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a) Cách 1: y' = (9 -2x)'(2x3- 9x2 +1) +(9 -2x)(2x3- 9x2 +1)' = -2(2x3- 9x2 +1) +(9 -2x)(6x2 -18x) = -16x3 +108x2 -162x -2.
Cách 2: y = -4x4 +36x3 -81x2 -2x +9, do đó
y' = -16x3 +108x2 -162x -2.
b) y' = .(7x -3) +
(7x -3)'=
(7x -3) +7
.
c) y' = (x -2)'√(x2 +1) + (x -2)(√x2 +1)' = √(x2 +1) + (x -2) = √(x2 +1) + (x -2)
= √(x2 +1) +
=
.
d) y' = 2tanx.(tanx)' - (x2)' =
.
e) y' = sin
=
sin
.
a: \(y=\left(x+2\right)^2=x^2+4x+4\)
=>\(y'=2x+4\)
Đặt y'>0
=>2x+4>0
=>x>-2
Đặt y'<0
=>2x+4<0
=>x<-2
Vậy: Hàm số đồng biến trên \(\left(-2;+\infty\right)\) và nghịch biến trên \(\left(-\infty;-2\right)\)
b: \(y=\left(x^2-1\right)\left(x+2\right)\)
=>\(y'=\left(x^2-1\right)'\cdot\left(x+2\right)+\left(x^2-1\right)\left(x+2\right)'\)
\(=2x\left(x+2\right)+x^2-1=2x^2+4x+x^2-1=3x^2+4x-1\)
Đặt y'>0
=>\(3x^2+4x-1>0\)
=>\(\left[{}\begin{matrix}x>\dfrac{-2+\sqrt{7}}{3}\\x< \dfrac{-2-\sqrt{7}}{3}\end{matrix}\right.\)
Đặt y'<0
=>\(3x^2+4x-1< 0\)
=>\(\dfrac{-2-\sqrt{7}}{3}< x< \dfrac{-2+\sqrt{7}}{3}\)
Vậy: Hàm số đồng biến trên các khoảng \(\left(-\infty;\dfrac{-2-\sqrt{7}}{3}\right);\left(\dfrac{-2+\sqrt{7}}{3};+\infty\right)\)
Hàm số nghịch biến trên khoảng \(\left(\dfrac{-2-\sqrt{7}}{3};\dfrac{-2+\sqrt{7}}{3}\right)\)
c: \(y=\left(x+2\right)\left(2x^2-3\right)\)
=>\(y'=\left(x+2\right)'\left(2x^2-3\right)+\left(x+2\right)\left(2x^2-3\right)'\)
\(=2x^2-3+4x\left(x+2\right)\)
\(=6x^2+8x-3\)
Đặt y'>0
=>\(6x^2+8x-3>0\)
=>\(\left[{}\begin{matrix}x>\dfrac{-4+\sqrt{34}}{6}\\x< \dfrac{-4-\sqrt{34}}{6}\end{matrix}\right.\)
Đặt y'<0
=>\(6x^2+8x-3< 0\)
=>\(\dfrac{-4-\sqrt{34}}{6}< x< \dfrac{-4+\sqrt{34}}{6}\)
Vậy: hàm số đồng biến trên các khoảng \(\left(-\infty;\dfrac{-4-\sqrt{34}}{6}\right);\left(\dfrac{-4+\sqrt{34}}{6};+\infty\right)\)
Hàm số nghịch biến trên khoảng \(\left(\dfrac{-4-\sqrt{34}}{6};\dfrac{-4+\sqrt{34}}{6}\right)\)
d: \(y=\left(x-1\right)^2\left(x+2\right)\)
\(=\left(x^2-2x+1\right)\left(x+2\right)\)
\(=x^3+2x^2-2x^2-4x+x+2\)
=>\(y=x^3-3x+2\)
=>\(y'=3x^2-3\)
Đặt y'>0
=>\(3x^2-3>0\)
=>\(x^2>1\)
=>\(\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
Đặt y'<0
=>\(3x^2-3< 0\)
=>x^2<1
=>-1<x<1
Vậy: Hàm số đồng biến trên các khoảng \(\left(1;+\infty\right);\left(-\infty;-1\right)\)
Hàm số nghịch biến trên khoảng (-1;1)
\(L_1=\lim\limits_{x\rightarrow0}\frac{x\left(x^2+3x-2\right)}{x\left(x^4+4\right)}=\lim\limits_{x\rightarrow0}\frac{x^2+3x-2}{x^4+4}=-\frac{1}{2}\)
