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Find all the values of x for which the function is differentiable. f(x)= (x^3-8)/(x^2-4x-5)
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A function is differentiable when The function is continuous. \[\frac{x^3-2^3}{x^2 - 4x -5}\]Just check x for which denominator is equal to zero exclude those values and you have a differentiable function in Real Domain.
\[x^2 - 4x - 5 \neq 0 \]
It would be differentiable at all real numbers except 5 and -1?
Yea, you got that right.
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