Bunuel wrote:
If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values?
A. 5
B. 10
C. 11
D. 20
E. 21
For this question, it's useful to know that
|a - b| represents the DISTANCE from point a to point b on the number line.
For example, |3 - 10| = the DISTANCE from 3 to 10 on the number line.
Since |3 - 10| = |-7| = 7, we know that 7 is the distance from 3 to 10 on the number line
So, in this case, |x - 5| = the distance from x to 5
Likewise, since x + 5 = x - (-5), we know that |x + 5| = |x - (-5)|
So, |x + 5| = the distance from x to -5
The equation y = |x + 5| - |x - 5| has two
critical points. These are x-values that MINIMIZE the value of |x + 5| and MINIMIZE the value of |x - 5|
First, |x + 5| is minimized when
x = -5. That is, when x = -5, |x + 5| = |(-5) + 5| = |0| = 0
Next, |x - 5| is minimized when
x = 5. That is, when x = 5, |x - 5| = |5 - 5| = |0| = 0
Let's add these critical points ( x = 5 and x = -5) to the number line.
Notice that these critical points divide the number line into 3 regions.
Let's see what happens to the value of y when x lies in each region.
Let's start with region 1
When x lies in region 1 (e.g., x = -7), notice that the
blue bar represents the value of
|x - 5|When x lies in region 1 (e.g., x = -7), notice that the
red bar represents the value of
|x + 5| (aka |x - (-5)|
Since the distance between -5 and 5 is 10, we can see that the length of the
blue bar must be 10 units LONGER than the
red barSo, it must be true that
|x + 5| -
|x - 5| = -10
Now let's generalize.
For ANY value of x in region 1, the length of the
blue bar will be 10 units LONGER than the
red bar.
So,
for ANY value of x in region 1, |x + 5| - |x - 5| = -10---------------------------------------------------------------------
Now let's see what's going on in region 3
When x lies in region 3 (e.g., x = 8), notice that the
blue bar represents the value of
|x - 5|When x lies in region 3 (e.g., x = 8), notice that the
red bar represents the value of
|x + 5| (aka |x - (-5)|
Since the distance between -5 and 5 is 10, we can see that the length of the
red bar must be 10 units LONGER than the
blue barSo, it must be true that
|x + 5| -
|x - 5| = 10
To generalize, we can say that,
for ANY value of x in region 3, |x + 5| - |x - 5| = 10IMPORTANT ASIDE: At this point, we've shown that
|x + 5| -
|x - 5| can equal 10 and -10
---------------------------------------------------------------------
Now onto region 2
Let's see what happens when x = -4
If x = -4, then
|x + 5| (the red bar) = 1 and
|x - 5| = 9
So,
|x + 5| -
|x - 5| = 1 - 9 = -8
Now let's see what happens when x = 3.5
[img]https://i.imgur.com/1SyK8Ii.png[/img
If x = 3.5, then
|x + 5| = 8.5 and
|x - 5| = 1.5
So,
|x + 5| -
|x - 5| = 8.5 - 1.5 = 7
---------------------------------------------------------------------
CONCLUSION: At this point, we've shown that |x + 5| - |x - 5| can equal 10, -10, and all values BETWEEN 10 and -10There are 21 INTEGERS, from -10 to 10 inclusive.
So, y can have 21 different integer values
Answer: E
Cheers,
Brent
why are you using only strict inequality sign, for second region i used -5 ≤ x <5