\(L_2=\lim\limits_{x\rightarrow+\infty}\frac{1-\frac{3}{x^2}+\frac{2}{x^3}}{\left(\frac{4}{x}-2\right)^3}=\frac{1}{\left(-2\right)^3}=-\frac{1}{8}\)
\(L_3=\lim\limits_{x\rightarrow-1}\frac{\left(2x+1\right)\left(x+1\right)}{x\left(x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{2x+1}{x}=1\)
\(L_4=\lim\limits_{x\rightarrow2}\frac{x^2-4x+1}{4-x^2}=\frac{1}{0}=+\infty\)
\(L_5=\lim\limits_{x\rightarrow3}\frac{\sqrt{x+1}-2}{x-2}=\frac{0}{1}=0\)
\(L_6=\lim\limits_{x\rightarrow1}\frac{x+3-\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{-\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{-\left(x+2\right)}{\left(x+1\right)\left(\sqrt{x+3}+x+1\right)}=\frac{-3}{2.4}=-\frac{3}{8}\)
\(L_7=\lim\limits_{x\rightarrow+\infty}\frac{x^2+x+1-\left(x-1\right)^2}{\sqrt{x^2+x+1}+x-1}\lim\limits_{x\rightarrow+\infty}\frac{3x}{\sqrt{x^2+x+1}+x-1}=\lim\limits_{x\rightarrow+\infty}\frac{3}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+1-\frac{1}{x}}=\frac{3}{2}\)
\(L_8=\lim\limits_{x\rightarrow-\infty}\frac{x^2+x+1-\left(3x-2\right)^2}{\sqrt{x^2+x+1}+3x-2}=\lim\limits_{x\rightarrow-\infty}\frac{-8x^2+13x-3}{\sqrt{x^2+x+1}+3x-2}=\lim\limits_{x\rightarrow-\infty}\frac{-8+\frac{13}{x}-\frac{3}{x^2}}{-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+3-\frac{2}{x}}=\frac{-8}{-1+3}=-4\)
a: \(y=\left(5x-10\right)^4\)
=>\(y'=4\cdot\left(5x-10\right)'\cdot\left(5x-10\right)^3\)
\(=4\cdot5\cdot\left(5x-10\right)^3=20\left(5x-10\right)^3\)
Đặt y'>0
=>\(20\left(5x-10\right)^3>0\)
=>\(\left(5x-10\right)^3>0\)
=>5x-10>0
=>x>2
Đặt y'<0
=>\(20\left(5x-10\right)^3< 0\)
=>\(\left(5x-10\right)^3< 0\)
=>5x-10<0
=>x<2
Vậy: hàm số đồng biến trên \(\left(2;+\infty\right)\)
Hàm số nghịch biến trên \(\left(-\infty;2\right)\)
c: \(y=\left(x^3-1\right)^3\)
=>\(y'=3\left(x^3-1\right)'\cdot\left(x^3-1\right)^2\)
\(=9x^2\left(x^3-1\right)^2>=0\forall x\)
=>Hàm số luôn đồng biến trên R
d: \(y=\left(x^2-1\right)\left(x+2\right)\)
=>\(y'=\left(x^2-1\right)'\left(x+2\right)+\left(x^2-1\right)\left(x+2\right)'\)
\(=2x\left(x+2\right)+x^2-1\)
\(=2x^2+4x+x^2-1=3x^2+4x-1\)
Đặt y'>0
=>\(3x^2+4x-1>0\)
=>\(\left[{}\begin{matrix}x< \dfrac{-2-\sqrt{7}}{3}\\x>\dfrac{-2+\sqrt{7}}{3}\end{matrix}\right.\)
Đặt y'<0
=>\(3x^2+4x-1< 0\)
=>\(\dfrac{-2-\sqrt{7}}{3}< x< \dfrac{-2+\sqrt{7}}{3}\)
Vậy: Hàm số đồng biến trên các khoảng \(\left(-\infty;\dfrac{-2-\sqrt{7}}{3}\right);\left(\dfrac{-2+\sqrt{7}}{3};+\infty\right)\)
Hàm số nghịch biến trên khoảng \(\left(\dfrac{-2-\sqrt{7}}{3};\dfrac{-2+\sqrt{7}}{3}\right)\)
