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Figure $5(\mathrm{c})$ shows the graph of a piecewise linear function $f$ Find the algebraic formulation of $f(x) .$ Find $f(-2), f(1 / 2),$ and $f(5)$.

$$f(x)=\left\{\begin{array}{cc}1 & x \leq-1 \\-x & -1<x \leq 1 f(-2)=1, f(1.5)=-1.5, f(5)=7 \\2 x-3 & x>1\end{array}\right.$$

Algebra

Chapter 1

Functions and their Applications

Section 3

Applications of Linear Functions

Functions

Campbell University

Harvey Mudd College

Idaho State University

Lectures

01:43

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).

03:18

04:51

Figure $5(a)$ shows the gr…

03:47

Figure $5(b)$ shows the gr…

01:25

Write the linear function …

02:20

Writing a Linear Function.…

00:35

Find the slope of the grap…

01:38

Find a linear function $f …

00:40

Sketch the graph of $f$

using piece wise functions, we're gonna be determining the function of ffx. Given the graph of F of X that we have here to start. Let's segment this into three separate lines. Let's segment this into line a line. Be online, see at each of the brakes. Let's now identify the functions for each one of those lines, starting by finding the slope given our soap formula of y tu minus y one all of our x two minus x one Doing that for a We find that we have a slope equal to zero and you can see that graphically this horizontal line, which will always mean we have a slope equal to zero plugging ourself of zero and our coordinate point negative 11 of that line into our point slope form. We find that we have a function that is equal to one moving on for segment. Be doing the same thing by plugging are to coordinate points into ourself formula. We find that we have a slope equal to negative one plugging that and one of our coordinate points into our point slope form again, we find that we have a function equal to negative X and for segment. See, we're going to do the same thing, finding that we have a slope equal to two and a function equal to two X minus three. We now want to find what are bounds are for each of these looking at it graphically, we can see that for our segment A that exists for any value that is less than or equal to negative one. If X is less than or equal to negative one that is anywhere going in that direction. Giving us X is less than or equal to 91. For our segment B, we find that that exists. It exercise anywhere which is greater than negative one, but less than or equal to one. We find that between these two points here for a segment, see, we find that this is valid if X is greater than one as found here as X increases. Now we can write this within our as a piece wise function which looks something like this. We have ffx is equal to one. If X is less than or equal to negative. One half of X is equal to negative X If X lies between negative one and positive one and that it is equal to two X minus three. If X is greater than one using this, we can practice by finding some values of X. Let's suppose that X is equal to negative two in supposing that we find which function we want to use. So the bounds which negative two would fall. That's going to be our function. A because X of negative two is less than or equal to negative one, which gives us a value of one if F is equal to one half. We plugged that into our bi function because one half lies between negative one and one. In doing that, we have unequal thio in negative one half. We can simplify by dropping those parentheses, having people to negative one half. If X is equal to five, we're gonna be using our functions. See, because five is greater than one plugging that in two times five minus three, we get a value of 10 minus three, which equals seven

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