Modulus problem with a negative sign

In summary, the second method he discussed in that video is plotting the function $f(x) = |x+4| - |x| - |x-3|$ using the coordinate values $(-4,f(-4))$, $(0,f(0))$, and $(3, f(3))$.
  • #1
In a video, a person discussed how to solve modulus problems with a negative sign. This is the link of that video lecture.

He showed two methods to solve the problem. The first method is commonly used. Later he showed another method where he used a number line and a graph.

Unfortunately, I couldn't understand his explanation properly. What is the original name of the second method he discussed in that video? If I could learn the actual name of the second method, I could googled to learn more about that.
 
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  • #2
He is just plotting the function $f(x) = |x+4| - |x| - |x-3|$ using the coordinate values $(-4,f(-4))$, $(0,f(0))$, and $(3, f(3))$. He selects two more x-values, one > 3 and the other < -4, and evaluates the function at those other two to predict the behavior of the graph.

Note the graph's highest y-value is at $(3, f(3))$ ... it never gets as high as 12, indicating no solution to the equation.

Unfortunately, he doesn't say what to do if, say, $f(x) = 2$ instead of $f(x)=12$. Still, one can see by the graph there would be two possible solutions.

abs_func.jpg
 
  • #3
@skeeter , Will you please say what we could do if f(x)=2 instead of f(x)=12?

And probably for f(x)=12, still there might be two possible solutions. Will you please say what those solutions are, and how to find out those solutions?
 
  • #4
f(x) = 12 has no solutions ... note the graph of f(x) does not intersect the horizontal line y = 12.

If f(x) = 2, I would find the linear equation of the line between the points $(0,f(0))$ and $(3,f(3))$, set it equal to 2 and solve for x.

Do the same for the linear equation of the line between the points $(3,f(3))$ and say, $(5,f(5))$ and follow the same procedure.
 
  • #5
skeeter said:
f(x) = 12 has no solutions ... note the graph of f(x) does not intersect the horizontal line y = 12.

If f(x) = 2, I would find the linear equation of the line between the points $(0,f(0))$ and $(3,f(3))$, set it equal to 2 and solve for x.

Do the same for the linear equation of the line between the points $(3,f(3))$ and say, $(5,f(5))$ and follow the same procedure.
Thanks a lot for your cordial help.
 

1. What is a modulus problem with a negative sign?

A modulus problem with a negative sign is an operation that involves finding the remainder after dividing a negative number by another number. It is denoted by the symbol "|" and is also known as the absolute value.

2. How do you solve a modulus problem with a negative sign?

To solve a modulus problem with a negative sign, you can follow these steps:

  1. Divide the negative number by the other number.
  2. If the result is positive, then the answer is the remainder.
  3. If the result is negative, then add the divisor to the result to make it positive, and the answer is the remainder.

3. What is the difference between a modulus problem with a positive sign and a negative sign?

The only difference between a modulus problem with a positive sign and a negative sign is the result. In a modulus problem with a positive sign, the result will always be positive, while in a modulus problem with a negative sign, the result can be either positive or negative.

4. Can a modulus problem with a negative sign have a negative remainder?

Yes, a modulus problem with a negative sign can have a negative remainder. This happens when the result of dividing the negative number by the other number is negative, and the divisor is not a factor of the negative number.

5. What is the purpose of using modulus with a negative sign in mathematics?

The purpose of using modulus with a negative sign in mathematics is to find the remainder after dividing a negative number by another number. It is also used to simplify mathematical expressions and solve equations involving negative numbers.

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