Let r(x)=q(x)p(x), where p and q are differentiable on an open interval I and q(x)=0 for all x∈I. Which of the following best explains why the equation
p′(x)q(x)−p(x)q′(x)=0
identifies all interior points of I at which r could attain a local extremum?
a) Because the equation forces p′(x) and q′(x) to have opposite signs, meaning one function is increasing while the other is decreasing, so their ratio must reach a local maximum or minimum at that point.
b) Because the product rule gives (pq)′=p′(x)q(x)+p(x)q′(x); if p′(x)q(x)−p(x)q′(x)=0, then (pq)′=0, so p(x)q(x) is locally constant and therefore the quotient r has an extremum.
c) Because r′(x)=[q(x)]2p′(x)q(x)−p(x)q′(x), setting the numerator equal to 0 forces r′(x)=0 whenever the denominator is non-zero, giving the only possible horizontal tangents.
d) Because r′(x)=[q(x)]2p′(x)q(x)−p(x)q′(x), the given equation makes the denominator of r′(x) equal to 0, so the derivative is undefined and r must change direction there, producing an extremum.