- + - +
______₀______₀______₀_______
- 1 0 1
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ответ : x ∈ (- 1 ; 0 ) ∪ (1 ; + ∞)
![f(x)=(x-1)^{2}\cdot(x+1)^{2}=(x^{2} -1)^{2}\\\\f'(x)=[(x^{2}-1)^{2} ]'=2(x^{2}-1)\cdot(x^{2} -1)'=4x(x^{2} -1)\\\\f'(x)0\\\\4x(x^{2}-1)0\\\\x(x-1)(x+1)0](/tpl/images/2001/9187/5d5ee.png)
