Question

СРОЧНООООООО ХЕЛППППП РЕБЯТАА

Answer 1

$1)\; \; y=11x^4-\frac{2}{x^3}+3tg3x+4e^{2x}-3^{x}+5\\\\y'=44x^3+\frac{6}{x^4}+\frac{9}{cos^23x}+8e^{2x}-3^{x}\, ln3\\\\\\2)\; \; f(x)=tg(3x^4-2)-ln(7x+5)\\\\f'(x)=\frac{12x^3}{cos^2(3x^4-2)}-\frac{7}{7x+5}$

$3)\; \; y=\frac{x^5}{5}-\frac{x^3}{3}-6x+1\\\\y'=x^4-x^2-6=0\; \; \to \; \; (x^2+2)(x-\sqrt3)(x+\sqrt3)=0\\\\znaki\; y'\, :\; \; \; +++(-\sqrt3)---(\sqrt3)+++\\\\.\qquad \qquad \qquad \nearrow \; (-\sqrt3)\; \; \; \searrow \quad (\sqrt3)\; \; \nearrow \\\\x_{max}=-\sqrt3\; ,\; \; x_{min}=\sqrt3\\\\y_{max}=-\frac{9\sqrt3}{5}+7\sqrt3+1=\frac{26\sqrt3+5}{5}\\\\y_{min}=\frac{9\sqrt3}{5}-7\sqrt3+1=\frac{5-26\sqrt3}{5}$

$y\; vozrzstzet:\; \; x\in (-\infty ;-\sqrt3\, ]\cup [\; \sqrt3;+\infty )\\\\y\; ybuvaet:\; \; x\in [-\sqrt3\, ;\, \sqrt3\; ]$