Tag Archives: absolute value

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 11

October 5, 2014 etoix Leave a comment

Problem

Find all numbers $x$ for which

$|x-3|=8$
$|x -3| < 8$
$|x+4|< 2$
$|x-1|+|x-2| > 1$
$|x-1| + |x+1| < 2$
$|x-1|+|x+1| < 1$
$|x-1| \cdot |x+1| =0$
$|x-1| \cdot |x+2| = 3$

Solution

$|x-3|=8$

$x-3 =8 \to x = 11$

$3-x =8 \to x = -5$

$|x -3| < 8$

$x-3< 8 = x < 11$

$3-x< 8 = x > -5$

So $-5 < x < 11$ .

$|x+4|< 2$

$x+4 < 2 \to x < -2$

$-x-4< 2\to x > -6$

So $-6 < x < -2$ .

$|x-1|+|x-2| > 1$

$x-1 + x-2>1 \to 2x-3>1 \to 2x > 4 \to x > 2$ .

$1-x + 2-x>1 \to 3-2x>1 \to 2x < 2 \to x < 1$ .

$1-x+x-2>1 \to -1>1$ . This is nonsense.

$x-1+2-x > 1 \to 1 > 1$ . This is nonsense.

So $x < 1$ or $x > 2$ .

$|x-1| + |x+1| < 2$

$x-1 + x+1 < 2 \to 2x < 2 \to x < 1$

$1-x -x -1 < 2 \to -2x < 2 \to x > -1$

$1-x + x+1 < 2 \to 2 < 2$ . Nonsense.

$x-1 -x -1 < 2 \to -2 < 2$ . No use here.

So you might think that $-1 < x < 1$ . However, try letting $x = 0$ . You get $|0-1| + |0+1| < 2 \to |-1| + 1 < 2 \to 2 < 2$ . Doesn’t work.

Think about it this way, for any $x$ , the distance from $x-1$ to $x+1$ must be 2. So there is no number in $\mathbb{R}$ that satisfies this inequality.

$|x-1|+|x+1| < 1$

Following the logic from above. If the distance from $x-1$ to $x+1$ must be 2, it certainly cannot be less than 1.

$|x-1| \cdot |x+1| =0$

We need either $x-1 = 0$ or $x+1=0$ . The absolute value doesn’t even matter in this problem. So $x= 1$ or $x = -1$ .

$|x-1| \cdot |x+2| = 3$

3 is positive. For that to happens our two factors must be both positive or both negative. So $x > 1$ or $< -2$ . In this case $(x-1) (x+2) = 3$ . Expand to get $x^2 + x - 5 = 0$ . Using the quadratic formula $\frac{-1 \pm \sqrt{1-(4)(-5)}}{2} = \frac{-1 \pm \sqrt{21}}{2}$ . $\frac{-1 + \sqrt{21}}{2} > 1$ and $\frac{-1 - \sqrt{21}}{2} < -2$ so that fits what we’re looking for.

For $-2 < x < 1$ , we know that $(1-x)(x+2) = 3 \to -x^2 -x -1 = 0 \ to x^2 + x + 1 = 0$ . The solution here is $\frac{-1 \pm \sqrt{-3}}{2}$ , which is not $\mathbb{R}$ .

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 10

October 4, 2014 etoix Leave a comment

Problem

Express each of the following without absolute value signs, treating various cases separately when necessary.

$|a+b|-|b|$ .
$|(|x|-1)|$ .
$|x| - |x^2|$ .
$a-|(a-|a|)|$ .

Solution

$|a+b|-|b|$

There are two possibilities for $|a+b|$ : $a+b$ and $-a-b$ .

There are two possibilities for $|b|$ : $b$ and $-b$ .

So there are four possibilities altogether.

$a+b-b = a$
$a+b+b = a+2b$
$-a-b+b = -a$
$-a-b-b = -a-2b$

$|(|x|-1)|$

This becomes $|x|-1$ and $1-|x|$ .

$|x|-1$ can be either $x-1$ or $-x-1$ .

$1-|x|$ can be either $1-x$ or $x+1$ .

Four possibilities.

$|x| - |x^2|$

$x^2$ will always be positive so this is really $|x| - x^2$ .

This gives two possibilities: $x-x^2$ and $-x-x^2$ .

$a-|(a-|a|)|$

There is just one variable here. There is a case where $a \geq 0$ and a case where $\leq 0$ .

$a \geq 0$ . In this case we just drop the bars and get $a - ( a -a ) = a - 0 = a$ .

$a \leq 0$ . In this case we negative all the bars and get $a - - (a - - a) = a - - 2a = 3a$ .

So $a \geq 0 \to a-|(a-|a|)| = a$ and $a \leq 0 \to a-|(a-|a|)| = 3a$ .

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 9

October 3, 2014 etoix Leave a comment

Problem

Express each of the following with at least one less pair of absolute value signs.

$|\sqrt{2} + \sqrt{3} - \sqrt{5} + \sqrt{7}|$ .
$|(|a+b|-|a|-|b|)|$ .
$|(|a+b|+|c|-|a+b+c|)|$ .
$|x^2-2xy+y^2|$ .
$|(|\sqrt{2}+\sqrt{3}|-|\sqrt{5}-\sqrt{7}|)|$ .

Solution

$|\sqrt{2} + \sqrt{3} - \sqrt{5} + \sqrt{7}|$

$\sqrt{7} > \sqrt{5}$ so the whole expression will be positive without absolute value, so we can simply drop the bars and write $\sqrt{2} + \sqrt{3} - \sqrt{5} + \sqrt{7}$ .

$|(|a+b|-|a|-|b|)|$

We can rewrite the above as $|(|a+b|- ( |a|+|b|))|$ .

In the text we learned that $|a+b| \leq |a|+|b|$ .

So we know that the difference between $|a+b|$ and $|a|+|b|$ will be negative. So we can swap them to make the difference positive and drop the bars.

$|a| + |b| - |a+b|$

$|(|a+b|+|c|-|a+b+c|)|$

The this similar to the previous part. Let $d = a + b$ . Now you’re dealing with $|(|d|+|c|-|d+c|)|$ . Since $|d+c| \leq |d| + |c|$ , we know that that subtraction will always be positive, as is. So we can simply drop the bars.

$|a+b|+|c|-|a+b+c|$

$|x^2-2xy+y^2|$

This is simply $|(x-y)^2|$ . Since square is always positive, we can simply drop the bars.

$x^2 - 2xy + y^2$

$|(|\sqrt{2}+\sqrt{3}|-|\sqrt{5}-\sqrt{7}|)|$

It should be clear that $\sqrt{2} + sqrt{3} > 0$ and $\sqrt{5} - \sqrt{7} < 0$ . So the subtraction above will be positive, as is. We can simply drop the bars.