Answer to Question #345265 in Math for Geoffroy yawogan

Question

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).

 

Answer 1

1.

$$x^2+4𝑥−5e^x−1>0$$

$$x<-4.252$$

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

2.

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

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

The slope of the tangent line

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

$$𝑓(𝑥)=\sqrt{x}+2𝑥$$

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

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

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

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

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

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

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

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

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

3.

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

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

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

$$x=2$$

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

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

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

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

4.

$$\int x^2(x^3+4)^2 dx$$

$$u=x^3+4, du=3x^2dx$$

$$\int x^2(x^3+4)^2 dx=\dfrac{1}{3}\int u^2du=\dfrac{u^3}{9}+C$$

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

5.

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

Integrate

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

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

$$\dfrac{1}{2}\ln|2y+1|=\ln|\cos(x)|+\dfrac{1}{2}\ln C$$

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

Given $y(\dfrac{\pi}{4})=12$

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

$$C=50$$

The particular solution of the differential equation

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

6.

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

$$A=\begin{pmatrix} -4 & -3 \\ 3 & 2 \end{pmatrix}, B=\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}$$

Let

$$X=\begin{pmatrix} x_{11}& x_{12} \\ x_{21} & x_{22} \end{pmatrix}$$

Then

$$2X=\begin{pmatrix} 2x_{11}& 2x_{12} \\ 2x_{21} & 2x_{22} \end{pmatrix}$$

$$AX=\begin{pmatrix} -4 & -3 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} x_{11}& x_{12} \\ x_{21} & x_{22} \end{pmatrix}$$

$$=\begin{pmatrix} -4x_{11}-3x_{21}& -4x_{12}-3x_{22} \\ 3x_{11}+2x_{21} & 3x_{12}+2x_{22} \end{pmatrix}$$

$$2X+AX=\begin{pmatrix} -2x_{11}-3x_{21}& -2x_{12}-3x_{22} \\ 3x_{11}+4x_{21} & 3x_{12}+4x_{22} \end{pmatrix}$$

$$3B-A=\begin{pmatrix} 7 & 0 \\ 3 & 1 \end{pmatrix}$$

$$\begin{pmatrix} -2x_{11}-3x_{21}& -2x_{12}-3x_{22} \\ 3x_{11}+4x_{21} & 3x_{12}+4x_{22} \end{pmatrix}=\begin{pmatrix} 7 & 0 \\ 3 & 1 \end{pmatrix}$$

$$-2x_{11}-3x_{21}=7$$ $$3x_{11}+4x_{21}=3$$

$$-2x_{12}-3x_{22}=0$$ $$3x_{12}+4x_{22}=1$$

$$x_{21}=27$$ $$-2x_{11}-3x_{21}=7$$

$$-x_{22}=2$$ $$3x_{12}+4x_{22}=1$$

$$x_{11}=37, x_{21}=-27, x_{22}=-2, x_{12}=3$$

$$X=\begin{pmatrix} 37& -27 \\ 3 & -2 \end{pmatrix}$$