Ccdot+%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B%5Cnot3%7D+%2B+1+++ "\displaystyle f \left( \frac{\sqrt[]{2}}{3} \right) = 2 \cdot \left( \frac{2}{9} \right) - \not3 \cdot \frac{\sqrt[]{2}}{\not3} + 1")
![\displaystyle f \left( \frac{\sqrt[]{2}}{3} \right) = \frac{4}{9} - \sqrt[]{2} + 1](https://tex.z-dn.net/?f=+%5Cdisplaystyle+f+%5Cleft%28+%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B3%7D+%5Cright%29++%3D+%5Cfrac%7B4%7D%7B9%7D+-+%5Csqrt%5B%5D%7B2%7D+%2B+1+ "\displaystyle f \left( \frac{\sqrt[]{2}}{3} \right) = \frac{4}{9} - \sqrt[]{2} + 1")
![\displaystyle f \left( \frac{\sqrt[]{2}}{3} \right) = \frac{4}{9} - \frac{9 \hspace{0,07cm} \sqrt[]{2} }{9} + \frac{9}{9}](https://tex.z-dn.net/?f=+%5Cdisplaystyle+f+%5Cleft%28+%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B3%7D+%5Cright%29++%3D+%5Cfrac%7B4%7D%7B9%7D+-+%5Cfrac%7B9+%5Chspace%7B0%2C07cm%7D+%5Csqrt%5B%5D%7B2%7D+%7D%7B9%7D+%2B+%5Cfrac%7B9%7D%7B9%7D++++ "\displaystyle f \left( \frac{\sqrt[]{2}}{3} \right) = \frac{4}{9} - \frac{9 \hspace{0,07cm} \sqrt[]{2} }{9} + \frac{9}{9}")
![\displaystyle f \left( \frac{\sqrt[]{2}}{3} \right) = \frac{4 + 9 - 9 \hspace{0,07cm} \sqrt[]{2} }{9}](https://tex.z-dn.net/?f=+%5Cdisplaystyle+f+%5Cleft%28+%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B3%7D+%5Cright%29++%3D+%5Cfrac%7B4+%2B+9+-+9+%5Chspace%7B0%2C07cm%7D+%5Csqrt%5B%5D%7B2%7D+%7D%7B9%7D+++ "\displaystyle f \left( \frac{\sqrt[]{2}}{3} \right) = \frac{4 + 9 - 9 \hspace{0,07cm} \sqrt[]{2} }{9}")
![\displaystyle f \left( \frac{\sqrt[]{2}}{3} \right) =\frac{5 - 9 \hspace{0,07cm} \sqrt[]{2}}{9}](https://tex.z-dn.net/?f=+%5Cdisplaystyle+f+%5Cleft%28+%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B3%7D+%5Cright%29++%3D%5Cfrac%7B5+-+9+%5Chspace%7B0%2C07cm%7D+%5Csqrt%5B%5D%7B2%7D%7D%7B9%7D++ "\displaystyle f \left( \frac{\sqrt[]{2}}{3} \right) =\frac{5 - 9 \hspace{0,07cm} \sqrt[]{2}}{9}")
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Só precisamos substituir x por √2/3:
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