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

Mathematics

2 answers:

Anni [7]3 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]3 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 (-(-x) = +x) and -f(x) is a negative number (negative of a positive number is a negative number). Since a positive number cannot equal a negative number, then this is not possible.

Answer: No

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Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and

TiliK225 [7]

Answer:

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Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$