Question

Решите пожалуйста, даю много баллов​

Answer 1

$1)\; \; f(x)=x^3+6x^2-x^5+5\; \; ,\; \; f'(x)=3x^2+12x-5x^4\\\\2)\; \; f(x)=\frac{4x^3}{x^2-1}\\\\f'(x)=\frac{12x^2(x^2-1)-4x^3\cdot 2x}{(x^2-1)^2}=\frac{4x^4-12x^2}{(x^2-1)^2}=\frac{4x^2(x^2-3)}{(x^2-1)^2}\\\\3)\; \; \frac{16x^2-x}{12-x}\leq 0\; \; ,\; \; \frac{x(16x-1)}{-(x-12)}\leq 0\; \; ,\; \; \frac{x(16x-1)}{x-12}\geq 0\\\\x_1=0\; ,\; x_2=\frac{1}{16}\; ,\; x_3=12\\\\znaki:\; \; ---[\, 0\, ]+++[\, \frac{1}{16}\, ]---(12)+++\\\\x\in [\, 0,\frac{1}{16}\, ]\cup (12,+\infty )\\\\4)\; \; (4-x)(5+x)\cdot x^2\geq 0$

$-(x-4)(x+5)\cdot x^2\geq 0\\\\(x-4)(x+5)\cdot x^2\leq 0\; \; ,\; \; x_1=4\; ,\; x_2=-5\; ,\; x_3=0\\\\znaki:\; \; \; +++[-5\, ]---[\, 0\, ]---[\, 4\, ]+++\\\\x\in [-5,0\, ]\cup [\, 0,4\, ]\\\\x\in [-5,4\, ]$