b: \(y=\left(-x-1\right)\left(x+2\right)^4\)
=>\(y'=\left(-x-1\right)'\left(x+2\right)^4+\left(-x-1\right)\left[\left(x+2\right)^4\right]'\)
\(=-\left(x+2\right)^4+\left(-x-1\right)\cdot4\left(x+2\right)'\left(x+2\right)^3\)
\(=-\left(x+2\right)^4+4\left(x+2\right)^3\cdot\left(-x-1\right)\)
\(=-\left(x+2\right)^3\left[\left(x+2\right)+4\left(x+1\right)\right]\)
\(=-\left(x+2\right)^2\cdot\left(x+2\right)\left(5x+6\right)\)
Đặt y'<0
=>\(-\left(x+2\right)^2\left(x+2\right)\left(5x+6\right)< 0\)
=>(x+2)(5x+6)>0
TH1: \(\left\{{}\begin{matrix}x+2>0\\5x+6>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>-2\\x>-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow x>-\dfrac{6}{5}\)
TH2: \(\left\{{}\begin{matrix}x+2< 0\\5x+6< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< -2\\x< -\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow x< -2\)
Đặt y'>0
=>(x+2)(5x+6)<0
TH1: \(\left\{{}\begin{matrix}x+2>0\\5x+6< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>-2\\x< -\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow-2< x< -\dfrac{6}{5}\)
TH2: \(\left\{{}\begin{matrix}x+2< 0\\5x+6>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< -2\\x>-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy: HSĐB trên các khoảng \(\left(-\infty;-2\right);\left(-\dfrac{6}{5};+\infty\right)\)
HSNB trên khoảng \(\left(-2;-\dfrac{6}{5}\right)\)
\(\text{1) Đ}K:\left\{{}\begin{matrix}sinx\ne0\\1-sinx\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne m\pi\\x\ne\frac{\pi}{2}+n2\pi\end{matrix}\right.\)
\(2\text{) }ĐK:\left\{{}\begin{matrix}cos\left(2x+\frac{\pi}{3}\right)\ne0\\sinx\ne0\end{matrix}\right.\Leftrightarrow\\ \left\{{}\begin{matrix}2x+\frac{\pi}{3}\ne\frac{\pi}{2}+m\pi\\x\ne n\pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{12}+\frac{m\pi}{2}\\x\ne n\pi\end{matrix}\right.\)
\(3\text{) }ĐK:\left\{{}\begin{matrix}\frac{5-3cos2x}{1+sin\left(2x-\frac{\pi}{2}\right)}\ge0\\1+sin\left(2x-\frac{\pi}{2}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5-3cos2x\ge0\\sin\left(2x-\frac{\pi}{2}\right)\ne-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}cos2x\le\frac{5}{3}\left(T/m\right)\\2x-\frac{\pi}{2}\ne\frac{3\pi}{2}+k2\pi\end{matrix}\right.\Leftrightarrow x\ne\pi+k\pi\)
