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Consider the function given by$$f(x)=\left\{\begin{aligned}x+3 & \text { if } x \leq 0 \\-x+3 & \text { if } 0 < x\end{aligned}\right.$$Find $f(-5), f(0),$ and $f(3) .$ Plot the graph of $y=f(x)$

$$-2,3,0$$

Algebra

Chapter 1

Functions and their Applications

Section 3

Applications of Linear Functions

Functions

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In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x^2. The output of a function f corresponding to an input x is denoted by f(x).

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we are going to be working with peace fives functions the one has given and working to plot those. So we are given a piece wise function where F of X is equal to X plus three. If X is less than or equal to zero and f of X is equal to negative X plus three if X is greater than zero. If we want to plot these, we first want to find some coordinate points that we can use. So I've created a couple of tables here that established both functions and their boundaries there restrictions. Let's start with doing one of them, which is this here is gonna be our first one and our second one equation one here and equation to here. So for one, let's say let's pick values. Keeping in mind that these X values have to be less than or equal to zero, how about we do zero negative two and negative four. Those all satisfied that restriction plugging each of these values into our function of X plus three. We end up getting why values of three one and negative one doing the same with our second function. Keeping in mind that are X values have to be greater than zero. Let's choose values, say 12 and four. Each of these satisfying the restriction. Plugging these values into our function of negative X plus three. We end up with y values. He called it too one and negative one. So these here satisfy what would what are now our coordinate points We can begin to plot these looks like they're all within five. So we could just drops, um, tick marks down. All right, So starting with our first quarter point of 03 at zero go up to three, we have negative to one and negative for negative one. Those give us our coordinate points of our first piece of this function. And keep it in mind. We also are X values have to be less than or equal to zero, meaning that this boundary point here it can be a closed circle. If it had been X must be less than zero. We would want to make sure that that's an open circle to signify on the graph that that is not included in the graph. But we can then draw line through these look something like that moving on to our second one. Let's graph are points 12 21 and negative or and four negative one Here. Now here We can also draw a straight line through all these these circles. They can be filled in keeping in mind, though, if we consider that exit just has to be greater than zero, we could even go toe 0.1, and that could be included in and however, it can't be zero. So that is what our graph would look like for this particular function and now suggests suppose we want to find values for why, if X is equal to negative five doing that, we just want to figure out which function which piece of dysfunction we want to be plugging negative five into. So negative five satisfies the condition that X must be less than or equal to zero, which means we're gonna be using our X plus three. Piece of the function goes to negative five plus three, which is equal to negative two. Supposing that now we have an X value of zero that's going to satisfy the same function of X being less than or equal to zero plugging that into our X Plus three function, we have zero plus three, giving us a value equal to three. And suppose F X is equal to three. That's going to satisfy the second piece of our function where X must be greater than zero. Plugging that into our negative X plus three function, we have negative three plus three, which is equal to zero, and that is how you would graph it and find certain points along the function.

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