Cho y=x^3-3x^2+2. Tìm x để: y' < 3
Giải thích
y = x3 – 3x2 + 2.
⇒ y’ = (x3 – 3x2 + 2)’
= (x3)’ – (3x2)’ + (2)’
= 3x2 – 3.2x + 0
= 3x2 – 6x.
y’ < 3
⇔ 3x2 – 6x < 3
⇔ 3x2 – 6x – 3 < 0
⇔ 1- √2 < x < 1 + √2.
y = x3 – 3x2 + 2.
⇒ y’ = (x3 – 3x2 + 2)’
= (x3)’ – (3x2)’ + (2)’
= 3x2 – 3.2x + 0
= 3x2 – 6x.
y’ < 3
⇔ 3x2 – 6x < 3
⇔ 3x2 – 6x – 3 < 0
⇔ 1- √2 < x < 1 + √2.