Stanton D. answered • 10/20/20
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Hi Justin B., (assuming you meant another ) in your question!)
So what you must do (in this case as well as in general!), is unify the expression for the function. Mutiply the first term by (x-3)/(x-3) so that you now have (7x-21)+(21-6x), all over (x)(x-3). That's easily reduced to x/((x)(x-3)) or 1/(x-3), except for x=0 . But then 1/(0-3) = = -1/3 is the obvious candidate for the discontinuity, right?
-- Cheers, -- Mr. d.
Stanton D.
I think what was meant was, IF you defined the function with f(x)=4 assigned for x=0, is that the correct way of removing the removable discontinuity? And it isn't, what is the correct value to assign? The point to be grasped here, is that you COULD define a function as given in the question, with f(0)=4. The function would then have a unique value across its entire domain, but it would still have a discontinuity at x=0. A discontinuity wreaks havoc on certain math operations (derivatives) at that value. Nevertheless, discontinuous functions are useful in many contexts -- you might look up the "step function" to begin with, used in electrical circuit behavior problems (among other areas!).
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10/22/20
Justin B.
Yes, that was correct. What does the 4 if x=0 mean then? That's what kept throwing me off10/20/20