Question

Решите неравенство f'(x)≤0, если f(x)=4x+2x^2

Answer 1

Ответ:

f ' (x) = (4x + 2x^2) ' = (4x) ' + (2x^2) ' = 4 (x) ' + 2 (x^2) ' =

= 4*1 + 2*2x = 4 + 4x

f ' (x) ≤ 0

4x + 4 ≤ 0 / : 4

x + 1≤ 0

x ≤ - 1

x ∈ ( - ∞; - 1]