Ask question

Ask question

All categories

3 years ago

11

Let f be a function such that f(x)=2x-4 is defined on the domain.2 is less than or equal to x and x is lss than or equal to 6. T

he range of the function

2 answers:

STALIN [3.7K]3 years ago

8 0

F(x) = 2x - 4
f(2 ≤ x) = 2(2 ≤ x) - 4
f(x ≥ 2) = 2(x ≥ 2) - 4
f(x ≥ 2) = 2(x) ≥ 2(2) - 4
f(x ≥ 2) = 2x ≥ 4 - 4
f(x ≥ 2) = 2x ≥ 0
f(x ≥ 2) = x ≥ 0

f(x) = 2x - 4
f(x ≤ 6) = 2(x ≤ 6) - 4
f(x ≤ 6) = 2(x) ≤ 2(6) - 4
f(x ≤ 6) = 2x ≤ 12 - 4
f(x ≤ 6) = 2x ≤ 8
f(x ≤ 6) = x ≤ 4

iris [78.8K]3 years ago

7 0

The range of the function is the set of values it can achieve. The minimum value plugs in 2 to get 2*2-4, or 0. The maximum value uses 6, and is 6*6-4, or 8. Thus, the range is 0 to 8, inclusive.

You might be interested in

Is this a Percent Increase or Decrease Problem?

nydimaria [60]

Answer:

Percent Decrease

<h2>Please mark as brainliest for further answers :)</h2>

3 0

2 years ago

Center: (9,0), Radius: 5√3

densk [106]

Answer:

ากทก่หวไใหท่ปาก

Step-by-step explanation:

่ก่หาห่กาไย

7 0

3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago

What is the greatest common factor in 30 and 50

Contact [7]

10 because 30 divided by 10 equals 3 and 50 divided by 10 equals 5 and 3 and 5 only have a GCF of 1 so 10 is your answer

8 0

3 years ago

Read 2 more answers

Use the given pair of functions to find and simplify expressions for the following functions and state the domain of each using

Tamiku [17]

Answer:

Given functions,

$f(x) = 3 - x^2$

$g(x) = \sqrt{x}+1$

Since, by the compositions of functions,

1. (g◦f)(x) = g(f(x))

$=g(3-x^2)$

$=\sqrt{3-x^2}+1$

Since, (g◦f) is defined,

If 3 - x² ≥ 0

⇒ 3 ≥ x²

⇒ -√3 ≤ x ≤ √3

Thus, Domain = [-√3, √3]

2. (f◦g)(x) = f(g(x))

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

$=3-(\sqrt{x}+1)^2$

Since, (g◦f) is defined,

If x ≥ 0

Thus, Domain = [0, ∞)

3. (f◦f)(x) = f(f(x))

$=f(3-x^2)$

$=3-(3-x^2)^2$

$=3-9-x^4+6x^2$

$=-6+6x^2-x^4$

Since, (f◦f) is a polynomial,

We know that,

A polynomial is defined for all real value of x,

Thus, Domain = (-∞, ∞)

8 0

3 years ago