1.
f'(x) = 12*3x²+18*2x-7 = 36x²+36x-7
36x²+36x-7>0
D = 36²-4*36*(-7) = 1296+1008 = 2304
√D = 48
X1 = (-36-48) / 2*36 = -84/72 = -21/18 = -7/6 = -1Ц 1/6
Х2 = (-36+48)/72 = 12/72 = 1/6
Ответ: (-∞, -1ц 1/6) ∪ (1/6, +∞)
2.
f'(x) = cosx
cosx>0
(-π/2 + 2πк; π/2+2πк)
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Answers & Comments
1.
f'(x) = 12*3x²+18*2x-7 = 36x²+36x-7
36x²+36x-7>0
D = 36²-4*36*(-7) = 1296+1008 = 2304
√D = 48
X1 = (-36-48) / 2*36 = -84/72 = -21/18 = -7/6 = -1Ц 1/6
Х2 = (-36+48)/72 = 12/72 = 1/6
Ответ: (-∞, -1ц 1/6) ∪ (1/6, +∞)
2.
f'(x) = cosx
cosx>0
(-π/2 + 2πк; π/2+2πк)