Question

Calcule o valor máximo ou mínimo dá função f(x) = -3x^2 + x + 2

Answer 1

f(x) = -3x² + x + 2
Como a = 3 < 0, a função admite valor máximo, em x do vértice:
xv = -b/2a
xv = -1/[2.(-3)]
xv = -1/(-6)
xv = 1/6

O valor máximo pode ser determinada, calculando f(1/6)
Vmáx = -3(1/6)² + 1/6 + 2
Vmáx = -3.1/36 + 1/6 + 2
Vmáx = (-3 + 6 + 72)/36
Vmáx = 75/36
Vmáx = 25/12

Answer 2

Resolução da questão, veja como se procede:


Valor mínimo da função f(x) = -3x² + x +2:

$\textsf{Valor~m\'inimo~da~fun\c{c}\~ao}:~ \mathsf{f(x) = -3x^2+x+2}:\\\\\\\ \mathsf{X_{v} = \dfrac{-b}{2a}}}\\\\\\ \mathsf{X_{v} = \dfrac{-1}{2~\cdot~(-3)}}\\\\\\ \mathsf{X_{v} = \dfrac{-1}{-6}}~~\therefore~ ~\large\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{X_{v} = \dfrac{1}{6}}}}}}}}}}}}$

Valor máximo da função f(x) = -3x² + x +2:

$\textsf{Valor~m\'aximo~da~fun\c{c}\~ao}: ~~\mathsf{f(x) = -3x^2+x+2}:\\\\\\\ \mathsf{Y_{v} = \dfrac{-\Delta}{4a}}}\\\\\\ \mathsf{\Delta= b^2-4~\cdot~a~\cdot~c}}\\\\\ \mathsf{\Delta = 1^2-4~\cdot~(-3)~\cdot~2}}\\\\\ \mathsf{\Delta = 1+24}}\\\\\ \mathsf{\Delta = 25}}\\\\\\\\ \mathsf{Y_{v} = \dfrac{-\Delta}{4a}}}\\\\\\ \mathsf{Y_{v} = \dfrac{-25}{4~\cdot~(-3)}}\\\\\\ \mathsf{Y_{v} = \dfrac{-25}{-12}}~~\therefore~~ \large\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{Y_{v} = \dfrac{25}{12}}}}}}}}}}}}$

Para concluir temos que: 

$\large\boxed{\boxed{\boxed{\boxed{\boxed{\large\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{X_{v} = \dfrac{1}{6}}~~\textsf{e}~~\mathsf{Y_{v} = \dfrac{25}{12}}}}}}}}}}}}}}}}}}}}}}}}}$

Espero que te ajude. :-).

Dúvidas? Comente.