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 $f(x)$ for a given$, to determine which piece of the piecewise function to use based on the value of $x$.<br /><br />Given the function:<br />$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} $<br /><br />Let's evaluate the function for $x = -3$.<br /><br />Since $x = -3$ is greater than $-4$, we use the second piece of the function:<br />$f(x) = \frac{-1}{x+4} - 1(x+4)$<br /><br />Substituting $x = -3$ into the equation:<br />$f(-3) = \frac{-1}{-3+4} - 1(-3+4)$<br />$f(-3) = \frac{-1}{1} - 1(1)$<br />$f(-3) = -1 - 1$<br />$f(-3) = -2$<br /><br />Therefore, the value of the function $f(x)$ for $x = -3$ is $-2$.