HELPPPP ME PLEASEEE I NEED HELP :(

Find the range of f(x)=x+3 <br> when the domain (2,4,6)

jarptica [38.1K]

<u>f(x) = x + 3</u>
f(2) = 2 + 3
f(2) = 5

f(4) = 4 + 5
f(4) = 9

f(6) = 6 + 5
f(6) = 11

{(2, 5), (4, 9), (6, 11)}

5 0

3 years ago

Use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving

harina [27]

Answer:

The volume of the solid is 714.887 units³

Step-by-step explanation:

* Lets talk about the shell method

- The shell method is to finding the volume by decomposing

a solid of revolution into cylindrical shells

- Consider a region in the plane that is divided into thin vertical

rectangle

- If each vertical rectangle is revolved about the y-axis, we

obtain a cylindrical shell, with the top and bottom removed.

- The resulting volume of the cylindrical shell is the surface area

of the cylinder times the thickness of the cylinder

- The formula for the volume will be: V = $\int\limits^a_b {2\pi xf(x)} \, dx$ ,

where 2πx · f(x) is the surface area of the cylinder shell and

dx is its thickness

* Lets solve the problem

∵ y = $x^{\frac{5}{2}}$

∵ The plane region is revolving about the y-axis

∵ y = 32 and x = 0

- Lets find the volume by the shell method

- The definite integral are x = 0 and the value of x when y = 32

- Lets find the value of x when y = 0

∵ $y = x^{\frac{5}{2}}$

∵ y = 32

∴ $32=x^{\frac{5}{2}}$

- We will use this rule to find x, if $x^{\frac{a}{b}}=c, then=== x=c^{\frac{b}{a}}$ , where c

is a constant

∴ $x=(32)^{\frac{2}{5}}=4$

∴ The definite integral are x = 0 , x = 4

- Now we will use the rule

∵ $V = \int\limits^a_b {2\pi}xf(x) \, dx$

∵ y = f(x) = x^(5/2) , a = 4 , b = 0

∴ $V=2\pi \int\limits^4_0 {x}.x^{\frac{5}{2}}\, dx$

- simplify x(x^5/2) by adding their power

∴ $V = 2\pi \int\limits^4_0 {x^{\frac{7}{2}}} \, dx$

- The rule of integration of $x^{n} is ==== \frac{x^{n+1}}{(n+1)}$

∴ $V = 2\pi \int\limits^4_0 {x^{\frac{9}{2}}} \, dx=2\pi[\frac{x^{\frac{9}{2}}}{\frac{9}{2}}]$ from x = 0 to x = 4

∴ $V=2\pi[\frac{2}{9}x^{\frac{9}{2}}]$ from x = 0 to x = 4

- Substitute x = 4 and x = 0

∴ $V=2\pi[\frac{2}{9}(4)^{\frac{9}{2}}-\frac{2}{9}(0)^{\frac{9}{2}}}]=2\pi[\frac{1024}{9}-0]$

∴ $V=\frac{2048}{9}\pi=714.887$

* The volume of the solid is 714.887 units³

5 0

3 years ago

Let f(x) = xe^-x+ ce^-x, where c is a positive constant. For what positive value of c does f have an absolute

Illusion [34]

Answer:

$c=6$

Step-by-step explanation:

The absolute maximum of a continuous function $f(x)$ is where $f'(x)=0$ . Therefore, we must differentiate the function and then set $x=-5$ and $f'(x)=0$ to determine the value of $c$ :

$f(x)=xe^{-x}+ce^{-x}$

$f'(x)=-xe^{-x}+e^{-x}-ce^{-x}$

$0=-(-5)e^{-(-5)}+e^{-(-5)}-ce^{-(-5)}$

$0=5e^{5}+e^{5}-ce^{5}$

$0=e^5(5+1-c)$

$0=6-c$

$c=6$

Therefore, when $c=6$ , the absolute maximum of the function is $x=-5$ .

I've attached a graph to help you visually see this.

7 0

3 years ago

NEED HELP ASAP, WILL GIVE BRAINLIEST

Svetlanka [38]

Domain: [-5, 5)
range:(-4,4]
hope this helps

5 0

3 years ago

Which which value is not located between square root 9/3 2n

tangare [24]

3 and -3 are said to be the square roots of 9. A square root is written with a radical symbol √ and the number or expression inside the radical symbol, below denoted a, is called the radicand. ... Negative numbers don't have real square roots since a square is either positive or 0.

7 0

4 years ago

HELPPPP ME PLEASEEE I NEED HELP :(​

HELPPPP ME PLEASEEE I NEED HELP :(