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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} x + 1 & \mbox{if $ x \le -1 $}\\ x^2 & \mbox{if $ x > -1 $} \end{array} \right.$

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$f(-3)=-2, f(0)=0, f(2)=4$

01:42

Jeffrey Payo

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

Missouri State University

Campbell University

Oregon State University

Harvey Mudd College

Lectures

04:31

A multivariate function is a function whose value depends on several variables. In contrast, a univariate function is a function whose value depends on only one variable. A multivariate function is also called a multivariate expression, a multivariate polynomial, a multivariate series, or a multivariate function of several variables.

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

02:32

Evaluate $ f(-3) $ , $ f(0…

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02:18

Evaluate $f(-3), f(0)$, an…

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03:05

04:41

Evaluate $f(-3), f(0),$ an…

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02:22

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02:33

02:00

Given the function defined…

04:24

Sketch the graph of a func…

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01:24

\begin{equation}\begin…

here we have a piece vice function, and we want to find some function values and then sketch the graph. So let's find f of negative three. Negative three falls into this domain. X is less than or equal to negative one, so we have negative three plus one, and that would be negative, too. Let's find F of zero Zero Falls into this domain. X is greater than negative one, so we would have zero squared and that would be zero. And now let's find f of tube to also falls into the second domain so we would have two squared, and that would be four. Okay, now let's think about how we would draw the graph in general. So what I like to do is first visualize what I'm expecting to see. So the first piece would be a piece of a line, and that would be the line y equals X plus one. So would have a slope of one and a Y intercept of one. And the second piece would be a piece of a parabola y equals X squared so we can picture that parable opening up and having a vertex at 00 now I want to find the endpoint of each piece. I noticed that the pieces both end at negative one. So if X is negative one for the first piece, we would get a why coordinative zero. So that's where that piece ends. And we already know another point on that peace would be negative. Three negative to from the work we did earlier so we could drop point at negative 10 That's the point where the line ends, the peace of the line ends, and then we could draw the point. Negative three. Negative, too. So we started the end point and we go through the other point and keep going. Now for the piece of a proble, we know it's going to end at negative one. If you square negative one, you get positive one. Now that's not actually going to be a point on the problem because it says just greater than negative one. It doesn't say equal to. So we're gonna make this an open circle open circle at negative 11 and then we already know that it's going to go through 00 and 24 so we can plot those points. So remember it's going to look like a parabola started the open circle, go through those points and then draw that parabola.

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