Given the function y=(x+1)(x-2)^2 find i) dy/dx ii) Stationary points and determine their nature

Here we have a function made from the product of two functions, so we canuse the product differenciation rule.

y=uv  =>  dy/dx=udv/dx + vdu/dx

Therefore dy/dy=(x-2)^2 + 2(x-2)(x+1)

Stationary points occur when the gradient is zero, we solve for (x-2)^2 + 2(x-2)(x+1)=0 which gives (0,4), (2,0)

Solving for nature of stationary point we find the second derivative d^2y/dx^2=6x-6

When x=0 we get a maximum, when x=2 we get a minimum point.

RB
Answered by Russell B. Maths tutor

4973 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find values of y such that: log2(11y–3)–log2(3) –2log2(y) = 1


(4-2x)/(2x+1)(x+1)(x+3) = A/(2x+1)+B/(x+1)+C(x+3) Find the values of the constants A, B and C


Differentiate x^3 − 3x^2 − 9x. Hence find the x-coordinates of the stationary points on the curve y = x^3 − 3x^2 − 9x


How can I use the normal distribution table to find probabilities other than P(z<Z)?


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