F(x) = x^4 - 4x^2+1
f '(x) = 4x^3 - 8x
1) Если f '(x) = 0
4x^3 - 8x = 0
4x(x^2 - 2) = 0
x(x^2 - 2) = 0
x = 0 или x^2 - 2 = 0
x = √2, x = -√2
Ответ: f '(x) = 0 при x = √2, x = -√2; x = 0
2) Если f '(x) < 0
4x^3 - 8x < 0
4x(x^2 - 2) < 0
x = 0 или x^2 - 2 = 0
x = √2, x = -√<span>2
</span>x ∈ (- ∞ ; -√2)∪(√2 ; + ∞)
3) Если f '(x) > 0
4x^3 - 8x > 0
4x(x^2 - 2) > 0
x = 0 или x^2 - 2 = 0
x = √2, x = -√<span>2
</span>x ∈ (-√2; 0)∪(√2 ; + ∞<span>)</span>
2)0,5=5/10=1/2
4)0,84=84/100=21/25
6)0,59=59/100
8)0,96=96/100=24/25
10)0,975=975/1000=39/40
<span>Верное......................................... </span>