MHB Absolute Value with two expressions.

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To solve the absolute value problem, the expression is defined as a piecewise function with three segments. The only solution found so far is at x=1, but there should be two valid roots within the specified domains. To sketch the area between -1 and 2, calculate the endpoints of the middle segment, which are f(-1) = -2 and f(2) = 1. Plot the points (-1, -2) and (2, 1), then connect them with a line segment to complete the graph. Understanding the roots and their respective domains is crucial for accurate representation.
stuart4512
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How do I do this? I have tried a few methods and end up getting x values that don't work when placed back into the equation.

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I would write the expression as a piecewise function:

$$f(x)=\begin{cases}-3x-5, & x<-1 \\[3pt] x-1, & -1\le x\le2 \\[3pt] 3x-5, & 2<x \\ \end{cases}$$

Can you proceed?

edit: I have moved this thread here from our Linear Algebra subforum, as this is a better fit. :D
 
I got the only solution to be at x=1.
How do I sketch the area between -1 and 2?
 
stuart4512 said:
I got the only solution to be at x=1.
How do I sketch the area between -1 and 2?

You should find 2 valid roots. Check the root of each piece, and if it is in the given domain for that piece, then it is a valid root.

As for the middle piece, just compute the end points, and connect them with a line segment.

$$f(-1)=-2$$

$$f(2)=1$$

So, plot the points $(-1,-2),\,(2,1)$ and connect them.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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