What is the range of the function $F(X)=(X-1{)}^{2}+6$?

Grayson Pierce
2022-07-21
Answered

What is the range of the function $F(X)=(X-1{)}^{2}+6$?

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tiltat9h

Answered 2022-07-22
Author has **14** answers

The range of a function F(X) is the set of all numbers that can be produced by the function.Calculus gives you some better tools to answer this type of equation, but since it's algebra, we won't use them. In this case, the best tool is probably to graph the equation.

It is of quadratic form, so the graph is a parabola, opening up.

This means that it has a minimum point. This is at X=1, at which

F(X)=6

There is NO value of X for which the function produces a result less than 6.

Therefore the range of the function is all real numbers Y such that

$Y\ge 6$

It is of quadratic form, so the graph is a parabola, opening up.

This means that it has a minimum point. This is at X=1, at which

F(X)=6

There is NO value of X for which the function produces a result less than 6.

Therefore the range of the function is all real numbers Y such that

$Y\ge 6$

Darian Hubbard

Answered 2022-07-23
Author has **7** answers

Observe that, $\mathrm{\forall}x\in \mathbb{R},(x-1{)}^{2}\ge 0$.

Adding $6,(x-1{)}^{2}+6\ge 0=6=6$

$\therefore \mathrm{\forall}x\in \mathbb{R},f(x)\ge 6$

Hence, the Range of $f=[6,\mathrm{\infty})$.

Adding $6,(x-1{)}^{2}+6\ge 0=6=6$

$\therefore \mathrm{\forall}x\in \mathbb{R},f(x)\ge 6$

Hence, the Range of $f=[6,\mathrm{\infty})$.

asked 2022-06-06

A list of 11 positive integers has a mean of 10, a median of 9, and a unique mode of 8. What is the largest possible value of an integer in the list?

From the information, I got the following information:

11 integers with a mean of 10 means a total must be 110.

A unique mode means that there must be at least two 8's.

if there are two 8's, and the median is 9, there must be at least two numbers greater than 9 also.

From the information, I got the following information:

11 integers with a mean of 10 means a total must be 110.

A unique mode means that there must be at least two 8's.

if there are two 8's, and the median is 9, there must be at least two numbers greater than 9 also.

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complete k-partite graphs

I am trying to solve the following problem:

Let G be a nonempty graph with the property that whenever $uv\notin E(G)$ and $vw\notin E(G)$, then $uw\notin E(G)$. Prove that G has this property if and only if G is a complete k-partite graph for some $k\ge 2$. (Consider $\overline{G}$).

The converse is straightforward and is given by the definition of the complete k-partite graphs, however, the direct way is not trivial and I could not get it.

I am trying to solve the following problem:

Let G be a nonempty graph with the property that whenever $uv\notin E(G)$ and $vw\notin E(G)$, then $uw\notin E(G)$. Prove that G has this property if and only if G is a complete k-partite graph for some $k\ge 2$. (Consider $\overline{G}$).

The converse is straightforward and is given by the definition of the complete k-partite graphs, however, the direct way is not trivial and I could not get it.

asked 2022-07-13

I'm curious what the phrase "on average" means. Here is an example:

On average, $30\mathrm{\%}$ were further than ___ kilometers away when they had their accident.

Is $30\mathrm{\%}$ a z-score or is it a mean? Transport Canada was investigating accident records to find out how far from their residence people were to when they got into a traffic accident. They took the population of accident records from Ontario and measured the distance the drivers were from home when they had their accident in kilometers (km). The distribution of distances was normally shaped, with $\mu =30$ kilometers and $\sigma =8.0$ kilometers.

On average, $30\mathrm{\%}$ were further than ___ kilometers away when they had their accident.

Is $30\mathrm{\%}$ a z-score or is it a mean? Transport Canada was investigating accident records to find out how far from their residence people were to when they got into a traffic accident. They took the population of accident records from Ontario and measured the distance the drivers were from home when they had their accident in kilometers (km). The distribution of distances was normally shaped, with $\mu =30$ kilometers and $\sigma =8.0$ kilometers.

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Which of the following is true regarding effect size?

