Absolute value Definition and 365 Threads

Is this mathematical concept EVER used in real life or "higher" levels of math? I just find it to be a practically useless thing. It's like some guys sat around and invented this math concept just for the sake of it. Or, am I wrong?

I lost a mark for integrating $$\frac{1}{x} $$to $$ln(x)$$ instead of $$ln\left| x \right|$$. Since the derivative of $$ln(x)$$ is \frac{1}{x} not \frac{1}{\left| x \right|} why is this the case?

Homework Statement from -lnIxI to lnI x^-1I , I try to go from -lnIxI to lnI x^-1I by using some properties. Homework Equations - lnIxI The Attempt at a Solution First I write the -lnIxI as -1*lnIxI and then I use -1 as an exponent to absolute value of x in the natural log that is ln (...

I did this question two ways, each of which yielded a different answer. I'll post method one first. $$\left| (2|x-3|-5|x+4|) \right|$$ Sign changes occur when $x<-4$, $-4<x<3$, and $x>3$ If $x<-4$, then $|3x+26|>4$, and so $x>-22/3$ and $x<-10$. Clearly since $x$ much be less than $-4$, only...

Show that if $y_0\ne 0$ and $|y-y_0|<\min\left(\dfrac{|y_0|}{2},\dfrac{\varepsilon|y_0|^2}{2}\right)$, then $y_0\ne 0$ and $\left|\dfrac{1}{y}-\dfrac{1}{y_0}\right|<\varepsilon.$

1) True or Fale? |a-b| \geq ||a|-|b|| My solution: I broke this up into 4 different cases. 1. a> 0 b>0 2. a<0 b<0 and so on... For each case, I ended up with a more simplified version of the inequality. For instance, in case 1 where I used a<0 and b<0, I ended up with the simplified statement...

$$\d{}{x}\frac{1}{\sqrt{x}}$$ by the definition of the derivative. $$\lim_{{h}\to{0}}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}=\lim_{{h}\to{0}}\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x^2+2xh}}=\lim_{{h}\to{0}}\frac{x-(x+h)}{h\sqrt{x^2+2xh}\left(\sqrt{x}+\sqrt{x+h}\right)}$$ Setting $h=0$...

I have trouble solving this equation |1/2x+1|=|x| My answers are x=2 and x=-2/3

Homework Statement y = 1-abs(x) / abs(1-x) The Attempt at a Solution For x < 0, abs(x) = -x y = (1+x) / -(1-x) = -(1+x)/(1-x) I stopped here because this is the part I got wrong. For x < 0, my solutions manual got (1+x) / (1-x). What did I do wrong?

Hi, while solving differential equations problems we have to sometimes use absolute value while taking an integral of which 1 over a function. But in some problems I understand that we do not have to use absolute value sign for example if the function related with atom number which can...

Homework Statement In integration, we are allowed to use identities such as sinx = \sqrt{1-cos^2x}. Why does that work, and why doesn't make a difference in integration? Graphing \sqrt{1-cos^2x} is only equal to sinx on certain intervals such as(0, \pi) and (2\pi, 3\pi). More correctly...

In integration, we are allowed to use identities such as $$sinx = \sqrt{1-cos^2x}$$. Why does that work, and why doesn't make a difference in integration? Graphing $$\sqrt{1-cos^2x}$$ is only equal to sinx on certain intervals such as$$ (0, \pi) $$and $$(2\pi, 3\pi)$$. More correctly, shouldn't...

Problem: y = ln|sec(x) + tan(x)| Attempted Solution: See Attachment I was hoping someone could identify my error and possibly write it out for me. At first I thought my steps were correct and everything was in order. However, when I checked my answer on WolframAlpha it gave me a slightly...

Can someone explain to me why |2-x| would have a horizontal translation to the right? When I've always been taught that anytime you see a [+] it will translate to the left. The graph would be a regular |x| graph but it is shift 2 spots to the right. Thanks to anyone for help

☺'s question at Yahoo! Answers: a definite integral whose integrand has an absolute value factor. Here is the question: I have posted a link there to this thread so the OP can view my work.

