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

5232 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Integrate by parts x2^x


Find the angle between two lines with vector equations r1 = (2i+j+k)+t(3i-5j-k) and r2 = (7i+4j+k)+s(2i+j-9k)


The curve C has equation 4x^2 – y^3 – 4xy + 2^y = 0 The point P with coordinates (–2, 4) lies on C . Find the exact value of dy/dx at the point P .


(A) express 4^x in terms of y given that 2^x = y. (B) solve 8(4^x ) – 9(2^x ) + 1 = 0


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:

© 2026 by IXL Learning