f(x)=vmatrix x^3 2x^2+1 1+3x 3x^2+2 2x x^3+6 x^3-x 4 x^2-2 vmatrixx R2f(0)+f^'(0)

Question

$f(x)=\begin{vmatrix} {x}^{3} \quad  2{x}^{2}+1 \quad  1+3x \ 3{x}^{2}+2 \quad  2x \quad  {x}^{3}+6 \ {x}^{3}-x \quad  4 \quad  {x}^{2}-2 \end{vmatrix}x\in R2f(0)+{f}^{'}(0)$

Accepted Answer

$\begin{gathered} f(0)=|\begin{array}{ccc}0 \quad  1 \quad  1 \ 2 \quad  0 \quad  6 \ 0 \quad  4 \quad  -2\end{array}|=12 \ {f}^{'}(x)=|\begin{array}{ccc}3{x}^{2} \quad  4x \quad  3 \ 3{x}^{2}+2 \quad  2x \quad  {x}^{3}+6 \ {x}^{3}-x \quad  4 \quad  {x}^{2}-2\end{array}|+ \end{gathered}$

$\begin{gathered} |\begin{array}{ccc}{x}^{3} \quad  2{x}^{2}+1 \quad  1+3x \ 6x \quad  2 \quad  3{x}^{2} \ {x}^{3}-x \quad  4 \quad  {x}^{2}-2\end{array}|+ \ |\begin{array}{ccc}{x}^{3} \quad  2{x}^{2}+1 \quad  1+3x \ 3{x}^{2}+2 \quad  2x \quad  {x}^{3}+6 \ 3{x}^{2}-1 \quad  0 \quad  2x\end{array}| \ 	herefore {f}^{'}(0)=|\begin{array}{ccc}0 \quad  0 \quad  3 \ 2 \quad  0 \quad  6 \ 0 \quad  4 \quad  -2\end{array}|+|\begin{array}{ccc}0 \quad  1 \quad  1 \ 0 \quad  2 \quad  0 \ 0 \quad  4 \quad  -2\end{array}|+|\begin{array}{ccc}0 \quad  1 \quad  1 \ 2 \quad  0 \quad  6 \ -1 \quad  0 \quad  0\end{array}| \ =24-6=18 \ 	herefore 2f(0)+{f}^{'}(0)=42 \end{gathered}$

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