Question

$f(x)=\begin{cases} x^3 & 2x^2+1 & 1+3x \\ 3x^2+2 & 2x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{cases}$

$2f(0)+f'(0)=?$

Answer 1

42

Answer 2

To find the value of $2f(0)+f'(0)$, we need to first calculate $f(0)$ and $f'(0)$.

The given function is a determinant: $f(x)=\begin{vmatrix} x^3 & 2x^2+1 & 1+3x \\ 3x^2+2 & 2x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{vmatrix}$

Step 1: Calculate $f(0)$

Substitute $x=0$ into the determinant: $f(0)=\begin{vmatrix} 0^3 & 2(0)^2+1 & 1+3(0) \\ 3(0)^2+2 & 2(0) & 0^3+6 \\ 0^3-0 & 4 & 0^2-2 \end{vmatrix} = \begin{vmatrix} 0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{vmatrix}$

Expanding the determinant along the first column (C1): $f(0) = 0 \cdot \text{cofactor}(0) - 2 \cdot \begin{vmatrix} 1 & 1 \\ 4 & -2 \end{vmatrix} + 0 \cdot \text{cofactor}(0)$ $f(0) = -2 \cdot ((1)(-2) - (1)(4)) = -2 \cdot (-2 - 4) = -2 \cdot (-6) = 12$

Step 2: Calculate $f'(0)$

The derivative of a determinant is the sum of determinants obtained by differentiating one row (or column) at a time.

Let $A(x) = (a_{ij}(x))$. Then $A'(x)$ is the sum of three determinants, where in each determinant, one row is replaced by its derivatives, and the other rows remain unchanged.

First, find the derivatives of each element: $a_{11}(x) = x^3 \implies a'_{11}(x) = 3x^2$ $a_{12}(x) = 2x^2+1 \implies a'_{12}(x) = 4x$ $a_{13}(x) = 1+3x \implies a'_{13}(x) = 3$

$a_{21}(x) = 3x^2+2 \implies a'_{21}(x) = 6x$ $a_{22}(x) = 2x \implies a'_{22}(x) = 2$ $a_{23}(x) = x^3+6 \implies a'_{23}(x) = 3x^2$

$a_{31}(x) = x^3-x \implies a'_{31}(x) = 3x^2-1$ $a_{32}(x) = 4 \implies a'_{32}(x) = 0$ $a_{33}(x) = x^2-2 \implies a'_{33}(x) = 2x$

Now, evaluate the elements and their derivatives at $x=0$:

Original elements at $x=0$: Row 1: $(0, 1, 1)$ Row 2: $(2, 0, 6)$ Row 3: $(0, 4, -2)$

Derivatives of elements at $x=0$: Row 1: $(0, 0, 3)$ Row 2: $(0, 2, 0)$ Row 3: $(-1, 0, 0)$

Now, form the three determinants for $f'(0)$: $f'(0) = \begin{vmatrix} a'_{11}(0) & a'_{12}(0) & a'_{13}(0) \\ a_{21}(0) & a_{22}(0) & a_{23}(0) \\ a_{31}(0) & a_{32}(0) & a_{33}(0) \end{vmatrix} + \begin{vmatrix} a_{11}(0) & a_{12}(0) & a_{13}(0) \\ a'_{21}(0) & a'_{22}(0) & a'_{23}(0) \\ a_{31}(0) & a_{32}(0) & a_{33}(0) \end{vmatrix} + \begin{vmatrix} a_{11}(0) & a_{12}(0) & a_{13}(0) \\ a_{21}(0) & a_{22}(0) & a_{23}(0) \\ a'_{31}(0) & a'_{32}(0) & a'_{33}(0) \end{vmatrix}$

$f'(0) = \begin{vmatrix} 0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{vmatrix} + \begin{vmatrix} 0 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 4 & -2 \end{vmatrix} + \begin{vmatrix} 0 & 1 & 1 \\ 2 & 0 & 6 \\ -1 & 0 & 0 \end{vmatrix}$

Calculate each determinant:

1. $D_1 = \begin{vmatrix} 0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{vmatrix}$ Expand along R1: $D_1 = 3 \cdot \begin{vmatrix} 2 & 0 \\ 0 & 4 \end{vmatrix} = 3 \cdot (2 \cdot 4 - 0 \cdot 0) = 3 \cdot 8 = 24$

2. $D_2 = \begin{vmatrix} 0 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 4 & -2 \end{vmatrix}$ Since the first column (C1) consists entirely of zeros, the determinant is 0. $D_2 = 0$

3. $D_3 = \begin{vmatrix} 0 & 1 & 1 \\ 2 & 0 & 6 \\ -1 & 0 & 0 \end{vmatrix}$ Expand along R3: $D_3 = -1 \cdot \begin{vmatrix} 1 & 1 \\ 0 & 6 \end{vmatrix} - 0 \cdot \text{cofactor} + 0 \cdot \text{cofactor}$ $D_3 = -1 \cdot (1 \cdot 6 - 1 \cdot 0) = -1 \cdot 6 = -6$

Summing these determinants: $f'(0) = D_1 + D_2 + D_3 = 24 + 0 + (-6) = 18$

Step 3: Calculate $2f(0) + f'(0)$

Substitute the values of $f(0)$ and $f'(0)$: $2f(0) + f'(0) = 2(12) + 18 = 24 + 18 = 42$

Therefore, the final answer is 42.