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.
 
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