Questions tagged [absolute-value]

For questions about or involving the absolute value function.

2,024 questions
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How to plot absolute value graphs?

I have to plot graph of $$f(x)=|x|+2|x-1|+|x-4|$$ See I know graphs of individual $|x|,2|x-1|,|x-4|$ But how can I draw their sum. I have to find minimum value of the sum using graph.
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How to solve $|a+b|+|a-b|=c$?

It is intuitive that $a=\pm \frac{c}{2}$, with $-\frac{c}{2}\leq b\leq \frac{c}{2}$ or vice-versa are solutions to the problem. Can I get to these solutions without dividing the expression in all the ...
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Sum of two absolute values equal to a whole number

The following is the equation: $|x+1|+|x+2|=3$ How can I solve this problem? Do I have to reformat it to $|x+1|=3-|x-2|$? I would like a simple answer that by no means uses set theory. The answer ...
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Redefinition of a Constant Leading to Nullification of Absolute Value

I am currently taking a calculus 3 class in college, and my teacher did an intriguing problem about a tsunami in class which took up about 4 full white boards. Anyways, at a certain point in the ...
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solving equation involving absolute values [closed]

Given the equation $|x+2| + |3x+6| = 8,$ how can I find the sum of all its roots?
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How can I solve this absolute value equation?

This is the equation: $|\sqrt{x-1} - 2| + |\sqrt{x-1} - 3| = 1$ Any help would be appreciated. Thanks!
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Number of solutions of an equality with absolute value operator

Consider: $$\left|\left|\left|x-1\right|-2\right|-4\right|=4$$ What is the number of solutions for this equation? This one was particularly easy to me. If first observed that if this inequality ...
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Suppose $|z|\ge 2$, Prove $|z^8+135|\ge121$.

Suppose $|z|\ge 2$, Prove $|z^8+135|\ge121$. My work: $|z^8+135|=\sqrt{(z^8+135)(\bar{z}^8+135)}=\sqrt{|z|^{16}+135(z^8+\bar{z}^8)+135^2}\ge\sqrt{2^{16}+135(z^8+\bar{z}^8)+135}$ For the last term, ...
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Natural logarithm with absolute value: Can I cancel the absolute value?

I was calculating basic rational integrals and came up with this kind of problem. I have this expression: $$2\ln|x|$$ I can re-write it down like that: $$\ln{x^2}$$ and thus cancel the modulus. ...
How do I solve $|-2x^2+1+e^x+\sin x| = |2x^2-1|+e^x+|\sin x|$ where x belongs to [0,2π]?
How do I solve $|-2x^2+1+e^x+\sin x| = |2x^2-1|+e^x+|\sin x|,$ where $x$ belongs to [0,2π]? My book solves it in this way: since RHS is positive, it concludes that $1- 2x^2 \ge 0$ and $\sin x \ge 0$....