Show that $\sqrt{1+x}<1+\frac{x}{2}$ if $x>0$

Nathanial Levine
2022-09-29
Answered

Show that $\sqrt{1+x}<1+\frac{x}{2}$ if $x>0$

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Jordan Owen

Answered 2022-09-30
Author has **7** answers

Why not just square both sides?

$1+x<1+x+\frac{{x}^{2}}{4}$

is trivial.

$1+x<1+x+\frac{{x}^{2}}{4}$

is trivial.

gsragator9

Answered 2022-10-01
Author has **2** answers

We have that for all $x>0$, we have

$1+x<1+x+\frac{{x}^{2}}{4}={(1+\frac{x}{2})}^{2},$

and hence

$\sqrt{1+x}<1+\frac{x}{2}.$

$1+x<1+x+\frac{{x}^{2}}{4}={(1+\frac{x}{2})}^{2},$

and hence

$\sqrt{1+x}<1+\frac{x}{2}.$

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What is the value of $\frac{100}{0}$?

1) 0

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asked 2022-04-07

Suppose that $f:[a,b]\to \mathbb{R}$ is continuous and that $f(a)<0$ and $f(b)>0$. By the intermediate-value theorem, the set $S=\{x\in [a,b]:f(x)=0\}$ is nonempty. If $c=supS$, prove that $c\in S$.

My first thought was to show that $S$ is finite therefore $c\in S$, but there is no guarantee that $f$ doesn't have infinitely many zeros.

A thought I have now is to show that $c>max(S)$ can not be true, but I do not know how to show this, is this even the correct thing to show? Thank you in advance for any input.

My first thought was to show that $S$ is finite therefore $c\in S$, but there is no guarantee that $f$ doesn't have infinitely many zeros.

A thought I have now is to show that $c>max(S)$ can not be true, but I do not know how to show this, is this even the correct thing to show? Thank you in advance for any input.

asked 2022-08-17

Suppose that $(A,\mathrm{\Sigma},m)$ is a measure space and $H$ is a linear functional on ${L}^{\mathrm{\infty}}(A,\mathrm{\Sigma},m)$. If

$\mathcal{U}:=\{u:A\to \mathbb{R}\text{}{\textstyle |}\text{}u\text{is measurable, bounded and}{\int}_{A}u\text{}dm=1\}$

and there are functions ${u}_{1},{u}_{2}\in \mathcal{U}$ such that

$H({u}_{1})\le 0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}H({u}_{2})\ge 0,$

then my question is:

Is there a function ${u}_{3}\in \mathcal{U}$ such that $H({u}_{3})=0$?

$\mathcal{U}:=\{u:A\to \mathbb{R}\text{}{\textstyle |}\text{}u\text{is measurable, bounded and}{\int}_{A}u\text{}dm=1\}$

and there are functions ${u}_{1},{u}_{2}\in \mathcal{U}$ such that

$H({u}_{1})\le 0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}H({u}_{2})\ge 0,$

then my question is:

Is there a function ${u}_{3}\in \mathcal{U}$ such that $H({u}_{3})=0$?

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A number $a$ is called a fixed point of a function $f$ if $f(a)=a$. Consider the function $f(x)={x}^{87}+4x+2$, $x\in \mathbb{R}$.

(a) Use the Mean Value Theorem to show that $f(x)$ cannot have more than one fixed point.

(b) Use the Intermediate Value Theorem and the result in (a) to show that $f(x)$ has exactly one fixed point.

(a) Use the Mean Value Theorem to show that $f(x)$ cannot have more than one fixed point.

(b) Use the Intermediate Value Theorem and the result in (a) to show that $f(x)$ has exactly one fixed point.

asked 2022-09-20

Question: Sketch all the continuous functions $f:\mathbb{R}\to \mathbb{R}$ which satisfy

$(f(x){)}^{2}=(x-1{)}^{2}(x-2{)}^{2}.$

Justify your answers.

I have found the eight possible continuous functions as follows:

$\begin{array}{rl}f(x)& =(x-1)(x-2)\\ f(x)& =-(x-1)(x-2)\\ f(x)& =|x-1|(x-2)\\ f(x)& =-|x-1|(x-2)\\ f(x)& =(x-1)|x-2|\\ f(x)& =-(x-1)|x-2|\\ f(x)& =|x-1||x-2|\\ f(x)& =-|x-1||x-2|.\end{array}$

and have got the conclusion of this question which has at most eight possible such functions, but how to proof the "at most" statement.

$(f(x){)}^{2}=(x-1{)}^{2}(x-2{)}^{2}.$

Justify your answers.

I have found the eight possible continuous functions as follows:

$\begin{array}{rl}f(x)& =(x-1)(x-2)\\ f(x)& =-(x-1)(x-2)\\ f(x)& =|x-1|(x-2)\\ f(x)& =-|x-1|(x-2)\\ f(x)& =(x-1)|x-2|\\ f(x)& =-(x-1)|x-2|\\ f(x)& =|x-1||x-2|\\ f(x)& =-|x-1||x-2|.\end{array}$

and have got the conclusion of this question which has at most eight possible such functions, but how to proof the "at most" statement.