Respostas

2014-05-28T14:47:54-03:00
y=2x^2-6x+5\\\\\boxed{x=3 ;\Delta x=0,01}


temos que
\Delta_V =f(x+\Delta x) - f(x)\\\ \Delta_v =f(3+0,01) - f(3)\\\\\Delta_V=f(3,01) - f(3)

calculando as funções
f(3,01) = 2*(3,01)^2-6*(3,01)+5\\\\f(3,01) = 5,0602
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f(3) = 2*3^2 -6*3 +5\\\\f(3) =5

então:
\Delta_V=5,0602-5\\\Delta_V=0,0602

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y=6x^2-4\\\\\boxed{x=2;\Delta x=0,001}

;
\Delta_V =f(x+\Delta x) - f(x)\\\\\Delta_V =f(2+0,001) - f(2)\\\\ \Delta_V = f(2,001) - f(2)

calculando s funções
f(2,001)=6*(2,001)^2 -4\\\\f(2,001)=20,0240\\\\\\\\f(2) = 6*2^2-4\\\\f(2)=20

a o acréscimo será
\Delta_V= 20,0240-20  \approx0,240
2 5 2