URGENTEEE...PRA HJ
Se f e g são funções IR em IR tais que f(x)=2x-1 e f(g(x))=x2-1,então g(x) é igual a:
a)2X2+1 (DOIS X AO QUADRADO+UM)
b)(X/2)-1 (X SOBRE 2 MENOS 1)
c)x2/2 (X AO QUADRADO SOBRE 2)
d)x+1 (X MAIS !)
e)x+(1/2)
Soluções para a tarefa
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Olá Nilton, agora sim!
![\\ \mathsf{f(x) = 2x - 1} \\\\ \mathsf{f(g(x)) = 2 \cdot g(x) - 1} \\\\ \mathsf{x^2-1=2 \cdot g(x) - 1} \\\\ \mathsf{x^2 - 1 + 1 = 2 \cdot g(x)} \\\\ \mathsf{2 \cdot g(x) = x^2} \\\\ \boxed{\mathsf{g(x) = \frac{x^2}{2}}} \\ \mathsf{f(x) = 2x - 1} \\\\ \mathsf{f(g(x)) = 2 \cdot g(x) - 1} \\\\ \mathsf{x^2-1=2 \cdot g(x) - 1} \\\\ \mathsf{x^2 - 1 + 1 = 2 \cdot g(x)} \\\\ \mathsf{2 \cdot g(x) = x^2} \\\\ \boxed{\mathsf{g(x) = \frac{x^2}{2}}}](https://tex.z-dn.net/?f=%5C%5C+%5Cmathsf%7Bf%28x%29+%3D+2x+-+1%7D+%5C%5C%5C%5C+%5Cmathsf%7Bf%28g%28x%29%29+%3D+2+%5Ccdot+g%28x%29+-+1%7D+%5C%5C%5C%5C+%5Cmathsf%7Bx%5E2-1%3D2+%5Ccdot+g%28x%29+-+1%7D+%5C%5C%5C%5C+%5Cmathsf%7Bx%5E2+-+1+%2B+1+%3D+2+%5Ccdot+g%28x%29%7D+%5C%5C%5C%5C+%5Cmathsf%7B2+%5Ccdot+g%28x%29+%3D+x%5E2%7D+%5C%5C%5C%5C+%5Cboxed%7B%5Cmathsf%7Bg%28x%29+%3D+%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%7D)
NiltonSampaio:
muitoo obg...vc salvou minha vida
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