Tagged Questions
2
votes
1answer
36 views
Simplifying $\left|\left|\sqrt{-x^2}-1\right|-2\right|$
How do we simplify the expression $\left|\left|\sqrt{-x^2}-1\right|-2\right|$?
This is very confusing. Do they cancel out and become just simply $\sqrt{-x^2}-1-2$?
2
votes
3answers
36 views
Question about absolute value in inequalities
My book presents the following: $$7 \le x \le 9 $$ so $$ -1 \le x - 8 \le 1 $$ and $$ |x-8| \le 1$$
I usually get confused with the way that taking the absolute value of an expression works. Could ...
1
vote
2answers
86 views
How to evaluate the inequality $|x+1|<-1$?
Okay perhaps the title isn't specific enough, I didn't know how to word it exactly. I'm finding the interval of convergence for a power series and i know the answer to be (-2,0]
I end up with the ...
0
votes
1answer
26 views
Is this (or when) does this equality hold?
Let $a,b,c,d \in \mathbb{R}$ and $x,y$ are variables which are also real numbers
$$|ax + by|^2 + |cx + dy|^2 + 2|ax + by||cx + dy| = (ax + by)^2 + (cx + dy)^2 + 2(ax + by)(cx + dy)$$
Is this always ...
0
votes
1answer
56 views
Finding The Contour Maps Of A Function Of Two Variables
I am given the function $f(x,y) = \ln|y-x^2|$, and am suppose to find the contour maps.
Let $z = c = f(x,y)$. $c = \ln|y-x^2| \rightarrow e^c = e^{\ln|y-x^2|} \rightarrow e^c = |y-x^2|$
I know I ...
1
vote
3answers
67 views
Solving $2|x+1|>|x+4|$
I'm trying to solve the following equations and inequalities for $x\in\mathbb R$:
$$2|x+1|>|x+4|$$
I know I'm supposed to consider the intervals $(-\infty,-4), [-4,-1]$ and $(-1,\infty)$ but ...
-1
votes
3answers
103 views
Determine all solutions to $|x+12|+|x-5|=15$
Determine all solutions to the following.
$$ \lvert x+12\rvert +\lvert x-5\rvert =15.$$
3
votes
3answers
110 views
Exposition On An Integral Of An Absolute Value Function
At the moment, I am trying to work on a simple integral, involving an absolute value function. However, I am not just trying to merely solve it; I am undertaking to write, in detail, of everything I ...
1
vote
2answers
71 views
When does a absolute value equation have one unique solution?
Find $m \in \mathbb R$ for which the equation $|x-1|+|x+1|=mx+1$ has only one unique solution. When does a absolute value equation have only 1 solution?
I solved for $x$ in all 4 cases and got ...
1
vote
2answers
56 views
Confusion solving $\sqrt{4m^2-4m+1}+|1-2m|\leq2$, weird solution.
I am trying to solve $\sqrt{4m^2-4m+1}+|1-2m|\leq2$.
Since i know $|1-2m| = \pm(1-2m)$ i tough solving $\sqrt{4m^2-4m+1}+1-2m\leq2$ and $\sqrt{4m^2-4m+1}-1+2m\leq2$. As solutions i get $0\leq2$ and ...
1
vote
2answers
69 views
How to prove this inequality $x,y\in\Bbb R$, $|x|<1,|y|<1$ show that $\bigg|\frac{x-y}{1-xy}\bigg| < 1$ (and similar ones)
I have to show that the inequality below is true, i tried some thing but got stuck,
i tried to eliminate the absolute value $-1<\frac{x-y}{1-xy}<1$ and then solve for $x$ and $y$ with no ...
7
votes
2answers
157 views
Maximum of the difference
What is the maximum value of
$f(… f(f(f(x_{1} – x_{2}) – x_{3})-x_{4}) … – x_{2012})$
where $x_{1}, x_{2}, … , x_{2012}$ are distinct integers in the set ${1, 2, 3, …, 2012}$ and $f$ is the absolute ...
3
votes
2answers
92 views
Two solutions for $x$
I am trying to solve this. I already got the answer but my doubt is if I am doing it right.
There are two solutions for $x$ in the equation, to 2 decimal places.
What is the value of the greater ...
6
votes
4answers
435 views
Inequality with two absolute values
I'm new here, and I was wondering if any of you could help me out with this little problem that is already getting on my nerves since I've been trying to solve it for hours.
Studying for my next ...
1
vote
4answers
152 views
Calculate absolute values with unknown constant
I am to calculate all $x$ if $f(x) = g(x)$ and if
$$f(x)= |2x+2| + |3-2x|$$
$$g(x)= x + 14$$
How do I mix regular numbers with absolute values in such a sense? I thought I could calculate it like ...
2
votes
1answer
95 views
Simplifying $|a+b|^2 + |a-b|^2$
I want to simplify $|a+b|^2 + |a-b|^2$ where $a, b \in \mathbb{C}$. I've used Wolfram Alpha to get
$$
|a+b|^2 + |a-b|^2 = 2\left(|a|^2 + |b|^2\right)
$$
I'm trying to understand the steps involved in ...
1
vote
3answers
196 views
Quadratic equation with absolute value
Prepping for the GMAT, I came across the following question:
What is the product of all solutions of:
$$x^2 - 4x + 6 = 3 - |x - 1|?$$
First, I set up two equations, ie:
$$x^2 - 4x + 6 ...
2
votes
4answers
113 views
the solution set of $\left | \frac{2x - 3}{2x + 3} \right |< 1$
what is the solution set of $\left | \frac{2x - 3}{2x + 3} \right |< 1$ ?
