Question

Найдите значение функции f(x) в точку x0

Answer 1

$f(x) = {2x}^{2} - 3x + 7$

$f( \frac{1}{2} ) = {2( \frac{1}{2}) }^{2} - \frac{3}{2} + 7 = \frac{2}{4} - 1.5 + 7 = 0.5 - 1.5 + 7 = 6$

$f(1) = 2 \times 1 {}^{2} - 3 \times 1 + 7 = 2 - 3 + 7 = 6$

$f( - 2) = 2 \times ( - 2) {}^{2} - 3 \times ( - 2) + 7 = 8 + 6 + 7 = 21$

$f(x) = \frac{1}{ {x}^{3} } - x$

$f(1) = \frac{1}{1 {}^{3} } - 1 = 0$

$f(2) = \frac{1}{2 {}^{3} } - 2 = \frac{1}{8} - 2 = \frac{1 - 16}{8} = \frac{ - 15}{8} = - 1 \frac{7}{8}$

$f( - 3) = \frac{1}{ { - 3}^{3} } + 3 = - \frac{1}{27} + 3 = \frac{81 - 1}{27} = \frac{80}{27} = 2 \frac{26}{27}$