1. The problem statement, all variables and given/known data I have an equation f'(x) = (2x)/(3(x^2-4)^(2/3)) 2. Relevant equations Vertical cusps are where the one sided limits of the derivative at a point are infinities of opposite signs. Vertical tangent lines are where the one sided limits of the derivative at a point are infinities of the same sign. 3. The attempt at a solution I know that the derivative blows up (goes to infinity) at both x = 2 and x = -2 since that will zero the denominator (and nothing cancels with the denominator in the numerator.) However, it is not possible for the limit of the derivative at x = 2 or -2 to change signs because the denominator will always be positive; in the denominator there is a constant multiplied by a squared term. Whether x is approaching 2 from the right or left hand side and whether the term becomes a very small positive number close to 0 or a very small negative number close to zero is irrelevant as the term will be squared, making it positive. Basically, the numerator will control the sign of the limit. Therefore, it is impossible for the graph of f(x) to have vertical cusps at x = 2 or x = -2. It's impossible for the one sided limits at x = 2 or x = -2 to change signs. Is my reasoning correct?