1) x² - x - 6 ≥ 0
(x - 3)(x + 2) ≥ 0
+ - +
______[- 2]______[3]_______
////////////// ////////////////
x ∈ (- ∞ ; - 2] ∪ [3 , + ∞)
2) x² - x - 30 < 0
(x - 6)(x + 5) < 0
+ - +
_______(- 5)______(6)______
///////////////////
x ∈ (- 5 ; 6)
Окончательно :
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ /////////////////////////
______(- 5)_____[- 2]_______[3]_____(6)_____
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ответ : x ∈ (- 5 ; - 2] ∪ [3 ; 6)
ответ:
y=x^3-2x^2+x+2 y'=3x^2-2\cdot 2x+1=3x^2-4x+1
y= \sqrt{x} (2\sin x+1) y'=( \sqrt{x})' (2\sin x+1)+ \sqrt{x} (2\sin x+1)'= = \dfrac{1}{2 \sqrt{x} } (2\sin x+1)+ \sqrt{x} \cdot 2\cos x= \dfrac{\sin x}{ \sqrt{x} } + \dfrac{1}{2 \sqrt{x} } + 2\sqrt{x} \cos x
y= \dfrac{1}{x^2} =x^{-2} y'=-2x^{-2-1}=-2x^{-3}=- \dfrac{2}{x^3}
y= \dfrac{1}{\cos x} =(\cos x)^{-1} y'=-(\cos x)^{-1-1}\cdot (\cos x)'=-(\cos x)^{-2}\cdot (-\sin x)= \dfrac{\sin x}{\cos ^2x}
y=3x^2- \dfrac{2}{x^3} =3x^2- 2x^{-3} y'=3\cdot 2x- 2\cdot(-3x^{-4})=6x+ 6x^{-4}=6x+ \dfrac{6}{x^4}
y=\mathrm{tg}x+ \dfrac{1}{x} y'= \dfrac{1}{\cos^2x}- \dfrac{1}{x^2}
объяснение:
я перепесал с интернета без обид
