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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} 3 - \frac{1}{2}x & \mbox{if $ x < 2 $}\\ 2x - 5 & \mbox{if $ x \ge 2 $} \end{array} \right.$

$\frac{3}{2}$$,3,-1$

01:48

Jeffrey P.

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Chapter 1

Functions and Models

Section 1

Four Ways to Represent a Function

Functions

Integration Techniques

Partial Derivatives

Functions of Several Variables

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here we have a piece voice function, and we want to find some function values and then sketch the graph. So let's find f of negative three. So negative three falls in this piece where X is less than two. So we're going to substitute negative three into three minus 1/2 X. We get three minus 1/2 times negative three, and that's going to be three plus three halves. So that's nine house okay f of zero for F of zero. We're going to plug zero into this piece as well, because zero is in the interval. That's less than two. So we have three minus one halftime zero, and that's going to be three. And then for F of two, we're going to substitute to into the bottom piece since two Falls and the X is greater than or equal to two category. So we have two times two minus five, and that's going to be negative one. Okay, so we have those function values and those will help us when it comes to graphing. But here's how I like to graph a piece Wise function. I like to start by thinking about what I'm expecting to see. So for the top piece, I'm expecting to see a line that has a slope of negative 1/2 and a Y intercept of three, but only part of that line. And then, for the bottom piece, I'm expecting to see a line with a slope of two and a wide receptive negative five Onley part of that line. Both lines are going to end at X equals two. So for the top piece I'm going to plug into now. I can't technically plug into because two is not in that domain. But I can make an open circle there. So if I substitute to and there I get three minus 1/2 times to, and that's going to be, too. We're gonna have an open circle at 22 And then if I substitute something less than two in there, Well, I've already got one negative 39 house so we can plot that point. So we started our open circle at 22 We go through negative 39 halves. It's up here somewhere, and we keep going, so that should look more or less like a line with a slope of negative 1/2 and a y intercept of three. I could probably make it a little bit more accurate. So let's find Ah, height of nine house. That would be 34 way up there a little bit higher. Okay, now for the other piece. So it starts at X equals two. We already know the point to negative one from the work we did earlier and let's find a second point. So another point that is greater than two. We could use three, for example, so two times three minus five would be one. So now we start with a closed circle at the point to negative one, and we go through the 0.31 and keep going. So does it look like we ended up with the line with a slope of negative 1/2 in a Y intercept of three? That would be that one. And does it look like we ended up with the line with the Y intercept of negative five and a slope of to You can imagine that when continuing down further and that seems reasonable

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