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Mathematics 13 Online
OpenStudy (anonymous):

Algebra 2! Create your own piecewise function and graph it on the coordinate plane. Explain whether your piecewise function is continuous or discontinuous.

OpenStudy (anonymous):

ok create you own do you know that a "piecewise " function is?

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

what one did you make up?

OpenStudy (anonymous):

i cant figure out how to come up with one

OpenStudy (anonymous):

oops

OpenStudy (anonymous):

\[f(x) = \left\{\begin{array}{rcc} x + 4 & \text{if} & x < 2 \\ x^2& \text{if} & x \geq 2 \end{array} \right.\]

OpenStudy (anonymous):

here is another \[f(x) = \left\{\begin{array}{rcc} \frac{1}{x+3} & \text{if} & x < 2 \\ x^2+x& \text{if} & x \geq 2 \end{array} \right.\]

OpenStudy (anonymous):

make up any you like

OpenStudy (anonymous):

for the first one is it continuous because it never ends

OpenStudy (anonymous):

lol no

OpenStudy (anonymous):

continuous would mean they agree at the point where the function changes definition

OpenStudy (anonymous):

for example for the first one, \[f(x) = \left\{\begin{array}{rcc} x + 4 & \text{if} & x < 2 \\ x^2& \text{if} & x \geq 2 \end{array} \right.\]replace \(x\) by \(2\) in both expressions

OpenStudy (anonymous):

the top one gives \(2+4=6\) and the bottom \(2^2=4\) and since \(6\neq 4\) it is not continuous at \(x=2\)

OpenStudy (anonymous):

thank you so much

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