Question #344749

1. Find the domain of the function 𝑓(𝑥)=ln(−2𝑥2−𝑥−6)+√𝑥2−1.

 

2. Find the inverse function of the function 𝑓(𝑥)=𝑥2−4𝑥+5,𝑥∈〈3,4⟩. Find the domain and the range of the inverse function.

 

3. Construct the tangent line to the graph of the function 𝑓(𝑥)=4𝑥⋅√𝑥−2⋅√𝑥 which is parallel to the line 𝑦=𝑥.

 

4. Find the maximal intervals of monotonicity of the function 𝑓(𝑥)=𝑒𝑥+3𝑥2+2𝑥+6.

 

5. Find the integral ∫6⋅𝑥3⋅𝑒𝑥2+2𝑑𝑥.

 

6. Find the general solution of the differential equation 𝑥2+1+𝑦′⋅cos(𝑦)=0. 


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

1.

a)


f(x)=ln(2x2x6)+x21f(x)=\ln(-2x^2-x-6)+\sqrt{x^2-1}


2x2x6>0-2x^2-x-6>0x210x^2-1\ge0


D=(1)24(2)(6)=47<0D=(-1)^2-4(-2)(-6)=-47<0

Then 2x2x6<0,xR-2x^2-x-6<0, x\in \R


Domain:{}Domain:\{\}


b)


f(x)=ln(2x2x6)+x21f(x)=\ln(-\dfrac{2}{x^2-x-6})+\sqrt{x^2-1}


x2x6<0x^2-x-6<0x210x^2-1\ge0

(x+2)(x3)<0(x+2)(x-3)<0

x1 or x1x\le -1\ or\ x\ge1

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


2.


𝑓(𝑥)=x24x+5,x[3,4]𝑓(𝑥)=x^2−4x+5,x\in [3, 4]

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

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


f(3)=(3)24(3)+5=2f(3)=(3)^2−4(3)+5=2

f(4)=(4)24(4)+5=5f(4)=(4)^2−4(4)+5=5

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

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


y=x24x+5,3x4y=x^2-4x+5, 3\le x\le4

Change xx and yy

x=y24y+5,3y4x=y^2-4y+5, 3\le y\le 4

Solve for yy

y24y+4=x1y^2-4y+4=x-1(y2)2=x1(y-2)^2=x-1

Since 3y43\le y\le 4


y2=x1y-2=\sqrt{x-1}




y=2+x1y=2+\sqrt{x-1}

Then


f1(x)=2+x1f^{-1}(x)=2+\sqrt{x-1}


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

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


3.


f(x)=4xx2xf(x)=4x\sqrt{x}-2\sqrt{x}

Domain: [0,)[0, \infin)


f(x)=4(32)x22x=6x1xf'(x)=4(\dfrac{3}{2})\sqrt{x}-\dfrac{2}{2\sqrt{x}}=\dfrac{6x-1}{\sqrt{x}}

slope=f(x)=6x1x=1slope=f'(x)=\dfrac{6x-1}{\sqrt{x}}=1

6xx1=06x-\sqrt{x}-1=0

(3x+1)(2x1)=0(3\sqrt{x}+1)(2\sqrt{x}-1)=0

Since x0,\sqrt{x}\ge0, we take 2x1=02\sqrt{x}-1=0


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

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

f(14)=4(14)14214=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=x14y+\dfrac{1}{2}=x-\dfrac{1}{4}


y=x34y=x-\dfrac{3}{4}

4.


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

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


f(x)=ex+6x+2f'(x)=e^x+6x+2

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

(ex+6x+2)=ex+6>0,xR(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 (,).(-\infin, \infin).

The equation ex+6x+2=0e^x+6x+2=0 has the only solution x0.4406.x\approx-0.4406.


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

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


5.


6x3ex2+2dx\int 6x^3e^{x^2+2}dx

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


6x3ex2+2dx=3e2tetdt\int 6x^3e^{x^2+2}dx=3e^2\int te^tdt

tetdt\int te^t dt

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

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

tetdt=tetetdt=tetet+C1\int te^t dt=te^t -\int e^t dt=te^t-e^t+C_1


6x3ex2+2dx=3ex2+2(x21)+C\int 6x^3e^{x^2+2}dx=3e^{x^2+2}(x^2-1)+C

 6.


x2+1+ycos(y)=0x^2+1+y'\cos(y)=0

cos(y)dy=(x2+1)dx\cos(y)dy=-(x^2+1)dx

Integrate


cos(y)dy=(x2+1)dx\int \cos(y)dy=-\int (x^2+1)dx

sin(y)=x33x+C\sin (y)=-\dfrac{x^3}{3}-x+C

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


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