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‘Graphing piece wise function part 2

Make a table and graph the piecewise function State its domain and range You can use your graphing calculator to generate the table

5, -6 < x <-3 W(x) = -7, -3 < X <5 2, 5 < x< 8’

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04:24

Graph each piecewise-defined function. Use the graph to determine the domain and range of the function.$$h(x)=\left\{\begin{array}{lll}{5 x-5} & {\text { if }} & {x<2} \\{-x+3} & {\text { if }} & {x \geq 2}\end{array}\right.$$

02:43

use a graphing utility to graph the function. Then determine the domain and range of the function.$$f(x)=5 x^{3}+6 x^{2}-1$$

01:57

Graph the piecewise-defined functions. State the domain and range in interval notation. Determine the intervals where the function is increasing, decreasing, or constant.$$f(x)=\left\{\begin{array}{ll}5-2 x & x<2 \\3 x-2 & x>2\end{array}\right.$$

02:49

use a graphing utility to graph the function. Then determine the domain and range of the function.$$f(x)=2 x^{2}-5 x+1$$

01:22

Graph each piecewise-defined function and state its domain and range. Use transformations of the toolbox functions where possible.$$h(x)=\left\{\begin{array}{ll}-\frac{1}{2} x-1 & x<-3 \\-|x|+5 & -3 \leq x \leq 5 \\3 \sqrt{x-5} & x>5\end{array}\right.$$

Transcript

And here we have a step function, or a piecewise function. We’re asked to make a table and then graph the function. You can use your graphing calculator to generate the table, but this one’s pretty easy. The first step or the first part of the graph is between negative six and negative three, inclusive of negative three but not including negative six, and all those values are five. So negative six, five is here, but that’s an open circle because it’s not including that. And then we’re going to go all the way to negative three, five. So here’s that first section of the graph. And let’s do this. Let me highlight that and then we’ll switch to red for the next section. So when it’s between negative three and five, our value is seven, negative seven. So we go negative seven all the way to here. We have an open circle at negative three because of that less than sign. And then we…

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