Homework Statement Prove that for every two real numbers x and y ##|x+y| \leq |x| + |y| ##Homework Equations The Attempt at a Solution There are three cases. The easiest ones is when they are both positive and negative. The third one I have problems with. The numbers have different sign. Say...

Homework Statement Find if continuous and differentiable. I am having problems with the differentiable part. Homework Equations f(x) = x² + 3, |x| ≤ 1 f(x) = |x| + 3, |x| > 1 The Attempt at a Solution (1)^2 + 3 = 4 |1| + 3 = 4 ∴ It is continuous Now, rewriting it, you...

The problem is ∫x^2 - 3x - 5 with the lower limit being -4 and the upper limit 7. I broke the integrals into three parts from [-4, -1.1926], [-1.1926, 4.1926], [4.1926, 7] I did the integral and got (x^3)/3 - (3/2)x^2 - 5x I subbed in the lower and upper limits and got 32.861 for [-4...

When do you check the limit from the right and left of a limit with an absolute value in the numerator or denominator? For example, why do you check the limit from both sides of: Lim x -> 3/2 (2x^2-3x)/absolute value(2x-3) But only the left side of: limit as x approaches -2...

Homework Statement That is prove that |a•c|≤|a||c| for any vector a=<a1,a2,a3> & c=<c1,c2,c3> Homework Equations The Attempt at a Solution I really don't have much of an attempt at the solution. I am not sure where to start. I can kind of justify it in my mind by saying the...

Solve |x^2-5|=4|x|. I tried to rewrite it as: (sqrt(x^2-5))^2=4*(sqrt(x))^2 Is this the right way to solve the equation?

Suppose I have a variable separable ODE, e.g., \frac{dy}{dx} = 3y. We all know that the solution is y=Ae^{3x} where A is a constant. My question is as follows. To actually find this solution we rearrange the equation and integrate to get \int \frac{dy}{y} = 3 \int dx, which gives \ln...

∫tan(x) dx = -ln lcos(x)l + C = f(x) So is f'(x) = -sin(x)/lcos(x)l ? When taking the derivative, do we only consider f'(x) to only exist on the intervals where cos(x) > 0 instead?

Homework Statement Find the CDF of f(x) = |\frac{x}{4}| if -2<x<2 \\ 0 otherwise Homework Equations The Attempt at a Solution I have to integrate the pdf and to do so, I have to split it into two parts \int_{-x}^{0}\frac{-t}{4}dt + \int_{0}{x}\frac{t}{4}dt integrating I get \frac{x^2}{8} +...

Homework Statement Given the function f(x) = (abs(x))*x +6, find f^-1(x) Homework Equations The Attempt at a Solution for x≥ 0, f(x) = x^2 + 6 y=x^2 +6 x = √(y-6) for y≥6 → f^-1(x) = √(x-6) for x≥6 for x< 0, f(x) = -x^2 + 6 y= -x^2 +6 x = √(6-y) for y<6 → f^-1(x) = √(6-x) for x<6 But...

Here's a claim: Assume that a function f:[a,b]\to\mathbb{R} is differentiable at all points in its domain. Then the inequality |f(b) - f(a)| \leq \int\limits_{[a,b]}|f'(x)|dm(x) holds. The integral is the Lebesgue integral. Looks simple, but I don't know if this is true. There exists...

Solve 2x - |x+1| < 4.

Homework Statement Y=abs -4(x+3) +1 Note: the 1 is outside the absolute value Homework Equations Switch y and x The Attempt at a Solution So you subtract the 1 to bring it to the other side After that do I put: x-1= -4(y+3) And x-1= 4(y+3) And solve for y? Am I doing...

Homework Statement abs(x+y+z)≤abs(x)+abs(y)+abs(z) indicate when this equality holds and prove this statement Homework Equations Triangle inequality? The Attempt at a Solution I have nothing :/

When I learned about derivatives I was taught to put the absolute value sign around the argument for ln and log. For example \log{|x|} and \ln{|x|} instead of log(x)ln(x). Does this make a difference? Should both brackets and the straight lines be used? When taking the derivative what is the...

Homework Statement The problem is in the context of a probability problem; however, my question in regards to a computation regarding a particular integral. All that is needed to know is that the probability density function is 1 in the range 0 < y < 1 , y-1 < x < 1 - y, and 0 otherwise. I...

