Official Solution:If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values? A. 5
B. 10
C. 11
D. 20
E. 21
Considering that the transition points for \(|x+5|\) and \(|x-5|\) are -5 and 5, respectively, let's examine the function \(y=|x+5|-|x-5|\) for the following three ranges
When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\). This implies that when \(x\leq{-5}\), \(y\) can take only 1 integer value, which is -10.
When \(-5 < x < 5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=x+5-(5-x)=2x\). For this range, \(-10 < (y=2x) < 10\). This implies that when \(-5 < x < 5\), \(y\) can take 19 integer values, ranging from -9 to 9, inclusive.
When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\). In this case, \(y=|x+5|-|x-5|=x+5-(x-5)=10\). This implies that when \(x\geq{5}\), \(y\) can take only 1 integer value, which is 10.
Total number of integer values for \(y\) = 1 + 19 + 1 = 21.
If you are interested, here is a graph of \(y=|x+5|−|x−5|:\)
As illustrated, \(y\) is a continuous function (its values change smoothly without any gaps when moving along the domain), ranging from -10 to 10, inclusive. Consequently, \(y\) can take 21 integer values from -10 to 10, inclusive.
Answer: E