First, what is an absolute value function? It is a function that will always yield positive results. It is written as such: | f(x)|. So |f(x)| = f(x) if the value of f(x) is positive, but is -f(x) if the value of f(x) is negative.
Let me illustrate with an example.
Say I have an absolute value function |x - 1|. Now, x - 1 will be positive (or zero) if x = 1 or more. But if x is less than 1, then it will return a negative value. Correct? So, then
|x - 1| = x - 1 when x ≥1; but is -(x – 1) when x < 1
Hence, |x - 1| will return two functions, depending on what the value of x is.
Understand so far?
If you do, then let us look at the first equation to solve: |x + 3| = 5
Then, it means that x + 3 = 5 or - (x + 3) = 5
Since the negative sign can be brought over to the other side of the equation, then the solution for
|x + 3| = 5 is x + 3 = ± 5
Which means that x + 3 = 5 or x + 3 = -5
Hence, x = 2 or x = -8.
So far so good, right?
Now, let’s go one step further. What happens when |f(x)| = g(x)?
Well, the same argument still applies.
The solution will be f (x) = g(x) or f(x) = -g(x) and we solve for x for each of these cases.
For example, solve for |3x – 2| = 1 – x
In this case,
3x – 2 = 1 – x or 3x – 2 = - (1 – x)
3x + x = 3 3x – 2 = - 1 + x
4x = 3 2x = 1
x = ¾ x = ½
Hence,
x = ½ or ¾.
So far still here?
I hope so, because this is not rocket science.
Now, what happens if our question is |f(x)| = |g(x)|?
Well in this case, the solution still remains the same way as above; i.e.
f (x) = g(x) or f(x) = -g(x)
So there won’t be any examples for this.
But watch what happens with we have a constant term added to the above equation with two absolute value function, like this |f(x)| = a + |g(x)|? This will make things a little more troublesome.
In this case, we will need to identify which range of values of x there are and apply the right combination of positive or negative functions. Let’s see an example of this to understand what I am trying to say…
Example we have
|3x – 2| = |1 – x | - 2
In this case, we will first need to see what combinations of positive and negative functions to use. We will take each of the functions in turn and determine the range of values of x for which they are positive.
So, if 3x – 2 > 0, then x > ⅔
And if 1 – x > 0, then x < 1
So, in this case we have three regions:
Less than ⅔, between ⅔ and 1 and greater than 1.
We have seen that for 3x – 2, it will yield a positive function only when x > ⅔. It means that for the region where x < ⅔, it would be negative.
For the case of 1 – x, we have seen that it will yield a positive value only when x < 1. Hence, for the region for x > 1, it will yield a negative value.
This now gives us the possibilities of values of x.
So, for the region x < ⅔, we will have the case of –(3x – 2) but +(1 – x). Hence, we will solve the equation:
-(3x – 2) = (1 – x) – 2
-3x + 2 = 1 – x – 2
-2x = - 3
x = 3/2
For the region of x in between ⅔ and 1,
(3x – 2) = (1 – x) – 2
3x – 2 = - x – 1
4x = 1
x = ¼
And finally, for the region where x > 1, we have
3x – 2 = - (1 – x) – 2
3x – 2 = -1 + x – 2
2x = -1
x = -½
So now, we have three solutions: x = 3/2, ¼ or -½
This is not the end. We now need to confirm which of these is valid. We will need to put it back into the equation |3x – 2| = |1 – x | - 2
So, when x = 3/2, we have 5/2 = ½ - 2 and this is NOT valid.
When x = ¼ , we have 1¼ = - 1 ¼ so this is also not valid.
Lastly, when x = - ½ we have 5/2 = ½ and this is ALSO not valid.
In this case, there are no valid solutions to the equation, even if we thought we could find it. To confirm this, check out wolfram alpha at http://www.wolframalpha.com/input/?i=solve+abs%283x-1%29%3D%281-x%29-2
So here you have it. How to solve absolute value functions. The mathematics is easy but the understanding needs time to sink it. So let it sink in…
If you have any queries, please ask away…
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