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# Evaluate $f(-3)$ , $f(0)$ and $f(2)$ for the piecewise defined function. Then sketch the graph of the function.$f(x) = \left\{ \begin{array}{ll} -1 & \mbox{if$ x \le 1 $}\\ 7 - 2x & \mbox{if$ x >1 $} \end{array} \right.$

## $-1,-1,3$

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

all right, we have this piece voice function, and we're going to find some function values and then sketch the graph. So let's find f of negative three. So negative three falls into this piece of the domain. X is less than or equal the one. And so the Y value is always going to be negative one for that Now let's find F of zero zero also falls into this piece. X is less than or equal to one, so F zero is also going to be negative one. Now let's find out of two two falls into this category here. X is greater than one, so we're going to substitute to into seven minus two X, and we get seven minus two times two, and that will be three. OK, now it's time to figure out how to sketch this graph, and so, at least at this point, we have three points on the graph, so that's a good start. It would also be a good idea to think about what each piece should look like. So if you think about the top peace, it should look like the line y equals negative one. So that's just a horizontal line, and if you think about the bottom piece, it should look like a line with a slope of negative two and a Y intercept of seven. Now each line is going to stop at some point because we only have a piece of each money and the line stop at one. So for the first piece, I'm going to substitute one in and see what I get when I substitute one in. I get negative one. We get negative one for everything there, so we know that peace is going to stop at the 0.1 negative one. So I'm going to plot that point, and we know that that peace is also going to go through the point. Negative three negative one that we found earlier and zero negative one that we found earlier so we could plot those points. Negative. Three. Negative 10 Negative one. So here we have that piece, that horizontal line. The other piece is also going to stop at X equals one. It's not going to go through one, so we're just going to have an open circle at one, and the open circle will be at one seven, minus two times one. So that would be one comma, five open circle there and then we know it's also going to go through the point to three since we found that point earlier. So we have an open circle at the 0.15 It's a little lock. My screen. I'm gonna have to just fudge it there, and then we have the 0.23 So we're going to get a line that looks like this for the other piece.

Oregon State University
##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp