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Question Number 145947 by Khalmohmmad last updated on 09/Jul/21

Commented by hknkrc46 last updated on 09/Jul/21
${: ((∫f(x)dx = F(x) + c)),((∫g(x)dx = G(x) + c)) } determinant (((G(2) = 3)),((f(5) = 1))) • f(x) = F ′(x) ∧ g(x) = G ′(x) • ((d[(x^2 + 2)G(x)])/dx) = ((d[F(3x − 1)])/dx) ⇒ 2xG(x) + (x^2 + 2)G ′(x) = 3F ′(3x − 1) ⇒ 2xG(x) + (x^2 + 2)g(x) = 3f(3x − 1) ▶ x = 2 ⇒ 4G(2) + 6g(2) = 3f(5) ⇒ 12 + 6g(2) = 3 ⇒ g(2) = −(3/2)$
$\begin{matrix} \int f (x) d x = F (x) + c \\ \int g (x) d x = G (x) + c \end{matrix}} \begin{matrix} G (2) = 3 \\ f (5) = 1 \end{matrix}$ $∙ f (x) = F^{'} (x) \land g (x) = G^{'} (x)$ $∙ \frac{d [(x^{2} + 2) G (x)]}{d x} = \frac{d [F (3 x - 1)]}{d x}$ $\Rightarrow 2 x G (x) + (x^{2} + 2) G^{'} (x) = 3 F^{'} (3 x - 1)$ $\Rightarrow 2 x G (x) + (x^{2} + 2) g (x) = 3 f (3 x - 1)$ $▸ x = 2 \Rightarrow 4 G (2) + 6 g (2) = 3 f (5)$ $\Rightarrow 12 + 6 g (2) = 3 \Rightarrow g (2) = - \frac{3}{2}$
	Answered by puissant last updated on 09/Jul/21

	$2 xG (x) + (x^{2} + 1) g (x) = 3 f (3 x - 1)$ $x = 2 \Rightarrow 4 G (2) + 5 g (2) = 3 f (5)$ $\Rightarrow 5 g (2) = 3 f (5) - 4 G (2)$ $\Rightarrow g (2) = \frac{1}{5} (3 - 12) = - \frac{9}{5} . .$ $\Rightarrow g (2) = - \frac{9}{5}$
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