Question #345265

1.Find the domain of the function 𝑓(𝑥)=log4(𝑥2+4𝑥−5𝑒𝑥−1).

 

2. Construct the tangent line to the graph of the function 𝑓(𝑥)=√𝑥+2𝑥 which is perpendicular to the line 𝑥+3𝑦−2=0.

 

3. Find the maximal intervals of monotonicity of the function 𝑓(𝑥)=arctg(𝑥2−4𝑥).

 

4. Calculate the integral ∫𝑥2(𝑥3+4)2 𝑑𝑥2−1.

 

5. Find the particular solution of the differential equation 𝑦′=−(2𝑦+1)⋅tg(𝑥) which fulfills the initial condition 𝑦(𝜋4)=12.

 

6. Solve the matrix equation 2𝒳+𝒜=3ℬ−𝒜𝒳 if 𝒜=(−4−332), ℬ=(1−121).

 


1
Expert's answer
2022-05-27T12:52:58-0400

1.


x2+4𝑥5ex1>0x^2+4𝑥−5e^x−1>0

x<4.252x<-4.252

Domain: (,4.252)(-\infin, -4.252)


2.


x+3y2=0x+3y-2=0

y=13x+23y=-\dfrac{1}{3}x+\dfrac{2}{3}

The slope of the tangent line

slope=m=113=3slope=m=-\dfrac{1}{-\dfrac{1}{3}}=3

𝑓(𝑥)=x+2𝑥𝑓(𝑥)=\sqrt{x}+2𝑥

f(x)=12x+2f'(x)=\dfrac{1}{2\sqrt{x}}+2

slope=m=3=12x+2slope=m=3=\dfrac{1}{2\sqrt{x}}+2

12x=1\dfrac{1}{2\sqrt{x}}=1

x=14x=\dfrac{1}{4}

y(14)=14+2(14)=1y(\dfrac{1}{4})=\sqrt{\dfrac{1}{4}}+2(\dfrac{1}{4})=1

The equation of the tangent line in point-slope form is


y1=3(x14)y-1=3(x-\dfrac{1}{4})

The equation of the tangent line in slope-intercept form is


y=3x+14y=3x+\dfrac{1}{4}



3.

Domain:(,)(-\infin, \infin)


f(x)=(arctan(x24x))=2x41+(x24x)2f'(x)=(\arctan(x^2-4x))'=\dfrac{2x-4}{1+(x^2-4x)^2}

f(x)=0=>2x41+(x24x)2=0f'(x)=0=>\dfrac{2x-4}{1+(x^2-4x)^2}=0

x=2x=2

If x<2,f(x)<0,f(x)x<2, f'(x)<0, f(x) decreases.

If x>2,f(x)>0,f(x)x>2, f'(x)>0, f(x) increases.

The function ff is monotone decreasing on (,2).(-\infin, 2).

The function ff is monotone increasing on (2,).(2,\infin).


4.


x2(x3+4)2dx\int x^2(x^3+4)^2 dx

u=x3+4,du=3x2dxu=x^3+4, du=3x^2dx

x2(x3+4)2dx=13u2du=u39+C\int x^2(x^3+4)^2 dx=\dfrac{1}{3}\int u^2du=\dfrac{u^3}{9}+C

=(x3+4)39+C=\dfrac{(x^3+4)^3}{9}+C

5.


dy2y+1=tan(x)dx\dfrac{dy}{2y+1}=-\tan (x)dx

Integrate


dy2y+1=tan(x)dx\int \dfrac{dy}{2y+1}=-\int\tan (x)dx

tan(x)dx=sin(x)cos(x)dx=lncos(x)+12lnC\int\tan (x)dx=\int\dfrac{\sin (x)}{\cos(x)}dx=-\ln|\cos(x)|+\dfrac{1}{2}\ln C

12ln2y+1=lncos(x)+12lnC\dfrac{1}{2}\ln|2y+1|=\ln|\cos(x)|+\dfrac{1}{2}\ln C

2y+1=Ccos2(x)2y+1=C\cos ^2 (x)

Given y(π4)=12y(\dfrac{\pi}{4})=12


2(12)+1=Ccos2(π4)2(12)+1=C\cos ^2 (\dfrac{\pi}{4})

C=50C=50

The particular solution of the differential equation

2y+1=50cos2(x)2y+1=50\cos ^2 (x)

6.


2X+A=3BAX2X+A=3B-AX


A=(4332),B=(1121)A=\begin{pmatrix} -4 & -3 \\ 3 & 2 \end{pmatrix}, B=\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}

Let


X=(x11x12x21x22)X=\begin{pmatrix} x_{11}& x_{12} \\ x_{21} & x_{22} \end{pmatrix}

Then


2X=(2x112x122x212x22)2X=\begin{pmatrix} 2x_{11}& 2x_{12} \\ 2x_{21} & 2x_{22} \end{pmatrix}

AX=(4332)(x11x12x21x22)AX=\begin{pmatrix} -4 & -3 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} x_{11}& x_{12} \\ x_{21} & x_{22} \end{pmatrix}

=(4x113x214x123x223x11+2x213x12+2x22)=\begin{pmatrix} -4x_{11}-3x_{21}& -4x_{12}-3x_{22} \\ 3x_{11}+2x_{21} & 3x_{12}+2x_{22} \end{pmatrix}


2X+AX=(2x113x212x123x223x11+4x213x12+4x22)2X+AX=\begin{pmatrix} -2x_{11}-3x_{21}& -2x_{12}-3x_{22} \\ 3x_{11}+4x_{21} & 3x_{12}+4x_{22} \end{pmatrix}

3BA=(7031)3B-A=\begin{pmatrix} 7 & 0 \\ 3 & 1 \end{pmatrix}

(2x113x212x123x223x11+4x213x12+4x22)=(7031)\begin{pmatrix} -2x_{11}-3x_{21}& -2x_{12}-3x_{22} \\ 3x_{11}+4x_{21} & 3x_{12}+4x_{22} \end{pmatrix}=\begin{pmatrix} 7 & 0 \\ 3 & 1 \end{pmatrix}


2x113x21=7-2x_{11}-3x_{21}=73x11+4x21=33x_{11}+4x_{21}=3

2x123x22=0-2x_{12}-3x_{22}=03x12+4x22=13x_{12}+4x_{22}=1

x21=27x_{21}=272x113x21=7-2x_{11}-3x_{21}=7


x22=2-x_{22}=23x12+4x22=13x_{12}+4x_{22}=1

x11=37,x21=27,x22=2,x12=3x_{11}=37, x_{21}=-27, x_{22}=-2, x_{12}=3


X=(372732)X=\begin{pmatrix} 37& -27 \\ 3 & -2 \end{pmatrix}

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