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roselin
 3 years ago
f(x) x^38/x^24 , x not equal to 2, X not equal to 2
3, x=2
4, x= 2
at which point is the function continuous?
roselin
 3 years ago
f(x) x^38/x^24 , x not equal to 2, X not equal to 2 3, x=2 4, x= 2 at which point is the function continuous?

This Question is Closed

tcarroll010
 3 years ago
Best ResponseYou've already chosen the best response.12 questions 1) is your equation\[f(x) = \frac{ x ^{3}  8 }{ x ^{2}  4 }\] with x not = 2 or 2? And 2) what do the second 2 lines mean? (3, x=2 and 4, x = 2)?

tcarroll010
 3 years ago
Best ResponseYou've already chosen the best response.1Start by factoring the numerator and denominator. Hint: you will then be able to get the common factor of (x2) to cancel.

tcarroll010
 3 years ago
Best ResponseYou've already chosen the best response.1What do you get after factoring? Show your work.

roselin
 3 years ago
Best ResponseYou've already chosen the best response.0(x2) ( x^2+2x+4)/ (x+2)(x2)

tcarroll010
 3 years ago
Best ResponseYou've already chosen the best response.1So, the first line has the restriction of x not = 2 or 2 because the first line is not defined for those 2 values of x. So far, we have a max of 2 points of discontinuity (before going to lines 2 and 3). For line 2, looking at how x2 cancelled out (you did that correctly! Good job!) We DEFINE f(2) = 3 and that is what the "factored out" equation would have given, so the overall function IS continuous at x=2. This leaves line 3 and x = 2 as the only possible point of discontinuity. The "factored out" equation would give 4/0, so the overall function is discontinuous only at x = 2.

tcarroll010
 3 years ago
Best ResponseYou've already chosen the best response.1You're welcome! Medals are nice.
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