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Oksi-84 [34.3K]
4 years ago
9

Is it possible for an odd function to have the interval 0 infinity) as its domain?

Mathematics
2 answers:
Anni [7]4 years ago
7 0

Answer with Step-by-step explanation:

Yes, it is possible for an odd function to have the interval 0 to infinity as its domain.

Let f(x) = -x

f(-x)= -(-x)

     = x

and -f(x)= -(-x)

            = x

Clearly, f(-x) = -f(x)

Hence, f(x) = -x  is an odd function.

and clearly the domain of f(x) = -x is the set of all real numbers which also contains the interval (0,∞)

Hence, it is possible for an odd function to have the interval 0 to infinity as its domain.

mojhsa [17]4 years ago
3 0

NO!

Prove by contradiction. First, define the the following terms:

odd function: f(-x) = -f(x)] and domain: all values of x

Proof: Suppose there f(-x) = -f(x) such that x is a negative integer, then f(-x) is a positive number <em>(-(-x) = +x)</em> and -f(x) is a negative number <em>(negative of a positive number is a negative number). </em>Since a positive number cannot equal a negative number, then this is not possible.

Answer: No

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Answer:

3x+4x+7

Step-by-step explanation:

First, a Trinomial consists of three terms.

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Next we need to add the coefficients.

<u>3</u>+<u>4</u>+<u>blank</u>

Add the variables.

3x+4x+<u>blank</u>

And lastly, add the constant.

3x+4x+7

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when a coin is flipped n times,what is the probability that the first head comes after exactly m tails
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Answer:

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Find the vertex of the graph of the function.
lapo4ka [179]

Answer: The correct options are  (1) (5,10), (2) (3,-3), (3) x = -1, (4) y=(x+2)^2+3, (5) 21s and (6) 0, -1, and 5.

Explanation:

Te standard form of the parabola is,

f(x)=a(x-h)^2+k        .....(1)

Where,  (h,k) is the vertex of the parabola.

(1)

The given equation is,

f(x)=(x-5)^2+10

Comparing this equation with equation (1),we get,

h=5 and k=10

Therefore, the vertex of the graph is (5,10) and the fourth option is correct.

(2)

The given equation is,

f(x)=3x^2-18x+24

f(x)=3(x^2-6x)+24

To make perfect square add (\frac{b}{2a})^2, i.e., 9. Since there is factor 3 outside the parentheses, so subtract three times of 9.

f(x)=3(x^2-6x+9)+24-3\times 9

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Comparing this equation with equation (1),we get,

h=3 and k=-3

Therefore, the vertex of the graph is (3,-3) and the fourth option is correct.

(3)

The given equation is

f(x)=4x^2+8x+7

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To make perfect square add (\frac{b}{2a})^2, i.e., 1. Since there is factor 4 outside the parentheses, so subtract three times of 1.

f(x)=4(x^2+2x+1)+7-4

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h=-1 and k=3

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h=-16t^2+672t

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