I solved it by first assuming: $-1 < \frac{2x - 3}{2x + 3 } < 1$
ended with: $x > 0 > -3/2$
Is that a ...
2
votes
3answers
140 views
Solve an absolute value equation simultaneously
My question is :
Solve simultaneously
$$\left\{\begin{align*}&|x-1|-|y-2|=1\\&y = 3-|x-1|\end{align*}\right.$$
What I did :
$y=3 - |x-1|$ is given.
Thus $y = 3-(x-1)$ or $y = ...
0
votes
2answers
82 views
about solving: Absolute value
How to solve: $|\sqrt{x-1}-2| + |\sqrt{x-1}-3|=1$.
I would like to know how to solve an absolute value equation when there is a square root sign inside.
0
votes
2answers
113 views
Solving $|x-2| + |x-5|=3$ [duplicate]
Possible Duplicate:
How could we solve $x$, in $|x+1|-|1-x|=2$?
How should I solve:
$|x-2| + |x-5|=3$
Please suggest a way that I could use in other problems of this genre too
Any help ...
0
votes
1answer
129 views
Absolute value of a real number
My question is:
Solve: $|x-4|< a$, where $a$ belongs to the real numbers. Solve this by considering various cases depending upon whether $a$ is negative, positive or zero.
What I have tried ...
1
vote
2answers
94 views
Equality with absolute values - Is this a valid solution?
For this problem
$|2 - |x-2|| = 2$
I've found the values $x = -2$ and $x = 2$. However, an third solution was presented to me, which I can't seem to find by myself: $x = -6$.
Is this solution ...
0
votes
5answers
161 views
The solution set of the equation $|2x - 3| = - (2x - 3)$
The solution set of the equation $\left | 2x-3 \right | = -(2x-3)$ is
$A)$ {$0$ , $\frac{3}{2}$}
$B)$ The empty set
$C)$ (-$\infty$ , $\frac{3}{2}$]
$D)$ [$\frac{3}{2}$, $\infty$ )
$E)$ All real ...
1
vote
2answers
108 views
How to set up the existence condition of an absolute value
$$
\frac{\sqrt{4 + \arccos\left|\frac{2-x}{x+3}\right|}}{\sqrt{x^2 - 4x + 5} - 3}
$$
I'm trying to find the natural domain of the function above. I set up this conditions:
$$
\begin{cases}\sqrt{x^2 ...
2
votes
1answer
260 views
How to manipulate absolute values when shifting parts in an inequality
I have the following inequality... $|4x - 2| \le 0.5$
I want to manipulate this so it is just $|x|$ on one side, and everything else on the other, but I'm not sure how the absolute value complicates ...
0
votes
6answers
272 views
How could we solve $x$, in $|x+1|-|1-x|=2$?
How could we solve $x$, in $|x+1|-|1-x|=2$?
Please suggest a analytical way that I could use in other problems too like this $ |x+1|+|1-x|=2$ and of this genre.
Thank you,
1
vote
1answer
152 views
How to solve equations with absolute value and using the Archimedean property
I'm trying to learn Real Analysis on my own, but I found that i'm a bit rusty with the elementary stuff.
How do I solve equations like $|x| + |x+1| = 1$ and $|x-1| + |x+1| = 2$? I don't want the ...
0
votes
5answers
162 views
Solving an equation with absolute values
The equation I am trying to solve is this : $\newcommand\abs[1]{|#1|}\abs{3y+7}=\abs{2y-1}$.
My conventional approach is to split this into three intervals with $1/2$ and $-7/3$ being the two "split" ...
0
votes
2answers
69 views
Domains and setbuilder notation
We're learning about domains and setbuilder notation in school at the moment, and I want to make sure what I did was right.
My thought process:
\begin{align*}
-\frac12|4x - 8| - 1 &< -1 \\
...
3
votes
1answer
102 views
Inequality with absolute value
I am unsure if have solved the following inequality correctly:
$ \dfrac{2x+3}{x+5} \leq \dfrac{x+1}{|x-1|} \tag{1}$
I've proceeded as follows.
If $x>1$ then $|x-1|=(x-1)$
If $x<1$ then ...
3
votes
3answers
110 views
Solving $ \left| \frac{-2x-6}{4} \right| \le 5$ for $x$
Say I have a statement like:
$$
\left| \frac{-2x-6}{4} \right| \le 5.
$$
And I want to find the closed interval form of $x$. i.e. I want to know what the maximum and minimum $x$ can be. How do I ...
2
votes
2answers
981 views
Square root of simple binomial function
Let's say I have the following formula:
$$\sqrt{a^2-2ab+b^2}=\sqrt{(a-b)^2}=\sqrt{(b-a)^2}$$
When do I know which one of the following I should use?:
$$\sqrt{(a-b)^2}=a-b\qquad\text{ or }\qquad ...
2
votes
1answer
204 views
Rules applying to nested absolute values
I'm trying to use some algebra get $||x-5|-10|<\epsilon$ into a more manageable form (I'd like it in terms of $0<|x+5|<\delta$) but I'm not sure where to begin. I don't really know the rules ...
2
votes
3answers
321 views
Trouble with absolute value in limit proof
As usual, I'm having trouble, not with the calculus, but the algebra. I'm using Calculus, 9th ed. by Larson and Edwards, which is somewhat known for racing through examples with little explanation of ...
1
vote
1answer
2k views
Solving inequality with two absolute values
Hey, !
In my pre-calculus class the teacher showed the solution of the following example:
\begin{align}
\vert x-3 \vert \lt \vert x - 4 \vert + x
\end{align}
He started by stated the domains ...
