TP 2.2.1 . Bap. 21 Ucc.re/ToBaTb 3a,TaHHYIO QyHKIIUIOO . IIOCTPOHTb 3CKI f(x)= ) -6.000sqrt [3]((x+5)^2)+4.000,&xleqslant -4 (-1)/(x+4)-1(x+4)&,&xgt -4

To find the value of the function for a given x f(x)=\{ \begin{matrix} -6.000\sqrt [3]{(x+5)^{2}}+4.000,&x\leqslant -4\\ \frac {-1}{x+4}-1(x+4)&,&x\gt -4\end{matrix} x = -3 x = -3 -4 f(x) = \frac{-1}{x+4} - 1(x+4) x = -3 f(-3) = \frac{-1}{-3+4} - 1(-3+4) f(-3) = \frac{-1}{1} - 1(1) f(-3) = -1 - 1 f(-3) = -2 f(x) x = -3 -2$.