show that y = (kx^2-1)/(kx^2+1) has exactly one stationary point when k is non-zero.

Stationary points are found by considering the points at which the gradient of the function equal zero. For the above, you need to employ the quotient rule, since both numerator and denominator are f(x), to find dy/dx (=y'). If y = u/v, then y' = (u'v - v'u) /v^2. and so y' = 4kx/(kx^2+1)^2. Stationary points are when y' = 0. The numerator is the only important term here, and 4kx = 0 only when x = 0 for all non-zero constants of k - i.e. one stationary point.

RM
Answered by Rob M. Maths tutor

5002 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Express x^2 - 7x + 2 in the form (x - p)^2 + q , where p and q are rational numbers.


How do you integrate xcos(x)?


Particle A mass 0.4kg and B 0.3kg. They move in opposite direction and collide. Before collision, A had speed 6m/s and B had 2m/s. After collision B had 3m/s and moved in opposite direction. Find speed of A after collision with direction and Impulse on B.


Solve 2^(3x-1) = 3


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning