Question

С, что-нибудь ! 1. найдите производную функции f(x)=(3x+2)^3*(2x-1)^4 2. вычислите производную функции f(x)=x^2-x-6 в точках пересечения графика этой функции с осями координат 3. решите неравенство (cos2x+3tgпи/8)'> =2cosx

Answer 1

$1)\; \; f(x)=(3x+2)^3\cdot (2x-1)^4\\\\f'(x)=3\cdot (3x+2)^2\cdot 3\cdot (2x-1)^4+(3x+2)^3\cdot 4\cdot (2x-1)^3\cdot 2=\\\\=9\cdot (3x+2)^2(2x-1)^4+8\cdot (3x+2)^3(2x-1)^3\\\\2)\; \; f(x)=x^2-x-6\\\\f(x)=0\; ,\; \; esli\; \; x^2-x-6=(x-3)(x+2)=0\; ,\; x_1=-2,\; x_2=3\\\\f'(x)=2x-1\\\\f'(-2)=2\cdot (-2)-1=-5\\\\f'(3)=2\cdot 3-1=5$

$3)\; \; (cos2x+\underbrace {3tg\frac{\pi}{8}}_{const})' \geq 2cosx\\\\-2sin2x+0 \geq 2cosx\; |:2\\\\cosx+sin2x \leq 0\\\\cosx+2sina\cdot cosx \leq 0\\\\cosx(1+2sinx) \leq 0\\\\a)\; \; \left \{ {{cosx \leq 0} \atop {1+2sinx \geq 0}} \right. \; \left \{ {{\frac{\pi}{2}+2\pi n \leq x \leq \frac{3\pi}{2}+2\pi n} \atop {sinx \geq -\frac{1}{2}}} \right. \; \left \{ {{\frac{\pi}{2}+2\pi n \leq x \leq \frac{3\pi}{2}+2\pi n} \atop { -\frac{\pi}{6}+2\pi k\leq x \leq \frac{7\pi}{6}+2\pi k}} \right. \; \Rightarrow$

$\frac{\pi}{2}+2\pi n \leq x \leq \frac{7\pi}{6}+2\pi n\; ,\; n\in Z\\\\b)\; \; \left \{ {{cosx \geq 0} \atop {sinx \leq -\frac{1}{2}}} \right. \; \left \{ {{-\frac{\pi}{2}+2\pi n \leq x \leq \frac{\pi}{2}+2\pi n} \atop { -\frac{5\pi}{6}+2\pi k\leq x \leq -\frac{\pi}{6}+2\pi k}} \right. \; \; \Rightarrow \\\\ -\frac{\pi}{2}+2\pi n\leq x \leq -\frac{\pi}{6}+2\pi n,\; n\in Z\\\\Otvet:\; \; x\in [-\frac{\pi}{2}+2\pi n;-\frac{\pi}{6}+2\pi n]\cup [\frac{\pi}{2}+2\pi n;\frac{7\pi}{6}+2\pi n]$

Answer 2

1.
f (x) = (3x + 2)³·(2x - 1)⁴
f'(x) = 3·(3x + 2)²·3·(2x - 1)⁴ + (3x + 2)³·4·(2x - 1)³·2 = (3x + 2)²·(2x - 1)³·(9·(2x - 1) + 8·(3x + 2)) = (3x + 2)²·(2x - 1)³·(18x - 9 + 24x + 16) = (3x + 2)²·(2x - 1)³·(42x + 7) = 7·(3x + 2)²·(2x - 1)³·(6x + 1)

2.
f (x) = x² - x - 6
f'(x) = 2x - 1
Координаты x точек пересечения с Oх:
x² - x - 6 = 0
По теореме Виета:
x₁ = -2
x₂ = 3
Координата x точки пересечения с Oy: x₃ = 0.
f'(-2) = 2·(-2) - 1 = -5
f'(3) = 2·3 - 1 = 5
f'(0) = 2·0 - 1 = -1

3.
(cos 2x + 3·tg π/8)' ≥ 2·cos x
-2·sin 2x ≥ 2·cos x
-sin 2x ≥ cos x
cos x + sin 2x ≤ 0
cos x + 2·sin x·cos x ≤ 0
cos x·(1 + 2·sin x) ≤ 0

cos x ≤ 0                  cos x ≥ 0
(1 + 2·sin x) ≥ 0       (1 + 2·sin x) ≤ 0

cos x ≤ 0                  cos x ≥ 0
sin x ≥ -1/2               sin x ≤ -1/2

x ∈ [π/2 + 2πn; 3π/2 + 2πn], n ∈ Z         x ∈ [-π/2 + 2πm; π/2 + 2πm], m ∈ Z
x ∈ [-π/6 + 2πk; 7π/6 + 2πk], k ∈ Z        x ∈ [7π/6 + 2πp; 11π/6 + 2πp], p ∈ Z

x ∈ [π/2 + 2πn; 7π/6 + 2πn], n ∈ Z         x ∈ [3π/2 + 2πk; 11π/6 + 2πk], k ∈ Z

x ∈ [π/2 + 2πn; 7π/6 + 2πn] ∪ [3π/2 + 2πn; 11π/6 + 2πn), n ∈ Z