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## Hot answers tagged absolute-value

14

Indeed, $\sqrt{a^2}=\lvert a\rvert$. But $\sqrt a^2=a$ (assuming that $a\geqslant0$), not $\lvert a\rvert$.

13

$$\left(\sqrt a\right)^2\ne\sqrt{a^2}.$$ Try with $a=-1$.

8

From the fact that you can take $\sqrt {5-x}$ you know that $5-x \ge 0$ so you don't need the absolute value signs.

6

The value of $|\sin(\cos\theta + i \sin\theta)|$ is indeed the image of mapping $f(z)=|\sin z|$ under unit circle $|z|=1$, which is a real number. One may write $$\sin(\cos\theta + i \sin\theta)$$ $$= \sin(\cos\theta)\cos(i\sin\theta) + \cos(\cos\theta)\sin(i\sin\theta)$$ $$= \sin(\cos\theta)\cosh(\sin\theta) + i\cos(\cos\theta)\sinh(\sin\theta)$$

5

Your function is only defined when $1-x^2 >0$. Therefore, on the domain of $f(x)$, it is always the case that $\sqrt{1-x^2}^2 = 1-x^2$. Also notice that the result of Symbolab is only defined when $1-x^2>0$, so your derivative and Symbolab's derivative are exactly the same.

4

No, just as we wouldn't call $x^4$'s graph a parabola, although the shape is very similar. A parabola as a purely geometrical shape, with a specific definition and/or construction and certain properties, is not necessarily the graph of a function: you could e.g. rotate $y=x^2$ in the plane: it would still be a parabola. However, if you're considering ...

4

$a=-1, b=-1$ looks like the counterexample here since $$||a|-b|=||-1|-(-1)| = |1+1|=2$$ and $$|a-b|=|-1-(-1)|=|0|=0$$ A more general hint: To find a counterexample in this and similar cases, I would first try the following four options: $$a=1,b=1\\ a=-1, b=1\\a=1, b=-1\\a=-1, b=-1$$

3

$|k|=|-a-k-b+k+a+b-k|\le |a+k| + |b-k| + |a+b-k|=1$ so $|k| \le 1$. Since $k \ne 0$ integer, it follows $|k| \ge 1$, hence $|k|=1$ $1=|a+k| + |b-k| + |-a-b+k| \ge |a+k|+|a| \ge |a+k-a| =|k|=1$ so we have equality in all the inequalities and in particular $|a+k|+|a|=|k|$ which means immediately $a$ real and of opposite sign to $k$ (either geometrically, ...

