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## Modulus functions

What is the difference between |f(x)| and f(|x|) ? Please provide an example.

|f(x)| takes the result of f(x) and applies modulus to it.

f(|x|) takes x and applies modulus to it before 'passing' x into the function.

Let's assume x = -2 and f(x) = 3x + 1:

1: |f(x)|

|f(-2)| = 3(-2) + 1
|f(-2)| = -6 + 1
|f(-2)| = -5

(Apply modulus to the result)

Therefore: |f(-2)| = 5

Let's again assume x = -2 and f(x) = 3x + 1 and we'll use the secondary example of modulus.

2: f(|x|)

Apply modulus to x = -2 before passing it in. |-2| becomes 2, hence:

f(|-2|) = 3x + 1
f(|2|) = 3(2) + 1
f(|2|) = 7

Let me know if the above makes sense.

Thanks that makes sense. I have another question. What are the exact steps to solving these sort of questions? For example, if I were told to sketch
y = |2/6-x| + 3
Would I sketch what is inside the modulus, then make the part of the graph that is below the x axis and reflect it into the x axis? And then apply any transformations that are not within the modulus (in this case translate it by 3 units up)?
If I could do solve this question using the method above, how would I solve this question using a hybrid approach? What does the hybrid tell me? The x-intercepts? When would I use the hybrid and how?
Good question. Best to post this as another question rather than a comment so it can be answered directly and in more detail.
Your approach works. Using a hybrid approach would involve finding when (2/6-x) is negative, and then splitting into two arcs: (2/6-x)+3 and -(2/6-x)+3. Using the hybrid function is better if you need an algebraic form of the function rather than just a graph. Algebraic form gives you the robustness to perform many other follow-on functions - differentiate it, solve for certain points exactly, etc.
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