0.11 = moderate effect size

0.12 = small effect size

0.18 - small effect size

0.58= large effect size

0.11 = moderate effect size

0.12 = small effect size

0.18 - small effect size

0.58= large effect size

asked 2022-07-21

chromatic number of a graph versus its complement

What can be said about the rate of growth of f(n), defined by

$f(n)=\underset{|V(G)|=n}{min}[\chi (G)+\chi (\overline{G})],$

where the minimum is taken over all graphs G on n vertices.

Two observations.

(1) Either G or $\overline{G}$ contains a clique on roughly logn vertices by Ramsey theory, so $f(n)\ge {c}_{1}\mathrm{log}n$ for some constant ${c}_{1}>0$.

(2) If G=G(n,1/2) is a random graph, then $\chi (G)\approx \chi (\overline{G})\approx n/\mathrm{log}n$ n/ log n almost surely, so we also have $f(n)\le {c}_{2}\phantom{\rule{thinmathspace}{0ex}}n/\mathrm{log}n$ n/log n for some constant ${c}_{2}>0$.

These bounds seem hopelessly far apart.

Can we improve on the bounds

${c}_{1}\mathrm{log}n\le f(n)\le {c}_{2}\phantom{\rule{thinmathspace}{0ex}}n/\mathrm{log}n$

for all sufficiently large n?

What can be said about the rate of growth of f(n), defined by

$f(n)=\underset{|V(G)|=n}{min}[\chi (G)+\chi (\overline{G})],$

where the minimum is taken over all graphs G on n vertices.

Two observations.

(1) Either G or $\overline{G}$ contains a clique on roughly logn vertices by Ramsey theory, so $f(n)\ge {c}_{1}\mathrm{log}n$ for some constant ${c}_{1}>0$.

(2) If G=G(n,1/2) is a random graph, then $\chi (G)\approx \chi (\overline{G})\approx n/\mathrm{log}n$ n/ log n almost surely, so we also have $f(n)\le {c}_{2}\phantom{\rule{thinmathspace}{0ex}}n/\mathrm{log}n$ n/log n for some constant ${c}_{2}>0$.

These bounds seem hopelessly far apart.

Can we improve on the bounds

${c}_{1}\mathrm{log}n\le f(n)\le {c}_{2}\phantom{\rule{thinmathspace}{0ex}}n/\mathrm{log}n$

for all sufficiently large n?

asked 2022-07-07

Given the mean, median and mode of a function and have to find the probability density function.

mean: $\gamma -\beta {\mathrm{\Gamma}}_{1}$

median: $\gamma -\beta (ln2{)}^{1/\delta}$

mode: $\gamma -\beta (1-1/\delta {)}^{1/\delta}$

Also given that

${\mathrm{\Gamma}}_{k}=\mathrm{\Gamma}(1+k/\delta )$

$\mathrm{\Gamma}(z)={\int}_{0}^{\mathrm{\infty}}{t}^{z-1}dt$

$-\mathrm{\infty}<x<\gamma ,\beta >0,\gamma >0$

Now I understand how to calculate the mean, mode and median when given a probability density function. However I'm struggling to go backwards. I initially tried to "reverse" the process by differentiating the mean or median however I know this is skipping the substitution over the given limit.

I then looked for patterns with known distributions and realised they are from Weibull distribution however $\gamma -$. Does this mean essentially this is a typical Weibull distribution however shifted by $\gamma $ and therefore the pdf will be $\gamma -Weibullpdf"$

mean: $\gamma -\beta {\mathrm{\Gamma}}_{1}$

median: $\gamma -\beta (ln2{)}^{1/\delta}$

mode: $\gamma -\beta (1-1/\delta {)}^{1/\delta}$

Also given that

${\mathrm{\Gamma}}_{k}=\mathrm{\Gamma}(1+k/\delta )$

$\mathrm{\Gamma}(z)={\int}_{0}^{\mathrm{\infty}}{t}^{z-1}dt$

$-\mathrm{\infty}<x<\gamma ,\beta >0,\gamma >0$

Now I understand how to calculate the mean, mode and median when given a probability density function. However I'm struggling to go backwards. I initially tried to "reverse" the process by differentiating the mean or median however I know this is skipping the substitution over the given limit.

I then looked for patterns with known distributions and realised they are from Weibull distribution however $\gamma -$. Does this mean essentially this is a typical Weibull distribution however shifted by $\gamma $ and therefore the pdf will be $\gamma -Weibullpdf"$