Odpowiedzi

2011-07-17T12:43:34+02:00
  • Roma
  • Community Manager
2011-07-17T13:05:10+02:00

f(x)=6(1-x^2)-3(2x^2)-2 = 6 -6x^2 - 6x^2 - 2 = - 12x^2 + 4

 

- 12x^2 + 4 = 0 \ / \ : (-4)

3x^2 - 1 = 0

(\sqrt{3}x)^2 - 1^2 = 0

(\sqrt{3}x - 1)(\sqrt{3}x + 1) = 0

\sqrt{3}x - 1 = 0 \vee \sqrt{3}x + 1= 0

 

\sqrt{3}x - 1 = 0

\sqrt{3}x = 1 \ / : \ \sqrt{3}

x = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{3}}{3}

lub

\sqrt{3}x + 1 = 0

\sqrt{3}x = -1 \ / : \ \sqrt{3}

x = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = -\frac{\sqrt{3}}{3}

Odp. Miejsca zerowe funkcji f to:

x = \frac{\sqrt{3}}{3} \ i \ x = -\frac{\sqrt{3}}{3}


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