Piecewise: Computation Practice 1

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This is a practice understanding of piecewise functions from an analytic viewpoint.

1 : Consider the following piecewise function: $f(x) = \begin {cases} \sage {p1f1} & \sage {p1l1}\leq x \leq \sage {p1r1} \\ \sage {p1f2} & \sage {p1r1} < x \leq \sage {p1r2} \\ \sage {p1f3} & \sage {p1r2} < x \leq \sage {p1r3} \end {cases}$

Evaluate $f(\sage {p1eval}) = \answer {\sage {p1ans}}$

To evaluate, find the row that has the input in the listed domain span. Once you find the correct row, plug the input in as the x-value into the function in that row to find the value of the piecewise function at that input.

2 : Consider the following piecewise function: $f(x) = \begin {cases} \sage {p2f1} & \sage {p2l1}\leq x \leq \sage {p2r1} \\ \sage {p2f2} & \sage {p2r1} < x \leq \sage {p2r2} \\ \sage {p2f3} & \sage {p2r2} < x \leq \sage {p2r3} \end {cases}$

Evaluate $f(\sage {p2eval}) = \answer {\sage {p2ans}}$

3 : Consider the following piecewise function: $f(x) = \begin {cases} \sage {p3f1} & \sage {p3l1}\leq x \leq \sage {p3r1} \\ \sage {p3f2} & \sage {p3r1} < x \leq \sage {p3r2} \\ \sage {p3f3} & \sage {p3r2} < x \leq \sage {p3r3} \end {cases}$

Evaluate $f(\sage {p3eval}) = \answer {\sage {p3ans}}$

4 : Consider the following piecewise function: $f(x) = \begin {cases} \sage {p4f1} & \sage {p4l1}\leq x \leq \sage {p4r1} \\ \sage {p4f2} & \sage {p4r1} < x \leq \sage {p4r2} \\ \sage {p4f3} & \sage {p4r2} < x \leq \sage {p4r3} \end {cases}$

Evaluate $f(\sage {p4eval}) = \answer {\sage {p4ans}}$

5 : Consider the following piecewise function: $f(x) = \begin {cases} \sage {p5f1} & \sage {p5l1}\leq x \leq \sage {p5r1} \\ \sage {p5f2} & \sage {p5r1} < x \leq \sage {p5r2} \\ \sage {p5f3} & \sage {p5r2} < x \leq \sage {p5r3} \end {cases}$

Evaluate $f(\sage {p5eval}) = \answer {\sage {p5ans}}$