Question

Решите неравенство f'(x)< 0 1) f(x)= 3/x-x^2 2) f(x)=x^4-4,5x^2+2 3) f(x)=-1/x^2-1/x-5 4)f(x)=x^4-4x-3

Answer 1

$1) \quad f(x) = \frac{3}{x} - x^2\\ \\ f'(x) = -\frac{3}{x^2}-2x; \quad f'(x) < 0\\ \\ -\frac{3}{x^2}-2x < 0 \\ \\ \frac{3}{x^2}+2x0 \\ \\ \frac{3 +2x^3}{x^2} 0 \\ \\ \\ ----(\sqrt[3]{-\frac 3 2})+++(0)++++_x \\ \\ x \in (\sqrt[3]{-\frac 3 2};0)\cup (0; +\infty)$

$2) \quad f(x) = x^4 - 4.5 x^2+2 \\ \\ f'(x) = 4x^3-9x;\\ \\ 4x^3 - 9x = 0 \\ \\ x(4x^2 - 9) = 0 \\ \\ x (2x-3)(2x+3) = 0 \\ x_1 = 0, ~x_2 = \frac 3 2, ~ x_3 = -\frac 3 2 \\ \\ f'(x) < 0 \\ \\ ---(-\frac 3 2)+++(0)---(\frac 3 2)+++_x \\ \\ x \in (-\infty;-\frac 3 2) \cup(0; \frac 3 2)$

$3) \quad f(x) = -\frac{1}{x^2} -\frac{1}{x} -5\\ \\ f'(x) = \frac{2}{x^3}+\frac{1}{x^2}; \\ \\ f'(x) < 0 \\ \\ \dfrac{2+x}{x^3} < 0 \\ \\ +++(-2)---(0)+++_x \\ \\ x \in (-2; 0)$

$4) \quad f(x) = x^4 -4x-3 \\ \\ f'(x) = 4x^3-4 \\ \\ f'(x) < 0 \\ \\ 4x^3 - 4 < 0 \quad \Big |:4 \\ \\ x^3 -1 < 0 \\ \\ ---(1)+++_x\\ \\ x\in (-\infty; 1)$