\(4\text{) }ĐK:\left\{{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)\ne0\\cos\left(3x-\frac{\pi}{4}\right)\ne0\\tan\left(3x-\frac{\pi}{4}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+\frac{\pi}{3}\ne a\pi\\3x-\frac{\pi}{4}\ne\frac{\pi}{2}+b\pi\\3x-\frac{\pi}{4}\ne c\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\frac{\pi}{3}+a\pi\\x\ne\frac{\pi}{4}+\frac{b\pi}{3}\\x\ne\frac{\pi}{12}+\frac{c\pi}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-\frac{\pi}{3}+a\pi\\x\ne\frac{\pi}{12}+\frac{k\pi}{6}\end{matrix}\right.\)
a: \(y=\left(x^2-1\right)^2\)
=>\(y'=2\left(x^2-1\right)'\left(x^2-1\right)\)
\(=4x\left(x^2-1\right)\)
Đặt y'>0
=>\(x\left(x^2-1\right)>0\)
TH1: \(\left\{{}\begin{matrix}x>0\\x^2-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>0\\x^2>1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\)
=>\(x>1\)
TH2: \(\left\{{}\begin{matrix}x< 0\\x^2-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 0\\-1< x< 1\end{matrix}\right.\Leftrightarrow-1< x< 0\)
Đặt y'<0
=>\(x\left(x^2-1\right)< 0\)
TH1: \(\left\{{}\begin{matrix}x>0\\x^2-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>0\\x^2< 1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\-1< x< 1\end{matrix}\right.\)
=>0<x<1
TH2: \(\left\{{}\begin{matrix}x< 0\\x^2-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 0\\x^2>1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0\\\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\)
=>x<-1
Vậy: Hàm số đồng biến trên các khoảng \(\left(1;+\infty\right);\left(-1;0\right)\)
Hàm số nghịch biến trên các khoảng (0;1) và \(\left(-\infty;-1\right)\)
b: \(y=\left(3x+4\right)^3\)
=>\(y'=3\left(3x+4\right)'\left(3x+4\right)^2\)
\(\Leftrightarrow y'=9\left(3x+4\right)^2>=0\forall x\)
=>Hàm số luôn đồng biến trên R
c: \(y=\left(x+3\right)^2\left(x-1\right)\)
=>\(y=\left(x^2+6x+9\right)\left(x-1\right)\)
=>\(y'=\left(x^2+6x+9\right)'\left(x-1\right)+\left(x^2+6x+9\right)\left(x-1\right)'\)
=>\(y'=\left(2x+6\right)\left(x-1\right)+x^2+6x+9\)
=>\(y'=2x^2-2x+6x-6+x^2+6x+9\)
=>\(y'=3x^2-2x+3\)
\(\Leftrightarrow y'=3\left(x^2-\dfrac{2}{3}x+1\right)\)
=>\(y'=3\left(x^2-2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{8}{9}\right)\)
=>\(y'=3\left(x-\dfrac{1}{3}\right)^2+\dfrac{8}{3}>=\dfrac{8}{3}>0\forall x\)
=>Hàm số luôn đồng biến trên R
d: \(y=\left(2x+2\right)\left(x^3-1\right)\)
=>\(y'=\left(2x+2\right)'\left(x^3-1\right)+\left(2x+2\right)\left(x^3-1\right)'\)
\(=2\left(x^3-1\right)+3x^2\left(2x+2\right)\)
\(=2x^3-2+6x^3+6x^2\)
\(=8x^3+6x^2-2\)
Đặt y'>0
=>\(8x^3+6x^2-2>0\)
=>\(x>0,46\)
Đặt y'<0
=>\(8x^3+6x^2-2< 0\)
=>\(x< 0,46\)
Vậy: Hàm số đồng biến trên khoảng tầm \(\left(0,46;+\infty\right)\)
Hàm số nghịch biến trên khoảng tầm \(\left(-\infty;0,46\right)\)