Dealing with absolute value functions

In summary, to find the integral, you need to find the difference between two functions, f(x) and g(x). To simplify the calculation, you can solve the problem twice, once with the contents of the modulus being positive and once with them being negative. In the first case, you can take out the absolute value of x and simplify the expression to - x. However, in the second case, you need to add an extra minus sign in front of the contents of the modulus. It is important to consider both cases in order to accurately solve the problem.
  • #1
sunfleck
6
0
In order to get an integral I need to find the difference between two functions, but I'm not sure how to deal with the absolute value...


[tex]f(x) = \left|x-1-1\right|[/tex]
[tex]g(x) = x^2 + 2x[/tex]

[tex]g(x) - f(x) = (\left|x-1\right|-1) - (x^2 + 2x) [/tex]
=...
I don't know if I can simplify it anymore... can I take that |x| out so - 2x + |x| = - x? If so what happens to the |-1| I feel like I probably can't simplify any further, but I'd like to know for sure
 
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  • #2
The easiest way I find to solve problems like this is to solve the problem twice - once where the contents of the modulus are positive anyway, and once when they're negative (in which case you need to put an extra minus sign in front of them).
 
  • #3
I understand your confusion with dealing with absolute value functions. Absolute value functions can be tricky to work with, but there are some strategies that can help simplify them.

In this case, you are correct that you can take the absolute value out of the expression -2x + |x|. This is because the absolute value of a number is always positive, so we can rewrite the expression as -2x + x = -x.

However, you are also correct in questioning what happens to the absolute value of -1. In this case, we can rewrite the expression as -2x + |x-1| - 1. This is because the absolute value of -1 is 1, so we can substitute that into the expression.

From here, we can simplify further by combining like terms. We have -2x and -x, so the final expression would be -3x + |x-1| - 1.

In general, when dealing with absolute value functions, it's important to consider the different cases that can arise. For example, in your original expression, we have two cases: x-1 is either positive or negative. This can help guide your simplification process and ensure that you are accounting for all possible values.

I hope this helps and gives you a better understanding of how to deal with absolute value functions. Don't hesitate to ask for clarification or further assistance if needed. Keep up the good work!
 

Related to Dealing with absolute value functions

1. What is an absolute value function?

An absolute value function is a mathematical expression that calculates the distance of a number from zero. It is represented by two vertical bars surrounding the input value, such as |x|.

2. How do I graph an absolute value function?

To graph an absolute value function, plot the points on a coordinate plane by substituting different values for x and solving for y. Then, connect the points with a V-shaped curve.

3. What is the domain and range of an absolute value function?

The domain of an absolute value function is all real numbers, since any value can be substituted for x. The range is the set of non-negative real numbers, as the absolute value of any number will always be positive.

4. How do I solve absolute value equations and inequalities?

To solve an absolute value equation, isolate the absolute value expression and then solve for both the positive and negative versions of the equation. To solve an absolute value inequality, first solve for the positive and negative versions of the inequality separately, and then combine the solutions.

5. Can I use absolute value functions in real life situations?

Absolute value functions can be used in a variety of real-world situations, such as calculating distances, finding the magnitudes of forces, and determining temperature changes. They can also be used to model and analyze data in fields such as economics and physics.

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