Question

f"(x)=-12x²+12x-2; f(0)=4; f'(0)=12

Answer 1

Temos a seguinte derivada segunda da função:

$f''_{(x)}=-12x^2+12x-2\\\\\\ Integrando, para\ encontrarmos\ a\ derivada\ primeira:\\\\ f'_{(x)}=\int (-12x^2+12x-2).dx\\\\\\ f'_{(x)}=-12\dfrac{x^{2+1}}{2+1}+12\dfrac{x^{1+1}}{1+1}-2\dfrac{x^0+1}{0+1}+constante\\\\\\ f'_{(x)}=-12\dfrac{x^{3}}{3}+12\dfrac{x^{2}}{2}-2x+constante\\\\\\ f'_{(x)}=-4x^{3}+6x^{2}-2x+constante\\\\\\ Sabemos\ que\ f'_{(0)}=12,\ logo:\\\\ f'_{(0)}=-4(0)^{3}+6(0)^{2}-2(0)+constante=12\\\\\\ f'_{(0)}=constante=12\\\\\\ constante = 12$


$Fun\c{c}\~ao\ completa:\\\\ f'_{(x)}=-4x^{3}+6x^{2}-2x+12\\\\\\ Integrando\ novamente\ para\ encontrar\ a\ fun\c{c}\~ao\ prim\'aria:\\\\ f_{(x)} = \int (-4x^{3}+6x^{2}-2x+12).dx\\\\ f_{(x)} = -4\dfrac{x^{3+1}}{3+1}+6\dfrac{x^{2+1}}{2+1}-2\dfrac{x^{1+1}}{1+1}+12\dfrac{x^{0+1}}{0+1}+constante\\\\ f_{(x)} = -4\dfrac{x^{4}}{4}+6\dfrac{x^{3}}{3}-2\dfrac{x^{2}}{2}+12x+constante\\\\ f_{(x)} = -x^{4}+2x^{3}-x^{2}+12x+constante$


$Sabemos\ que\ f_{(0)}=4,\ logo:\\\\ f_{(0)} = -(0)^{4}+2(0)^{3}-(0)^{2}+12(0)+constante=4\\\\ f_{(0)} = constante=4\\\\ constante = 4\\\\\\ Fun\c{c}\~ao\ completa:\\\\ \boxed{f_{(x)} = -x^{4}+2x^{3}-x^{2}+12x+4}$


Espero ter ajudado.
Bons estudos!