Find any points of discontinuity for the rational function.What are the points of discontinuity? Are they all removable?
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OpenStudy (anonymous):
OpenStudy (anonymous):
7 and 3 are the points of discontinuity of the function because here the the function sdenominator tends to 0 and the fonction itself tends to infinity|dw:1352531330237:dw|
OpenStudy (anonymous):
for discountinuity put denominator =0
x^2-10x+21=0
x^2-7x-3x+21=0
x(x-7)-3(x-7)=0
(x-7)(x-3)=0
at x=7 and x=3 it is discontinous
can be removed
y=x-7)(x-3)/x-7)(x-3)
y=1
OpenStudy (anonymous):
x = –7, x = –3; no would that be it ?
OpenStudy (anonymous):
no points of dicontinuity are 7 and 3 the way to remove them is to define the function in a new way by taking y=1
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OpenStudy (anonymous):
x = 1, x = –8, x = –2; yes ?
OpenStudy (anonymous):
or x = –1, x = 8, x = 2; no
OpenStudy (anonymous):
at x=7 and x=3
OpenStudy (anonymous):
no x=7,3 are points of discontinuity
OpenStudy (anonymous):
x = 7, x = 3; yes <-------------- this ?? one right ?
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