Answer to Question #344749 in Algebra for Geoffroy yawogan

Question #344749

1. Find the domain of the function ๐‘“(๐‘ฅ)=ln(โˆ’2๐‘ฅ2โˆ’๐‘ฅโˆ’6)+โˆš๐‘ฅ2โˆ’1.

 

2. Find the inverse function of the function ๐‘“(๐‘ฅ)=๐‘ฅ2โˆ’4๐‘ฅ+5,๐‘ฅโˆˆใ€ˆ3,4โŸฉ. Find the domain and the range of the inverse function.

 

3. Construct the tangent line to the graph of the function ๐‘“(๐‘ฅ)=4๐‘ฅโ‹…โˆš๐‘ฅโˆ’2โ‹…โˆš๐‘ฅ which is parallel to the line ๐‘ฆ=๐‘ฅ.

 

4. Find the maximal intervals of monotonicity of the function ๐‘“(๐‘ฅ)=๐‘’๐‘ฅ+3๐‘ฅ2+2๐‘ฅ+6.

 

5. Find the integral โˆซ6โ‹…๐‘ฅ3โ‹…๐‘’๐‘ฅ2+2๐‘‘๐‘ฅ.

 

6. Find the general solution of the differential equation ๐‘ฅ2+1+๐‘ฆโ€ฒโ‹…cos(๐‘ฆ)=0. 


1
Expert's answer
2022-05-26T07:03:25-0400

1.

a)


f(x)=lnโก(โˆ’2x2โˆ’xโˆ’6)+x2โˆ’1f(x)=\ln(-2x^2-x-6)+\sqrt{x^2-1}


โˆ’2x2โˆ’xโˆ’6>0-2x^2-x-6>0x2โˆ’1โ‰ฅ0x^2-1\ge0


D=(โˆ’1)2โˆ’4(โˆ’2)(โˆ’6)=โˆ’47<0D=(-1)^2-4(-2)(-6)=-47<0

Then โˆ’2x2โˆ’xโˆ’6<0,xโˆˆR-2x^2-x-6<0, x\in \R


Domain:{}Domain:\{\}


b)


f(x)=lnโก(โˆ’2x2โˆ’xโˆ’6)+x2โˆ’1f(x)=\ln(-\dfrac{2}{x^2-x-6})+\sqrt{x^2-1}


x2โˆ’xโˆ’6<0x^2-x-6<0x2โˆ’1โ‰ฅ0x^2-1\ge0

(x+2)(xโˆ’3)<0(x+2)(x-3)<0

xโ‰คโˆ’1 or xโ‰ฅ1x\le -1\ or\ x\ge1

Domain:(โˆ’2,โˆ’1]โˆช[1,3)(-2, -1]\cup[1,3)


2.


๐‘“(๐‘ฅ)=x2โˆ’4x+5,xโˆˆ[3,4]๐‘“(๐‘ฅ)=x^2โˆ’4x+5,x\in [3, 4]

xv=โˆ’โˆ’42(1)=2x_v=-\dfrac{-4}{2(1)}=2

The function ff increases on (3,4)(3, 4)


f(3)=(3)2โˆ’4(3)+5=2f(3)=(3)^2โˆ’4(3)+5=2

f(4)=(4)2โˆ’4(4)+5=5f(4)=(4)^2โˆ’4(4)+5=5

Domain: [3,4][3, 4]

Range: [2,5][2, 5]


y=x2โˆ’4x+5,3โ‰คxโ‰ค4y=x^2-4x+5, 3\le x\le4

Change xx and yy

x=y2โˆ’4y+5,3โ‰คyโ‰ค4x=y^2-4y+5, 3\le y\le 4

Solve for yy

y2โˆ’4y+4=xโˆ’1y^2-4y+4=x-1(yโˆ’2)2=xโˆ’1(y-2)^2=x-1

Since 3โ‰คyโ‰ค43\le y\le 4


yโˆ’2=xโˆ’1y-2=\sqrt{x-1}




y=2+xโˆ’1y=2+\sqrt{x-1}

Then


fโˆ’1(x)=2+xโˆ’1f^{-1}(x)=2+\sqrt{x-1}


Domain: [2,5][2, 5]