Hi! Need help in solving this double integral: 1 1 ∫ ∫ |x-y| dydx 0 0 Thanks in anticipation. Regards, Aby.

## x - |x-|x|| > 2 ## how would I go about solving something like this? my initial thoughts was to consider if x >= 0 I get 2-x < 0 then x > 2 in that case then consider if x < 0 which I get -|x+x| > 2-x then 2x > 2-x then x > 2/3 but I'm having troubles deciding which one is correct, and if...

Hey PF, I'm building a simple math calculator program in C for my Programming class. My problem is Xcode doesn't recognize the absolute value function (abs), even with math.h included: Any ideas on what I can do to fix this? Is there a way for me to define this on my own? Thank...

I have no problem that I am trying to solve but simply a question about the derivative of an absolute value equation. I know that the derivative of and absolute value function is (x*x')/(abs(x)) and I understand the process of reaching this equation through the process shown here...

Is |\vec{a}+\vec{b}|^{2} equal to the same thing as (\vec{a}+\vec{b})^{2}? And when is it equal to √(a^{2}+b^{2})? Thanks.

if you are given $$-6 \leq x \leq 14$$ from this how do you create an abs equation like $$|x-4| \leq 10$$ k

$\displaystyle\int|2x-1|dx$ please tell me what is the first step to solve this.

Solve $||||||| x^2 – x –1 |–3|–5|–7|–9| – 11|–13| = x^2 – 2x – 48$.

Basically I don't know anyone in real life that can help me with this, so I need help checking to see if my answers are correct :) PART A 9) Solve for all X: |2x-5| + 3 = 18 My Answer: x = 10, x = -5 10) Subtract and simplify: (5x^2 - 3x + 8) - (-4x^2 - x + 10) My Answer: 9x^2 - 2x - 2

Homework Statement Find the limit of this sequence: (-1)n(2n3)/(n3+1) Homework Equations The Attempt at a Solution At first glance I would attempt to use the absolute value theorem which gives an answer of 2 as n approaches infinite. My question is, when the theorem fails (...

I came across this algebra problem, can someone please help me solve this problem? Please show the steps as well. Much appreciated. |(x-1)| + |(y-3)| = 11 |(x- 3)| + |(y-17)| = 3 Find the nearest/possible x and y

I quote a question from Yahoo! Answers I have given a link to the topic there so the OP can see my response.

We have ## ||a+b|-|a|-|b|| ##. The only way I can think to eliminate one pair of absolute value signs is to consider the cases separately and determine which pair(s) can be removed without affecting the final answer. I'm trying to get better at doing these types of problems, so feel free to...

Define f(x)= |x+3|-|x-3| without absolute value bars piecewise in the following intervals (-∞,-3);[-3,3);[3,+∞). this is how i do the problem, I removed the absolute value bars first f(x)= x+3-x+3 = 6 now i don't know how to define it piecewise. can you show me how define it correctly. thanks!

Homework Statement ∫0-->x |t|dt Homework Equations // The Attempt at a Solution 1/2*x^2 for x>= 0 1/2*(-x)^2 for x<= 0 Not sure what to do to be honest. (the answer in the back of the book says 1/2*x|x|).

Homework Statement Find derivative of abs (2x^3 + 8x^2 + 5x +1).Homework Equations The Attempt at a Solution A little confused on how the absolute value changes the method of deriving that equation. When I derive it normally, I get (6x^2+16x+5).

Hello MHB, solve $$||x|-2|+|x+1|=3$$ and we find that $$x=0,-2,2,-1$$ I got problem to find the 'case', Regards, $$|\rangle$$

Hello all, I'm having trouble showing that |e^it|=1, where i is the imaginary unit. I expanded this to |cos(t)+isin(t)| and then used the definition of the absolute value to square the inside and take the square root, but I keep getting stuck with √(cos(2t)+sin(2t)). Does anyone have any...

Find the derivative of y = arctan(x^(1/2)). Using the fact that the derivative of arctanx = 1/(1+x^2) I got: dy/dx = 1/(1+abs(x)) * (1/2)x^(-1/2) But my textbook gives it without the absolute value sign. I don't understand why because surely x^(1/2) squared is the absolute value of x...