Where is y= 1/1-e^x discontinuous? and Why?
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When x = 0, because e^x or e^0 equals 1, and the function becomes y = 1/1-1 or y = 1/0. Well, in calculus, you cannot define a function where the denominator is equal to zero, that point on the graph becomes undefined, therefore breaking the continuity of the rest of the function.
Here is the visual representation of that graph:
http://www.wolframalpha.com/input/?i=y+%…
Hope this helps, Mike
Here is the visual representation of that graph:
http://www.wolframalpha.com/input/?i=y+%…
Hope this helps, Mike
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At x=0, e^0=1, so the denominator becomes 0.
y is undefined and becomes discontinuous.
y is undefined and becomes discontinuous.