Range: [3,4][3, 4]


3.


f(x)=4xxโˆ’2xf(x)=4x\sqrt{x}-2\sqrt{x}

Domain: [0,โˆž)[0, \infin)


fโ€ฒ(x)=4(32)xโˆ’22x=6xโˆ’1xf'(x)=4(\dfrac{3}{2})\sqrt{x}-\dfrac{2}{2\sqrt{x}}=\dfrac{6x-1}{\sqrt{x}}

slope=fโ€ฒ(x)=6xโˆ’1x=1slope=f'(x)=\dfrac{6x-1}{\sqrt{x}}=1

6xโˆ’xโˆ’1=06x-\sqrt{x}-1=0

(3x+1)(2xโˆ’1)=0(3\sqrt{x}+1)(2\sqrt{x}-1)=0

Since xโ‰ฅ0,\sqrt{x}\ge0, we take 2xโˆ’1=02\sqrt{x}-1=0


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

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

f(14)=4(14)14โˆ’214=โˆ’12f(\dfrac{1}{4})=4(\dfrac{1}{4})\sqrt{\dfrac{1}{4}}-2\sqrt{\dfrac{1}{4}}=-\dfrac{1}{2}

The tangent line to the graph is


y+12=xโˆ’14y+\dfrac{1}{2}=x-\dfrac{1}{4}


y=xโˆ’34y=x-\dfrac{3}{4}

4.


f(x)=ex+3x2+2x+6f(x)=e^x+3x^2+2x+6

Domain: (โˆ’โˆž,โˆž)(-\infin, \infin)


fโ€ฒ(x)=ex+6x+2f'(x)=e^x+6x+2

fโ€ฒ(x)=0=>ex+6x+2=0f'(x)=0=>e^x+6x+2=0

(ex+6x+2)โ€ฒ=ex+6>0,xโˆˆR(e^x+6x+2)'=e^x+6>0, x\in \R

The function g(x)=ex+6x+2g(x)=e^x+6x+2 is strictly increasing on (โˆ’โˆž,โˆž).(-\infin, \infin).

The equation ex+6x+2=0e^x+6x+2=0 has the only solution xโ‰ˆโˆ’0.4406.x\approx-0.4406.


The function f(x)f(x) is monotonic decreasing on (โˆ’โˆž,โˆ’0.4406).(-\infin, -0.4406).

The function f(x)f(x) is monotonic increasing on (โˆ’0.4406,โˆž).( -0.4406, \infin).


5.


โˆซ6x3ex2+2dx\int 6x^3e^{x^2+2}dx

t=x2,dt=2xdxt=x^2, dt=2xdx


โˆซ6x3ex2+2dx=3e2โˆซtetdt\int 6x^3e^{x^2+2}dx=3e^2\int te^tdt

โˆซtetdt\int te^t dt

u=t,du=dtu=t, du=dt

dv=etdt,v=etdv=e^tdt, v=e^t

โˆซtetdt=tetโˆ’โˆซetdt=tetโˆ’et+C1\int te^t dt=te^t -\int e^t dt=te^t-e^t+C_1


โˆซ6x3ex2+2dx=3ex2+2(x2โˆ’1)+C\int 6x^3e^{x^2+2}dx=3e^{x^2+2}(x^2-1)+C

 6.


x2+1+yโ€ฒcosโก(y)=0x^2+1+y'\cos(y)=0

cosโก(y)dy=โˆ’(x2+1)dx\cos(y)dy=-(x^2+1)dx

Integrate


โˆซcosโก(y)dy=โˆ’โˆซ(x2+1)dx\int \cos(y)dy=-\int (x^2+1)dx

sinโก(y)=โˆ’x33โˆ’x+C\sin (y)=-\dfrac{x^3}{3}-x+C

sinโก(y)+x33+x=C\sin (y)+\dfrac{x^3}{3}+x=C


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