Answer
$$\sqrt{f(-1) - f(0)} - [g(2)]^{2} + f(-2) \div g(2) \cdot g(-1) = -38$$
Work Step by Step
$$\sqrt{f(-1) - f(0)} - [g(2)]^{2} + f(-2) \div g(2) \cdot g(-1)$$
The values of $f(-1), f(0), g(2), f(-2), g(-1)$ can be obtained from the given table. The values in the parentheses indicate the value of $f$ or a given value of $x$
Based on the table:
$f(-1) = 3$
$f(0) = -1$
$g(2) = -6$
$f(-2) = 6$
$g(-1) = 4$
Plugging these values to the original equation:
$$=\sqrt{3 - (-1)} - [(-6)]^{2} + 6 \div (-6) \cdot 4$$ $$=\sqrt{4} - 36 + 6 \div (-6) \cdot 4$$ $$=2 - 36 + (-1) \cdot 4$$ $$=2 - 36 -4$$ $$=2 - 36 -4$$ $$=- 34 -4$$ $$=- 38$$
