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I am supposed to solve the inequality :)
This is kind of tricky. I might be giving way too much explanation, so if you want to skip to the steps, skip to *** in the next post. You want to find what values of x makes |x + 2| + |2 – x| < 8 a true statement. If you plug in random values for x, sometimes |x + 2| + |2 – x| is bigger than 8 (try x = 1000000), sometimes it’s smaller than 8 (try x = 0), maybe it can be equal to 8 for some x (it’s not obvious that there is such an x). The question is: For what values of x is the expression less than 8? Another way to say it is to ask: On what intervals of the number line do we have |x + 2| + |2 – x| is less than 8. I don’t know if you’ve learned functions yet. If you have, then you might have noticed that f(x) = |x| looks like a V. In particular, notice that the absolute value function is “continuous,” ie. you can draw it without lifting the pencil. One key idea about continuous functions is that if a continuous function is positive at some point x = A, and later it becomes negative at a point x = B, then there must be some point x = C between A and B at which the function was zero. This key idea tells us that finding the zeros of a continuous function breaks up the number line into different parts over which the function is either always positive or always negative. (For example, if our zeros are z1 and z2, and if f(x) > 0 for some x between z1 and z2, then f(x) > 0 for all x between z1 and z2.) The same idea can be applied here. We saw that |x + 2| + |2 – x| is sometimes bigger than 8, and sometimes it’s smaller than 8. Therefore, it must be equal to 8 for some x. If we can find all the values for those x’s, then we can find over which intervals the expression is strictly less than 8.
*** So we want to solve |x + 2| + |2 – x| = 8. This is tricky because there’s three cases to consider: x is less than -2, x is bigger than 2, or x is between -2 and 2. Choosing different x changes the way in which we can remove the absolute value bars. When x is less than -2 (so x < -2), then we have x + 2 < 0, so |x + 2| = -(x + 2). Also, when x is less than -2, then x is less than 2. Knowing |2 – x| = |x – 2| and x < 2 means that |x – 2| = -(x – 2). So when x is less than -2, we want to solve -(x + 2) – (x – 2) = 8. Solving this gives x = -4. We can check that |-4 + 2| + |2 – (-4)| = 8. ___ Similarly, when x is bigger than 2, then x is bigger than -2 (so x > -2). So we have x + 2 > 0, and so |x + 2| = (x + 2). Also, when x is bigger than 2, knowing |2 – x| = |x – 2| and x > 2 means that |x – 2| = (x – 2). So when x is bigger than 2, we want to solve (x + 2) + (x – 2) = 8. Solving this gives x = 4. We can check that |4 + 2| + |2 – 4| = 8. __ Finally, when x is between 2 and -2, we want to solve (x + 2) + (2 – x) = 8. But this reduces to 4 = 8, so there is no x between 2 and -2 which makes the expression 8. So only for x = -4 and 4 do we have |x + 2| + |2 – x| is equal to 8. __ Now we’re pretty much done. Draw out a number line and mark -4 and 4. Now there are three intervals we just broke the number line into: - infinity to -4, -4 to 4, and 4 to infinity. Now choose easy “test points” in each interval, like -100000 in the first, 0 in the second, and 100000 in the third. If we plug x = -100000 into |x + 2| + |2 – x|, we see that it’s way bigger than 8. If we plug x = 100000 into |x + 2| + |2 – x|, we also see that it’s way bigger than 8. If we plug x = 0 into 8, we see than the expression is less than 8. Therefore, our expression must be less than 8 over the entire interval -4 < x < 4, since that’s the interval that contains x = 0.
And we also see that the expression is bigger than 8 over the two intervals - infinity to -4 and 4 to infinity.