Question

1) Calcule o acréscimo ∆v para: 

a. y = 2x² − 6x + 5 ( x = 3; ∆x = 0,01). 

b. y = 6x² − 4 ( x = 2; ∆x = 0,001). 

Answer 1

$y=2x^2-6x+5\\\\\boxed{x=3 ;\Delta x=0,01}$

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
$\Delta_V= 20,0240-20 \approx0,240$