# How would you 'remove the discontinuity' of f? In other words, how would you define f(2) in order to make f continuous at 2. f(x) = (x^3 - 8)/(x^2 - 4)

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### 1 Answer

Given `f(x)=(x^3-8)/(x^2-4)`

We want to remove the discontinuity at x=2 if possible.

Factor the numerator and denominator:

`f(x)=((x-2)(x^2+2x+4))/((x+2)(x-2))`

Note that the factor (x-2) divides to 1 leaving:

`(x^2+2x+4)/(x+2)`

Now this function agrees with f(x) at every point except at x=2. The value of this function at x=2 is 3. Since the two functions are equivalent everywhere except at x=2, if we define f(2)=3, the functions will agree everywhere.

Since `(x^2+2x+4)/(x+2)`

is continuous at x=2, the new definition of f(x) will be continuous.

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Define f(2)=3

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`The graph:`

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The answer should be define f(2)=3.

The graph shows that the function will go through the point (2,3):