Say x^6 - 64 = f(x). The domain of this function is (-infty,infty). But f(x)/(x-2) is a continuous graph and does not have any breaks at x=2. Why is this? If it is because it is divisible by (x-2), then please explain why is that so? even in x^6 - 64/(x-2), x=2 will mean dividing by 0. How does this work?
f(x) is discontinuous. The domain of f(x) = (x⁶ - 64)/(x - 2) is (-∞,∞) where x ≠ 2. f(x) is undefined at x = 2 because division by zero is undefined.
(x⁶ - 64)/(x - 2) = (x³ - 8)(x³ + 8)/(x - 2) = (x - 2)(x² + 2x +4)(x³ + 8)/(x - 2) = (x² + 2x +4)(x³ + 8)
Therefore, the lim f(x) as x → 2 exists and takes the value (2² + 2∙2 + 4)(2³ + 8) = (12)(16) = 192. f(x) has a removable discontinuity at x = 2, which is an open point in the curve at (2,192).
Furthermore, the function g(x) = (x² + 2x +4)(x³ + 8) is continuous. The piecewise function h(x), defined as (x⁶ - 64)/(x - 2) where x ≠ 2 and h(x) = 192 where x = 2, is also continuous because the point discontinuity at (2,192) has been repaired.
Hope this helps!
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