3

So as StackID just explained in his comprehensive answer, no $x\to |x^3|$ is not a parabola, and neither is $x\to x^4$, because in terms of polynomials, only second degree polynomials can qualify as a parabola. But if what you want to know is actually whether $x\to |x^3|$ is still a polynomial, that is to say does there exist a polynomial $P\in \textbf{F}[x]... 3 It seems that you are doing things in the opposite order as is required. You should start with$|x|\le 2$. As you wrote, this means$|x|+2\leq 4$, or$|x|+|2|\leq 4$. Then use the triangle inequality, which implies that$|x+2|\leq |x|+|2|\leq 4$. Finally, multiply both sides by$|x-2|$. 2$\cos(2\arcsin x)$will be$\ge0,$if$-\dfrac\pi2\le2\arcsin x\le\dfrac\pi2\iff -\dfrac1{\sqrt2}\le x\le\dfrac1{\sqrt2}$In that case $$|2x^2-1|=-(2x^2-1)$$ Check if$x>\dfrac1{\sqrt2}$or$x<-\dfrac1{\sqrt2}$2 To compare$\sqrt{5-x}$to$\sqrt{a^2}$you must compare$5-x$to$a^2$. The problem is that$a^2 \ge 0$while$5-x$can be any real number. But if you add the restriction$5-x \ge 0$, then$25 \geq |5 - x|$becomes $$\text{25 \geq 5 - x \ \text{and} \ 5-x \ge 0}$$ 2 I like the following way. We need to solve $$2|x-3|\leq10-x^2,$$ which is $$-(10-x^2)\leq2(x-3)\leq10-x^2,$$ which is $$x^2+2x-16\leq0$$ and $$x^2-2x-4\leq0,$$ which is $$-1-\sqrt{17}\leq x\leq-1+\sqrt{17}$$ and $$1-\sqrt{5}\leq x\leq1+\sqrt5,$$ which gives the answer: $$[1-\sqrt5,\sqrt{17}-1].$$ 2 Well, if$x\geqslant 0$, we have$|x|=x$, so the inequality is$|x-1|<|1-x|$, but as$|t|=|-t|$always, there can be no solution here. Remaining case is$x< 0$, where the inequality is$|1+x|<|1-x|$, which says the distance of$x$from$-1$is less than the distance of$x$from$1$, which is true for all negatives, so all such$x$is in the ... 2 Note that $$|2x|=|(x+y-z) + (z+x-y)| \le |x+y-z| + |z+x-y|$$ $$|2y|=|(x+y-z) + (y+z-x)| \le |x+y-z| + |y+z-x|$$ $$|2z|=|(y+z-x) + (z+x-y)| \le |y+z-x| + |z+x-y|$$ The result follows from adding up the inequalities. 2$n$is not necessarily contained in$K$. But$n\in\mathbb{N}$acts on$K$in the usual way$n\cdot x=x+x+\dots+x$(sum$xn$-times) I guess you also have triangle inequality for your norm. So, for$n\geq 0$, you have $$|n\cdot x|=|x+\dots +x|\leq |x|+\dots +|x|=n|x|$$ Do the same thing for$n$negative. 1 Let$x\geq y\geq z$. Thus, $$(x+y-z,x+z-y,y+z-x)\succ(x,y,z)$$ and since$f(x)=|x|$is a convex function, our inequality it's just Karamata. 1 Neither $$|1-x^2|$$ nor $$1-x^2$$ are fully equivalent to $$\left(\sqrt{1-x^2}\right)^2,$$ just like neither $$|x|$$ nor $$x$$ are fully equivalent to $$\left(\sqrt x\right)^2.$$ Because the domains are different. 1 It's wrong. Try$a_1=b_1=1$and$a_2=b_2=-1$. 1 If$x>0$there is no solution since 0 is not strictly less than 0. Suppose$x<0$. You have to solve then:$|-x-1|=|x+1|<1-x$since$1-x>0$. Squaring both side yields: $$(x+1)^2<(1-x)^2.$$ I let you finish it... 1 If you look carefully, you'll notice your definition has the square inside the square root (not outside): $$|a| = \sqrt{a^2}$$ However, in your solution you seem to assume that: $$\sqrt{a^2} = (\sqrt{a})^2$$ However, $$\sqrt{a^2} \neq (\sqrt{a})^2$$ For example, as others have suggested, if$a = -1$we have: $$\sqrt{a^2} = \sqrt{(-1)^2} = \sqrt{1} = 1$$ but ... 1 For $$x\geq 3$$ we have to solve $$x^2+2(x-3)-10\le 0$$ this is fulfiiled for $$3\le x\le -1+\sqrt{17}$$ For $$x<3$$ we have to solve $$x^2-2(x-3)-10\le 0$$ this gives $$1-\sqrt{5}\le x<3$$ 1 Your friend's work is not correct for the reason that $$\Pr[Z \le 1.5] = \Phi(1.5),$$ but $$\Pr[|Z| \le 1.5] = \Phi(1.5) - \Phi(-1.5) = \Phi(1.5) - (1 - \Phi(1.5)) = 2\Phi(1.5) - 1.$$ Your computation is correct although perhaps unnecessarily complicated, since $$Z = \frac{\bar x - \mu}{\sigma/\sqrt{n}} \sim \operatorname{Normal}(0,1).$$ This assumes that ... 1 Asserting that$a+bi$is a logarithm of$z$is the same thing as asserting that$e^{a+bi}=z. But then\begin{align}\lvert z\rvert&=\lvert e^{a+bi}\rvert\\&=\lvert e^a\times e^{bi}\rvert\\&=\lvert e^a\rvert\times\lvert e^{bi}\rvert\\&=e^a,\end{align}since\lvert e^{bi}\rvert=e^{\operatorname{Re}(bi)}=e^0=1$. But then$\ln\lvert z\rvert=a$. 1 Let us assume$h$twice differentiable. Let$N$be the set of isolated zeros of$h$.$|h|$is differentiable on$\mathbb{R}\backslash N$, and$|h|'(x) = sgn(h(x)) h'(x)$where the sgn function is$0$on$0$,$1$on the positive reals and$-1$and the negative ones. Let$x_0 \in \mathbb{R}\backslash N$. For$x \in \mathbb{R} \backslash N$,$|h|'(x) - |h|'(...

1

We know that by the triangle inequality $$|a|+|b|\geq|a+b|$$ and the equality occurs for $ab\geq0.$ Now, we have $$|x^2-2x|+|x-4|=|x^2-2x|+|4-x|\geq|x^2-3x+4|.$$ Id est, our inequality is equivalent to $$(x^2-2x)(4-x)<0$$ or $$x(x-2)(x-4)>0,$$ which by the intervals method gives $$(4,+\infty)\cup(0,2).$$

1

First, note that $x^2-3x+4$ is always positive because discriminant $D=(-3)^2-4\times4<0$. Case 1: $x\in(-\infty, 0]$ In this range: $|x^2-2x|=x^2-2x$, $|x-4|=-(x-4)$ and the inequality becomes: $$x^2-2x-(x-4)>x^2-3x+4$$ $$x^2-3x+4>x^2-3x+4$$ ...which is never true. So the ineqality has no solution in the given range. Case 2: $x\in(0, 2]$ In ...

1

It can be interesting to notice that , in case x>0 or equal to 0, the two numbers inside the absolute value function are opposites. We know that, as a general rule: it is impossible for the absolute value of a number to be less than the absolute value of its opposite ( additive inverse). If A and B are opposites, then , their absolute value is necessarily ...

1

The best I think to solve such problems is to draw the graphs. I am attaching an image below stating the steps to draw the LHS. Leaving out drawing the RHS as an exercise to you and figuring out the answer. Hope this helps. Cheers!

1

At this point, we know $$\cos(2\sin^{-1}x) = \pm(2x^2-1).$$ Since both these functions are continuous, we can just check one value in each region where $2x^2-1 > 0$ and $2x^2-1 < 0$ to verify the sign to use in that region. For $|x|<1/\sqrt{2}$, check $x = 0$, and for $|x|>1/\sqrt{2}$, check $x = \pm 1$. You'll find they all work out to \$\cos(2\...

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