0\\\\x(x+24)\geq 0\\\\Metod\; intervalov:\\\\x_1=0\; ,\; \; x+24=0\; ,\; \; x_2=-24\\\\+++(-24)---(0)+++\\\\x\in (-\infty ,-24)\cup (0,+\infty ) " alt=" y=log_3(x^2+24x)\\\\OOF:\; \; \; x^2+24x>0\\\\x(x+24)\geq 0\\\\Metod\; intervalov:\\\\x_1=0\; ,\; \; x+24=0\; ,\; \; x_2=-24\\\\+++(-24)---(0)+++\\\\x\in (-\infty ,-24)\cup (0,+\infty ) " align="absmiddle" class="latex-formula">