F(x) = (x + 1)² what is the domain of f? please, №17887305, 07.12.2022 11:54

Mathematics

F(x) = (x + 1)² what is the domain of f? please help

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P Answered by PhD

1

The domain is all real numbers

Mathematics

F(x) = (x + 1)2 What is the domain of f? Choose 1 All real values of x such that x = -1 All real values of such that -1 All real values of x such that I + 0 All real values of x such that 2 0 غم عصتلحينلحصعالم

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P Answered by PhD

6

all real numbers

Step-by-step explanation:

f(x) = (x + 1)^2

There is no restriction on x, so the domain is:

all real numbers

Mathematics

F(x) = (x + 1)2 What is the domain of f? Choose 1 All real values of x such that x = -1 All real values of such that -1 All real values of x such that I + 0 All real values of x such that 2 0 غم عصتلحينلحصعالم

Step-by-step answer

P Answered by PhD

6

all real numbers

Step-by-step explanation:

f(x) = (x + 1)^2

There is no restriction on x, so the domain is:

all real numbers

Mathematics

1) for the function f(x)= 3(x-1)^2, identify the vertex, domain and range. a) the vertex is (1, 2), the domain is all real numbers, and the range is y _ 2 b) the vertex is (1, 2), the domain is all real numbers and the range is y _ 2 c) the vertex is ( -1, 2), the domain is all real numbers and the range is y _ 2 d) the vertex is (-1, 2), the domain is all real numbers, and the range is y _ 2 2) what is the equation of the following graph in vertex form? (picture included below) a) y= (x- 4)^2 - 4 b) y= (x + 4)^2 - 4 c) y= (x + 2)^2 + 6 d) y= (x

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P Answered by PhD

3

QUESTION 1

The given function is

f(x)=3(x-1)^2+2

This function is of the form:

f(x)=a(x-h)^2+k , where V(h,k) is the vertex of the function.

Hence the vertex is

(1,2)

The function is defined for all real values of . Hence the domain is all real numbers.

To find the range, we let

y=3(x-1)^2+2

$\Rightarrow y-2=3(x-1)^2$

$\Rightarrow \frac{y-2}{3}=(x-1)^2$

$\Rightarrow sqrt{\frac{y-2}{3}}=x-1$

$\Rightarrow x=sqrt{\frac{y-2}{3}}+1$

is defined for $\frac{y-2}{3}\geq 0$

is defined for y\geq 2[/tex]

The correct answer is A

QUESTION 2

Based on the description, I was able to picture the diagram as shown in the attachment.

This graph has the vertex (h,k)=(4,-4) , Hence the equation is of the form:

f(x)=a(x-h)^2+k

The equation is

y=(x-4)^2-4

The correct answer is A

QUESTION 3

Based on the description, the graph has vertices (2,1)

Since this is a minimum graph;

The equation is of the form;

f(x)=a(x-h)^2+k , where $a\:0$ .

Hence the equation is

y=(x-2)^2+1

The correct answer is A.

QUESTION 5

The given function is

f(x)= -(x+1)^2+4

This equation is of the form f(x)=a(x-h)^2+k where V(-1,4) is the vertex .

The function is defined for all real values of . Hence the domain is all real numbers.

To find the range, we let

y=-(x+1)^2+4

y-4=-(x+1)^2

$\Rightarrow 4-y=(x+1)^2$

$\Rightarrow \sqrt{4-y}=x+1$

$\Rightarrow x=\sqrt{4-y}-1$

is defined for $4-y\geq 0$

$\Rightarrow -y\geq -4$

$\Rightarrow y\le 4$

Hence the range is the range is $y\le 4$

B) The vertex is (–1, 4), the domain is all real numbers, and the range is $y\le 4$

1) for the function f(x)= 3(x-1)^2, identify the vertex, domain and range. a) the vertex is (1, 2),

Mathematics

1) for the function f(x)= 3(x-1)^2, identify the vertex, domain and range. a) the vertex is (1, 2), the domain is all real numbers, and the range is y _ 2 b) the vertex is (1, 2), the domain is all real numbers and the range is y _ 2 c) the vertex is ( -1, 2), the domain is all real numbers and the range is y _ 2 d) the vertex is (-1, 2), the domain is all real numbers, and the range is y _ 2 2) what is the equation of the following graph in vertex form? (picture included below) a) y= (x- 4)^2 - 4 b) y= (x + 4)^2 - 4 c) y= (x + 2)^2 + 6 d) y= (x

Step-by-step answer

P Answered by PhD

3

QUESTION 1

The given function is

f(x)=3(x-1)^2+2

This function is of the form:

f(x)=a(x-h)^2+k , where V(h,k) is the vertex of the function.

Hence the vertex is

(1,2)

The function is defined for all real values of . Hence the domain is all real numbers.

To find the range, we let

y=3(x-1)^2+2

$\Rightarrow y-2=3(x-1)^2$

$\Rightarrow \frac{y-2}{3}=(x-1)^2$

$\Rightarrow sqrt{\frac{y-2}{3}}=x-1$

$\Rightarrow x=sqrt{\frac{y-2}{3}}+1$

is defined for $\frac{y-2}{3}\geq 0$

is defined for y\geq 2[/tex]

The correct answer is A

QUESTION 2

Based on the description, I was able to picture the diagram as shown in the attachment.

This graph has the vertex (h,k)=(4,-4) , Hence the equation is of the form:

f(x)=a(x-h)^2+k

The equation is

y=(x-4)^2-4

The correct answer is A

QUESTION 3

Based on the description, the graph has vertices (2,1)

Since this is a minimum graph;

The equation is of the form;

f(x)=a(x-h)^2+k , where $a\:0$ .

Hence the equation is

y=(x-2)^2+1

The correct answer is A.

QUESTION 5

The given function is

f(x)= -(x+1)^2+4

This equation is of the form f(x)=a(x-h)^2+k where V(-1,4) is the vertex .

The function is defined for all real values of . Hence the domain is all real numbers.

To find the range, we let

y=-(x+1)^2+4

y-4=-(x+1)^2

$\Rightarrow 4-y=(x+1)^2$

$\Rightarrow \sqrt{4-y}=x+1$

$\Rightarrow x=\sqrt{4-y}-1$

is defined for $4-y\geq 0$

$\Rightarrow -y\geq -4$

$\Rightarrow y\le 4$

Hence the range is the range is $y\le 4$

B) The vertex is (–1, 4), the domain is all real numbers, and the range is $y\le 4$

Mathematics

If f(x) = (x+1)^-1 and g(x) = x - 2, what is the domain of f(x) ÷ g(x)? a. all values of x b. (∞, -1), (-1, 2), and (2,∞) c. (∞, 2) and (2,∞) d. (∞, -1] and [2,∞)

Step-by-step answer

P Answered by PhD

10

$a^{-1}=\dfrac{1}{a}\\\\\text{therefore:}\\\\f(x)=(x+1)^{-1}=\dfrac{1}{x+1}$

g(x)=x-2

$f(x)\div g(x)=\dfrac{1}{x+1}\div(x-2)=\dfrac{1}{x+1}\cdot\dfrac{1}{x-2}=\dfrac{1}{(x+1)(x-2)}\\\\\text{The domain:}\\\\(x+1)(x-2)\neq0\iff x+1\neq0\ \wedge\ x-2\neq0\\\\x\neq-1\ \wedge\ x\neq2$

$\text{Therefore your answer is B.}\\(-\infty;-1)\ \cup\ (-1;\ 2)\ \cup\ (2;\ \infty)$

Mathematics

Step-by-step answer

P Answered by PhD

15

The domain is all real numbers, and the range is all real

numbers less than or equal to 4.

Step-by-step explanation:

The domain of a function f(x) is the set of all values for which the function is defined

We are given f(x)= -(x+3)(x-1)

f(x) is defined for all real values of x since there are no restrictions on the value of x

So,The domain of the function is all real numbers

Range of the function is the set of all values that f takes.

So, Range of given function is all real numbers less than or equal to 4.

Hence The domain is all real numbers, and the range is all real numbers less than or equal to 4.

Mathematics

Step-by-step answer

P Answered by PhD

15

The domain is all real numbers, and the range is all real

numbers less than or equal to 4.

Step-by-step explanation:

The domain of a function f(x) is the set of all values for which the function is defined

We are given f(x)= -(x+3)(x-1)

f(x) is defined for all real values of x since there are no restrictions on the value of x

So,The domain of the function is all real numbers

Range of the function is the set of all values that f takes.

So, Range of given function is all real numbers less than or equal to 4.

Hence The domain is all real numbers, and the range is all real numbers less than or equal to 4.

Mathematics

If f(x) = (x+1)^-1 and g(x) = x - 2, what is the domain of f(x) ÷ g(x)? a. all values of x b. (∞, -1), (-1, 2), and (2,∞) c. (∞, 2) and (2,∞) d. (∞, -1] and [2,∞)

Step-by-step answer

P Answered by PhD

10

$a^{-1}=\dfrac{1}{a}\\\\\text{therefore:}\\\\f(x)=(x+1)^{-1}=\dfrac{1}{x+1}$

g(x)=x-2

$f(x)\div g(x)=\dfrac{1}{x+1}\div(x-2)=\dfrac{1}{x+1}\cdot\dfrac{1}{x-2}=\dfrac{1}{(x+1)(x-2)}\\\\\text{The domain:}\\\\(x+1)(x-2)\neq0\iff x+1\neq0\ \wedge\ x-2\neq0\\\\x\neq-1\ \wedge\ x\neq2$

$\text{Therefore your answer is B.}\\(-\infty;-1)\ \cup\ (-1;\ 2)\ \cup\ (2;\ \infty)$

Mathematics

For the function f(x) = –(x 1)^2 + 4, identify the vertex, domain, and range. a. the vertex is (–1, 4), the domain is all real numbers, and the range is y ≥ 4. b. the vertex is (–1, 4), the domain is all real numbers, and the range is y ≤ 4. c. the vertex is (1, 4), the domain is all real numbers, and the range is y ≥ 4. d. the vertex is (1, 4), the domain is all real numbers, and the range is y ≤ 4.

Step-by-step answer

P Answered by PhD

36

The vertex is at -b/(2A). if you use the foil method it will come out to -(x^2+2x+1)+4. then distribute the negative and simplify the equation (add the four) the equation comes out to be -x^2-2x+3. The vertex is at x=2/(2*-1), which is -1. plug -1 into the original equation to find y, (y=-(-1)^2-2*(-1)+3) which is y=4. so the vertex is at (-1,4). then since the parabola opens down (negative a value) and has a maximum at 4, y is always less than or equal to 4. x is all real numbers as it goes on forever. so b is your answer.

F(x) = (x + 1)²

what is the domain of f? please help

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F(x) = (x + 1)²what is the domain of f? please help

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F(x) = (x + 1)²

what is the domain of f? please help