Thread: Differentiability

Find all values of x for which the function is differentiable.

$\displaystyle y= \frac {x^3 - 8}{x^2 -4x-5}$

Easy way out is graphing.
Is there an algebraic way of figuring this out?

It looks to me that x=2 does not work, for the nominator equals 0.
x=5 appears to give on discontinuity, as the denominator is 0.

Anything I missed?

Thanks.

Originally Posted by Truthbetold

Find all values of x for which the function is differentiable.

$\displaystyle y= \frac {x^3 - 8}{x^2 -4x-5}$

Easy way out is graphing.
Is there an algebraic way of figuring this out?

It looks to me that x=2 does not work, for the nominator equals 0.
x=5 appears to give on discontinuity, as the denominator is 0.

Anything I missed?

Thanks.

note that one of the criteria of a function to be differentiable is continuity.
in that case, the values where the denominator is zero are not included, thus your function is differentiable at x=2; and x=5 and x=-1 are not included.